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NCERT Solutions for Class 12 Maths Chapter 7 Integrals

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals

Edited By Ramraj Saini | Updated on Sep 14, 2023 08:18 PM IST | #CBSE Class 12th

NCERT Integrals Class 12 Questions And Answers

NCERT Solutions for Class 12 Maths Chapter 7 Integrals are discussed here. This chapter deals with definite and indefinite integrals. Integration class 12 also includes elementary properties of integration including basic techniques of integration. NCERT Class 12 maths chapter 7 solutions will be very helpful when you are solving the questions from NCERT books for Class 12 Maths. These NCERT Class 12 Maths solutions chapter 7 are prepared by subject matter experts that are very easy to understand. students can practice integrals class 12 ncert solutions to get good hold on the concepts.

NCERT solutions for class 12 math chapter 7 integrals are important for board exams as well as for competitive examinations like JEE Main, VITEEE, BITSAT, etc but without command, on the concepts of integrals ncert solutions meritorious marks cant be scoured. Therefore chapter 7 maths class 12 is recommended to students. Also, you can check the NCERT solutions for other Classes here.

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Class 12 maths chapter 7 NCERT Solutions - Important Formulae

>> Integration as Inverse of Differentiation: Integration is the inverse process of differentiation.

In differential calculus, we find the derivative of a given function, while in integral calculus, we find a function whose derivative is given.

Indefinite Integrals:

∫f(x) dx = F(x) + C

These integrals are called indefinite integrals or general integrals.

C is an arbitrary constant that leads to different anti-derivatives of the given function.

Multiple Anti-Derivatives:

A derivative of a function is unique, but a function can have infinite anti-derivatives or integrals.

Properties of Indefinite Integral:

∫[f(x) + g(x)] dx = ∫f(x) dx + ∫g(x) dx

For any real number k, ∫k f(x) dx = k∫f(x)dx.

In general, if f1, f2, …, fn are functions and k1, k2, …, kn are real numbers, then ∫[k1f1(x) + k2f2(x) + … + knfn(x)] dx = k1 ∫f1(x) dx + k2 ∫f2(x) dx + … + kn ∫fn(x) dx

First Fundamental Theorem of Integral Calculus:

Define the area function A(x) = ∫[a, x]f(t)dt for x ≥ a, where f is continuous on [a, b].

Then A'(x) = f(x) for every x ∈ [a, b].

Second Fundamental Theorem of Integral Calculus:

If f is a continuous function on [a, b], then ∫[a, b]f(x)dx = F(b) - F(a), where F(x) is an antiderivative of f(x).

Standard Integral Formulas:

∫xn dx = xn+1/(n+1) + C (n ≠ -1)

∫cos x dx = sin x + C

∫sin x dx = -cos x + C

∫sec2 x dx = tan x + C

∫cosec2 x dx = -cot x + C

∫sec x tan x dx = sec x + C

∫cosec x cot x dx = -cosec x + C

∫ex dx = ex + C

∫ax dx = (ax)/ln(a) + C

∫(1/x) dx = ln|x| + C

Other Integral Formulas:

∫tan x dx = ln|sec x| + C

∫cot x dx = ln|sin x| + C

∫sec x dx = ln|sec x + tan x| + C

∫cosec x dx = ln|cosec x - cot x| + C

Free download Class 12 Maths Chapter 7 Question Answer for CBSE Exam.

NCERT Integrals Class 12 Questions And Answers (Intext Questions and Exercise)

Class 12 Integrals NCERT solutions Exercise: 7.1

Question:1 Find an anti derivative (or integral) of the following functions by the method of inspection. \sin 2x

Answer:

GIven \sin 2x ;

So, the anti derivative of \sin 2x is a function of x whose derivative is \sin 2x .

\frac{d}{dx}\left ( \cos 2x \right ) = -2\sin 2x

\implies \sin 2x =\frac{-1}{2} \frac{d}{dx}\left ( \cos 2x \right )

Therefore, we have \implies \sin 2x = \frac{d}{dx}\left ( \frac{-1}{2}\cos 2x \right )

Or, antiderivative of \sin 2x is \left ( \frac{-1}{2}\cos 2x \right ) .

Question:2 Find an anti derivative (or integral) of the following functions by the method of inspection. \cos 3x

Answer:

GIven \cos 3x ;

So, the antiderivative of \cos 3x is a function of x whose derivative is \cos 3x .

\frac{d}{dx}\left ( \sin 3x \right ) = 3\cos3x

\implies \cos 3x =\frac{1}{3} \frac{d}{dx}\left ( \sin 3x \right )

Therefore, we have the anti derivative of \cos 3x is \frac{1}{3}\sin 3x .

Question:3 Find an anti derivative (or integral) of the following functions by the method of inspection. e ^{2x}

Answer:

GIven e ^{2x} ;

So, the anti derivative of e ^{2x} is a function of x whose derivative is e ^{2x} .

\frac{d}{dx}\left ( e ^{2x}\right ) = 2e ^{2x}

\implies e ^{2x} = \frac{1}{2}\frac{d}{dx}(e ^{2x})

\therefore e ^{2x} = \frac{d}{dx}(\frac{1}{2}e ^{2x})

Therefore, we have the anti derivative of e^{2x} is \frac{1}{2}e ^{2x} .

Question:4 Find an anti derivative (or integral) of the following functions by the method of inspection. ( ax + b )^2

Answer:

GIven ( ax + b )^2 ;

So, the anti derivative of ( ax + b )^2 is a function of x whose derivative is ( ax + b )^2 .

\frac{d}{dx} (ax+b)^3 = 3a(ax+b)^2

\Rightarrow (ax+b)^2 =\frac{1}{3a}\frac{d}{dx}(ax+b)^3

\therefore (ax+b)^2 = \frac{d}{dx}[\frac{1}{3a}(ax+b)^3]

Therefore, we have the anti derivative of (ax+b)^2 is [\frac{1}{3a}(ax+b)^3] .

Question:5 Find an anti derivative (or integral) of the following functions by the method of inspection. \sin 2x - 4 e ^{3x}

Answer:

GIven \sin 2x - 4 e ^{3x} ;

So, the anti derivative of \sin 2x - 4 e ^{3x} is a function of x whose derivative is \sin 2x - 4 e ^{3x} .

\frac{d}{dx} (-\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x}) = \sin 2x -4e^{3x}

Therefore, we have the anti derivative of \sin 2x - 4 e ^{3x} is \left ( -\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x} \right ) .

Question:6 Find the following integrals

\int ( 4e ^{3x}+1) dx

Answer:

Given intergral \int ( 4e ^{3x}+1) dx ;

4\int e ^{3x} dx + \int 1 dx = 4\left ( \frac{e^{3x}}{3} \right ) +x +C

or \frac{4}{3} e^{3x} +x +C , where C is any constant value.

Question:7 Find the following integrals \int x ^2 ( 1- \frac{1}{x^2})dx

Answer:

Given intergral \int x ^2 ( 1- \frac{1}{x^2})dx ;

\int x^2 dx - \int1dx

or \frac{x^3}{3} - x +C , where C is any constant value.

Question:8 Find the following integrals \int ( ax ^2 + bx + c ) dx

Answer:

Given intergral \int ( ax ^2 + bx + c ) dx ;

\int ax^2\ dx + \int bx\ dx + \int c\ dx

= a\int x^2\ dx + b\int x\ dx + c\int dx

= a\frac{x^3}{3}+b\frac{x^2}{2}+cx +C

or \frac{ax^3}{3}+\frac{bx^2}{2}+cx +C , where C is any constant value.

Question:9 Find the following integrals intergration of \int \left ( 2x^2 + e ^x \right ) dx

Answer:

Given intergral \int \left ( 2x^2 + e ^x \right ) dx ;

\int 2x^2\ dx + \int e^{x}\ dx

= 2\int x^2\ dx + \int e^{x}\ dx

= 2\frac{x^3}{3}+e^{x} +C

or \frac{2x^3}{3}+e^{x} +C , where C is any constant value.

Question:10 Find the following integrals \int \left ( \sqrt x - \frac{1}{\sqrt x } \right ) ^2 dx

Answer:

Given integral \int \left ( \sqrt x - \frac{1}{\sqrt x } \right ) ^2 dx ;

or \int (x+\frac{1}{x}-2)\ dx

= \int x\ dx + \int \frac{1}{x}\ dx -2\int dx

= \frac{x^2}{2} + \ln|x| -2x +C , where C is any constant value.

Question:11 Find the following integrals intergration of \int \frac{x^3 + 5x^2 - 4}{x^2} dx

Answer:

Given intergral \int \frac{x^3 + 5x^2 - 4}{x^2} dx ;

or \int \frac{x^3}{x^2}\ dx+\int \frac{5x^2}{x^2}\ dx -4\int \frac{1}{x^2}\ dx

\int x\ dx + 5\int1. dx - 4\int x^{-2}\ dx

= \frac{x^2}{2}+5x-4\left ( \frac{x^{-1}}{-1} \right )+C

Or, \frac{x^2}{2}+5x+\frac{4}{x}+C , where C is any constant value.

Question:12 Find the following integrals \int \frac{x^3+ 3x +4 }{\sqrt x } dx

Answer:

Given intergral \int \frac{x^3+ 3x +4 }{\sqrt x } dx ;

or \int \frac{x^3}{x^{\frac{1}{2}}}\ dx+\int \frac{3x}{x^{\frac{1}{2}}}\ dx +4\int \frac{1}{x^{\frac{1}{2}}}\ dx

= \int x^{\frac{5}{2}}\ dx + 3\int x^{\frac{1}{2}}\ dx +4\int x^{-\frac{1}{2}}\ dx

=\frac{x^{\frac{7}{2}}}{\frac{7}{2}}+\frac{3\left ( x^{\frac{3}{2}} \right )}{\frac{3}{2}}+\frac{4\left ( x^{\frac{1}{2}} \right )}{\frac{1}{2}} +C

Or, = \frac{2}{7}x^{\frac{7}{2}} +2x^{\frac{3}{2}}+8\sqrt{x} +C , where C is any constant value.

Question:13 Find the following integrals intergration of \int \frac{x^3 - x^2 + x -1 }{x-1 } dx

Answer:

Given integral \int \frac{x^3 - x^2 + x -1 }{x-1 } dx

It can be written as

= \int \frac{x^2(x-1)+(x+1)}{(x-1)} dx

Taking (x-1) common out

= \int \frac{(x-1)(x^2+1)}{(x-1)} dx

Now, cancelling out the term (x-1) from both numerator and denominator.

= \int (x^2+1)dx

Splitting the terms inside the brackets

=\int x^2dx + \int 1dx

= \frac{x^3}{3}+x+c

Question:14 Find the following integrals \int (1-x) \sqrt x dx

Answer:

Given intergral \int (1-x) \sqrt x dx ;

\int \sqrt{x}\ dx - \int x\sqrt{x}\ dx or

\int x^{\frac{1}{2}}\ dx - \int x^{\frac{3}{2}} \ dx

= \frac{x^\frac{3}{2}}{\frac{3}{2}} - \frac{x^{\frac{5}{2}}}{\frac{5}{2}} +C

= \frac{2}{3}x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}}+C , where C is any constant value.

Question:15 Find the following integrals \int \sqrt x ( 3x^2 + 2x +3 )dx

Answer:

Given intergral \int \sqrt x ( 3x^2 + 2x +3 )dx ;

= \int 3x^2\sqrt{x}\ dx + \int 2x\sqrt{x}\ dx + \int 3\sqrt {x}\ dx or = 3\int x^{\frac{5}{2}}\ dx + 2\int x^{\frac{3}{2}} \ dx +3\int x^{\frac{1}{2}} \ dx

= 3\frac{x^\frac{7}{2}}{\frac{7}{2}} +2\frac{x^{\frac{5}{2}}}{\frac{5}{2}} +3\frac{x^{\frac{3}{2}}}{\frac{3}{2}} +C

= \frac{6}{7}x^{\frac{7}{2}} + \frac{4}{5}x^{\frac{5}{2}}+ 2x^{\frac{3}{2}}+C , where C is any constant value.

Question:16 Find the following integrals \int ( 2x - 3 \cos x + e ^x ) dx

Answer:

Given integral \int ( 2x - 3 \cos x + e ^x ) dx ;

splitting the integral as the sum of three integrals

\int 2x\ dx -3 \int \cos x\ dx +\int e^{x}\ dx

= 2 \frac{x^2}{2} - 3 \sin x + e^x+C

= x^2 - 3 \sin x + e^x+C , where C is any constant value.

Question:17 Find the following integrals \int ( 2 x ^2 - 3 \sin x + 5 \sqrt x ) dx

Answer:

Given integral \int ( 2 x ^2 - 3 \sin x + 5 \sqrt x ) dx ;

2\int x^2\ dx -3\int \sin x\ dx + 5\int \sqrt {x}\ dx

= 2 \frac{x^3}{3} - 3(-\cos x ) +5\left ( \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C

= \frac{2x^3}{3} +3\cos x +\frac{10}{3} x^{\frac{3}{2}}+C , where C is any constant value.

Question:18 Find the following integrals \int \sec x ( \sec x + \tan x ) dx

Answer:

Given integral \int \sec x ( \sec x + \tan x ) dx ;

\int (\sec^2x+ \sec x \tan x ) \ dx

Using the integral of trigonometric functions

= \int (sec^2 x )\ dx+ \int \sec x \tan x\ dx

= \tan x + \sec x +C , where C is any constant value.

Question:19 Find the following integrals intergration of \int \frac{sec ^2 x }{cosec ^2 x } dx

Answer:

Given integral \int \frac{sec ^2 x }{cosec ^2 x } dx ;

\int \frac{\frac{1}{\cos^2x}}{\frac{1}{\sin^2 x}}\ dx

= \int \frac{\sin^2 x }{\cos ^2 x } \ dx

=\int (\sec^2 x-1 )\ dx

=\int \sec^2 x\ dx-\int1 \ dx

= \tan x -x+C , where C is any constant value.

Question:20 Find the following integrals \int \frac{2- 3 \sin x }{\cos ^ 2 x } dx

Answer:

Given integral \int \frac{2- 3 \sin x }{\cos ^ 2 x } dx ;

\int \left ( \frac{2}{\cos^2x}-\frac{3\sin x }{\cos^2 x} \right )\ dx

Using antiderivative of trigonometric functions

= 2\tan x -3\sec x +C , where C is any constant value.

Question:21 Choose the correct answer
The anti derivative of \left ( \sqrt x + 1/ \sqrt x \right ) equals

A) \frac{1}{3}x ^{1/3} + 2 x ^{1/2}+ C \\\\ B) \frac{2}{3}x ^{2/3} + \frac{1}{2}x ^{2}+ C \\\\ C ) \frac{2}{3}x ^{3/2} + 2 x ^{1/2}+ C\\\\ D) \frac{3}{2}x ^{3/2} + \frac{1}{2} x ^{1/2}+ C

Answer:

Given to find the anti derivative or integral of \left ( \sqrt x + 1/ \sqrt x \right ) ;

\int \left ( \sqrt x + 1/ \sqrt x \right )\ dx

\int x^{\frac{1}{2}}\ dx + \int x^{-\frac{1}{2}}\ dx

= \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + \frac{x^{\frac{1}{2}}}{\frac{1}{2}}+C

= \frac{2}{3}x^{\frac{3}{2}} + 2x^{\frac{1}{2}} +C , where C is any constant value.

Hence the correct option is (C).

Question:22 Choose the correct answer The anti derivative of

If \frac{d}{dx}f(x) = 4 x ^3 - \frac{3}{x^4} such that f (2) = 0. Then f (x) is

A ) x ^ 4 + \frac{1}{x^3} - \frac{129 }{8} \\\\ B ) x ^ 3 + \frac{1}{x^4} - \frac{129 }{8} \\\\ C ) x ^ 4 + \frac{1}{x^3} + \frac{129 }{8}\\\\ D) x ^ 3 + \frac{1}{x^4} - \frac{129 }{8}

Answer:

Given that the anti derivative of \frac{d}{dx}f(x) = 4 x ^3 - \frac{3}{x^4}

So, \frac{d}{dx}f(x) = 4 x ^3 - \frac{3}{x^4}

f(x) = \int 4 x ^3 - \frac{3}{x^4}\ dx

f(x) = 4\int x ^3 - 3\int {x^{-4}}\ dx

f(x) = 4\left ( \frac{x^4}{4} \right ) -3\left ( \frac{x^{-3}}{-3} \right )+C

f(x) = x^4+\frac{1}{x^3} +C

Now, to find the constant C;

we will put the condition given, f (2) = 0

f(2) = 2^4+\frac{1}{2^3} +C = 0

16+\frac{1}{8} +C = 0

or C = \frac{-129}{8}

\Rightarrow f(x) = x^4+\frac{1}{x^3}-\frac{129}{8}

Therefore the correct answer is A .

Class 12 Integrals NCERT solutions Exercise: 7.2

Question:1 Integrate the functions \frac{2x}{1+ x ^2}

Answer:

Given to integrate \frac{2x}{1+ x ^2} function,

Let us assume 1+x^2 =t

we get, 2xdx = dt

\implies \int \frac{2x}{1+x^2} dx = \int \frac{1}{t} dt

= \log|t| +C

= \log|1+x^2| +C now back substituting the value of t = 1+x^2

as (1+x^2) is positive we can write

= \log(1+x^2) +C

Question:2 Integrate the functions \frac{( \log x )^2}{x}

Answer:

Given to integrate \frac{( \log x )^2}{x} function,

Let us assume \log |x| = t

we get, \frac{1}{x}dx= dt

\implies \int \frac{\left ( \log|x| \right )^2}{x}\ dx = \int t^2dt

= \frac{t^3}{3}+C

= \frac{(\log|x|)^3}{3}+C

Question:3 Integrate the functions \frac{1}{x+ x \log x }

Answer:

Given to integrate \frac{1}{x+ x \log x } function,

Let us assume 1+\log x = t

we get, \frac{1}{x}dx= dt

\implies \int \frac{1}{x(1+\log x )} dx = \int \frac{1}{t} dt

= \log|t| +C

= \log |1+ \log x | +C

Question:4 Integrate the functions \sin x \sin ( \cos x )

Answer:

Given to integrate \sin x \sin ( \cos x ) function,

Let us assume \cos x =t

we get, -\sin x dx =dt

\implies \int \sin x \sin(\cos x)dx = -\int \sin t dt

= -\left ( -\cos t \right ) +C

= \cos t +C

Back substituting the value of t we get,

= \cos (\cos x ) +C

Question:5 Integrate the functions \sin ( ax + b ) \cos ( ax + b )

Answer:

Given to integrate \sin ( ax + b ) \cos ( ax + b ) function,

\sin ( ax + b ) \cos ( ax + b ) = \frac{2\sin ( ax + b ) \cos ( ax + b )}{2} = \frac{\sin 2(ax+b)}{2}

Let us assume 2(ax+b) = t

we get, 2adx =dt

\int \frac{\sin 2(ax+b)}{2} dx = \frac{1}{2}\int \frac{\sin t}{2a} dt

= \frac{1}{4a}[-cos t] +C

Now, by back substituting the value of t,

= \frac{-1}{4a}[cos 2(ax+b)] +C

Question:6 Integrate the functions \sqrt { ax + b }

Answer:

Given to integrate \sqrt { ax + b } function,

Let us assume (ax+b) = t

we get, adx =dt

dx = \frac{1}{a}dt

\Rightarrow \int(ax+b)^{\frac{1}{2}} dx = \frac{1}{a}\int t^{\frac{1}{2}}dt

Now, by back substituting the value of t,

= \frac{1}{a}\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C

= \frac{2(ax+b)^\frac{3}{2}}{3a} +C

Question:7 Integrate the functions x \sqrt { x +2 }

Answer:

Given function x \sqrt { x +2 } ,

\int x\sqrt{x+2}

Assume the (x+2) = t 19634

\therefore dx =dt

\Rightarrow \int x\sqrt{x+2} dx = \int (t-2) \sqrt{t} dt

= \int (t-2) \sqrt{t} dt

= \int \left ( t^{\frac{3}{2}}-2t^{\frac{1}{2}} \right )dt

= \int t^{\frac{3}{2}}dt -2\int t^{\frac{1}{2}}dt

= \frac{t^{\frac{5}{2}}}{\frac{5}{2}} -2\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C

= \frac{2}{5}t^{\frac{5}{2}} -\frac{4}{3}t^{\frac{3}{2}} +C

Back substituting the value of t in the above equation.

or, \frac{2}{5}(x+2)^{\frac{5}{2}}- \frac{4}{3}(x+2)^\frac{3}{2} +C , where C is any constant value.

Question:8 Integrate the functions x \sqrt { 1+ 2 x^2 }

Answer:

Given function x \sqrt { 1+ 2 x^2 } ,

\int x \sqrt { 1+ 2 x^2 }\ dx

Assume the 1+2x^2= t

\therefore 4xdx =dt

\Rightarrow \int x\sqrt{1+2x^2}dx = \int \frac{\sqrt {t}}{4} dt

Or = \frac{1}{4}\int t^{\frac{1}{2}} dt = \frac{1}{4}\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C

= \frac{1}{6}(1+2x^2)^{\frac{3}{2}} +C , where C is any constant value.

Question:9 Integrate the functions ( 4x +2 ) \sqrt { x ^ 2 + x + 1 }

Answer:

Given function ( 4x +2 ) \sqrt { x ^ 2 + x + 1 } ,

\int ( 4x +2 ) \sqrt { x ^ 2 + x + 1 } dx

Assume the 1+x+x^2 = t

\therefore (2x+1)dx =dt

\Rightarrow \int (4x+2)\sqrt{1+x+x^2} dx

= \int 2\sqrt {t}dt = 2\int \sqrt{t}dt

= 2\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right ) +C

Now, back substituting the value of t in the above equation,

= \frac{4}{3}(1+x+x^2)^{\frac{3}{2}} +C , where C is any constant value.

Question:10 Integrate the functions \frac{1}{x - \sqrt x }

Answer:

Given function \frac{1}{x - \sqrt x } ,

\int \frac{1}{x - \sqrt x } dx

Can be written in the form:

\frac{1}{x - \sqrt x } = \frac{1}{\sqrt {x}(\sqrt{x}-1)}

Assume the (\sqrt{x}-1) =t

\therefore \frac{1}{2\sqrt{x}}dx =dt

\Rightarrow \int \frac{1}{\sqrt{x}(\sqrt{x}-1)}dx = \int \frac{2}{t}dt

= 2\log|t| +C

= 2\log|\sqrt{x}-1| +C , where C is any constant value.

Question:11 Integrate the functions \frac{x }{ \sqrt{ x +4} } , x > 0

Answer:

Given function \frac{x }{ \sqrt{ x +4} } ,

\int \frac{x }{ \sqrt{ x +4} }dx

Assume the x+4 =t so, x =t-4

\therefore dx=dt

\int \frac{x}{\sqrt{x+4}}dx = \int \frac{t-4}{\sqrt{t}}dt

\int t^\frac{1}{2}dt -4\int t^{\frac{-1}{2}}dt

= \frac{2}{3}t^{\frac{3}{2}} - 4\left ( 2t^{\frac{1}{2}} \right )+C

= \frac{2}{3}(x+4)^{\frac{3}{2}} -16(x+4)^{\frac{1}{2}}+C

, where C is any constant value.

Question:12 Integrate the functions ( x ^3 - 1 ) ^{1/3} x ^ 5

Answer:

Given function ( x ^3 - 1 ) ^{1/3} x ^ 5 ,

\int ( x ^3 - 1 ) ^{1/3} x ^ 5 dx

Assume the x^3-1 = t

\therefore 3x^2dx=dt

\implies \int(x^3-1)^{\frac{1}{3}} x^5 dx = \int (x^3-1)^{\frac{1}{3}}x^3.x^2dx

= \int t^{\frac{1}{3}}(t+1)\frac{dt}{3}

= \frac{1}{3} \int \left ( t^\frac{4}{3}+t^\frac{1}{3} \right )dt

= \frac{1}{3}\left [ \frac{t^{\frac{7}{3}}}{\frac{7}{3}}+\frac{t^{\frac{4}{3}}}{\frac{4}{3}} \right ]+C

= \frac{1}{3}\left [ \frac{3}{7}t^{\frac{7}{3}}+\frac{3}{4}t^{\frac{4}{3}} \right ]+C

= \frac{1}{7}(x^3-1)^{\frac{7}{3}} + \frac{1}{4}(x^3-1)^{\frac{4}{3}} +C , where C is any constant value.

Question:13 Integrate the functions \frac{x ^2 }{(2+3x^3)^3}

Answer:

Given function \frac{x ^2 }{(2+3x^3)^3} ,

\int \frac{x ^2 }{(2+3x^3)^3} dx

Assume the 2+3x^3 =t

\therefore 9x^2dx=dt

\implies \int\frac{x^2}{(2+3x^2)}dx = \frac{1}{9}\int \frac{dt}{t^3}

= \frac{1}{9}\left ( \frac{t^{-2}}{-2} \right ) +C

= \frac{-1}{18}\left ( \frac{1}{t^2} \right )+C

= \frac{-1}{18(2+3x^3)^2}+C , where C is any constant value.

Question:14 Integrate the functions \frac{1}{x (\log x )^m} , x > 0 , m \neq 1

Answer:

Given function \frac{1}{x (\log x )^m} , x > 0 , m \neq 1 ,

Assume the \log x =t

\therefore \frac{1}{x}dx =dt

\implies \int\frac{1}{x(logx)^m}dx = \int\frac{dt}{t^m}

=\left ( \frac{t^{-m+1}}{1-m} \right ) +C

= \frac{(log x )^{1-m}}{(1-m)} +C , where C is any constant value.

Question:15 Integrate the functions \frac{x}{9- 4 x ^2 }

Answer:

Given function \frac{x}{9- 4 x ^2 } ,

Assume the 9-4x^2 =t

\therefore -8xdx =dt

\implies \int\frac{x}{9-4x^2} = -\frac{1}{8}\int \frac{1}{t}dt

= \frac{-1}{8}\log|t| +C

Now back substituting the value of t ;

= \frac{-1}{8}\log|9-4x^2| +C , where C is any constant value.

Question:16 Integrate the functions e ^{ 2 x +3 }

Answer:

Given function e ^{ 2 x +3 } ,

Assume the 2x+3 =t

\therefore 2dx =dt

\implies \int e^{2x+3} dx = \frac{1}{2}\int e^t dt

= \frac{1}{2}e^t +C

Now back substituting the value of t ;

= \frac{1}{2}e^{2x+3}+C , where C is any constant value.

Question:17 Integrate the functions \frac{x }{e^{x^{2}}}

Answer:

Given function \frac{x }{e^{x^{2}}} ,

Assume the x^2=t

\therefore 2xdx =dt

\implies \int \frac{x}{e^{x^2}}dx = \frac{1}{2}\int \frac{1}{e^t}dt

= \frac{1}{2}\int e^{-t} dt

= \frac{1}{2}\left ( \frac{e^{-t}}{-1} \right ) +C

= \frac{-1}{2}e^{-x^2} +C

= \frac{-1}{2e^{x^2} }+C , where C is any constant value.

Question:18 Integrate the functions \frac{e ^{\tan ^{-1}x}}{1+ x^2 }

Answer:

Given,

\frac{e ^{\tan ^{-1}x}}{1+ x^2 }

Let's do the following substitution

\\ tan^{-1}x = t \\ \implies \frac{1}{1+x^2}dx = dt

\therefore \int \frac{e ^{\tan ^{-1}x}}{1+ x^2 }dx = \int e ^{t}dt = e^t + C

= e^{tan^{-1}x} + C

Question:19 Integrate the functions \frac{e ^{2x}-1}{e ^{2x}+1}

Answer:

Given function \frac{e ^{2x}-1}{e ^{2x}+1} ,

Simplifying it by dividing both numerator and denominator by e^x , we obtain

\frac{\frac{e^{2x}-1}{e^x}}{\frac{e^{2x}+1}{e^x}} = \frac{e^x-e^{-x}}{e^x+e^{-x}}

Assume the e^{x}+e^{-x} =t

\therefore (e^x-e^{-x})dx =dt

\implies \int \frac{e^{2x}-1}{e^{2x}+1}dx = \int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx

= \int \frac{dt}{t}

= \log |t| +C

Now, back substituting the value of t,

= \log |e^x+e^{-x}| +C , where C is any constant value.

Question:20 Integrate the functions \frac{e ^{2x}- e ^{-2x }}{e ^ {2x }+ e ^{ -2 x }}

Answer:

Given function \frac{e ^{2x}- e ^{-2x }}{e ^ {2x }+ e ^{ -2 x }} ,

Assume the e^{2x}+e^{-2x} =t

\therefore (2e^{2x}-2e^{-2x})dx =dt

\implies \int \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}dx = \int \frac{dt}{2t}

= \frac{1}{2}\int \frac{1}{t}dt

= \frac{1}{2}\log|t| +C

Now, back substituting the value of t,

= \frac{1}{2}\log|e^{2x}+e^{-2x}| +C , where C is any constant value.

Question:21 Integrate the functions \tan ^2 ( 2x-3 )

Answer:

Given function \tan ^2 ( 2x-3 ) ,

Assume the 2x-3 =t

\therefore 2dx =dt

\implies \int \tan^2(2x-3) dx = \frac{\int \tan^2(t)}{2}dt

=\frac{1}{2}\int (\sec^2t -1) dt \left [\because \tan^2t+1 = \sec^2 t \right ]

= \frac{1}{2}\left [ \tan t - t \right ] +C

Now, back substituting the value of t,

= \frac{1}{2}\left [ \tan(2x-3)-2x+3 \right ]+C

or \frac{1}{2} \tan(2x-3)-x+C , where C is any constant value.

Question:22 Integrate the functions \sec ^2 ( 7- 4x )

Answer:

Given function \sec ^2 ( 7- 4x ) ,

Assume the 7-4x=t

\therefore -4dx =dt

\implies \int \sec^2(7-4x)dx = \frac{-1}{4}\int \sec^2t dt

=-\frac{1}{4}(\tan t) +C

Now, back substituted the value of t.

=-\frac{1}{4}\tan(7-4x)+C , where C is any constant value.

Question:23 Integrate the functions \frac{\sin ^{-1}x}{\sqrt { 1- x^2 }}

Answer:

Given function \frac{\sin ^{-1}x}{\sqrt { 1- x^2 }} ,

Assume the \sin^{-1}x =t

\therefore \frac{1}{\sqrt{1-x^2}}dx = dt

\implies \int \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx =\int t dt

= \frac{t^2}{2}+C

Now, back substituted the value of t.

= \frac{(\sin^{-1}x)^2}{2}+C , where C is any constant value.

Question:24 Integrate the functions \frac{2 \cos x - 3\sin x }{6 \cos x + 4 \sin x }

Answer:

Given function \frac{2 \cos x - 3\sin x }{6 \cos x + 4 \sin x } ,

or simplified as \frac{2 \cos x - 3\sin x }{2(3 \cos x + 2 \sin x) }

Assume the 3\cos x +2\sin x =t

\therefore (-3\sin x + 2\cos x )dx =dt

\implies \int \frac{2\cos x - 3\sin x }{6\cos x +4\sin x }dx = \int \frac{dt}{2t}

= \frac{1}{2}\int \frac{dt}{t}

= \frac{1}{2}\log|t| +C

Now, back substituted the value of t.

= \frac{1}{2}\log|3\cos x +2\sin x| +C , where C is any constant value.

Question:25 Integrate the functions \frac{1 }{ \cos ^2 x (1-\tan x )^2}

Answer:

Given function \frac{1 }{ \cos ^2 x (1-\tan x )^2} ,

or simplified as \frac{1 }{ \cos ^2 x (1-\tan x )^2} = \frac{\sec^2x}{(1-\tan x)^2}

Assume the (1-\tan x)=t

\therefore -\sec^2xdx =dt

\implies \int \frac{\sec^2x}{(1-\tan x)^2}dx = \int\frac{-dt}{t^2}

= -\int t^{-2} dt

= \frac{1}{t} +C

Now, back substituted the value of t.

= \frac{1}{1-\tan x}+C

where C is any constant value.

Question:26 Integrate the functions \frac{\cos \sqrt x }{\sqrt x }

Answer:

Given function \frac{\cos \sqrt x }{\sqrt x } ,

Assume the \sqrt x =t

\therefore \frac{1}{2\sqrt x}dx =dt

\implies \int \frac{\cos \sqrt{x}}{\sqrt{x}}dx = 2\int \cos t dt

= 2\sin t +C

Now, back substituted the value of t.

= 2\sin \sqrt{x}+C , where C is any constant value.

Question:27 Integrate the functions \sqrt { \sin 2x } \cos 2x

Answer:

Given function \sqrt { \sin 2x } \cos 2x ,

Assume the \sin 2x = t

\therefore 2\cos 2x dx =dt

\implies \int \sqrt{\sin 2x }\cos 2x dx = \frac{1}{2}\int \sqrt t dt

= \frac{1}{2}\left ( \frac{t^{\frac{3}{2}}}{\frac{3}{2}} \right )+C

= \frac{1}{3}t^{\frac{3}{2}}+C

Now, back substituted the value of t.

= \frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+C , where C is any constant value.

Question:28 Integrate the functions \frac{\cos x }{\sqrt { 1+ \sin x }}

Answer:

Given function \frac{\cos x }{\sqrt { 1+ \sin x }} ,

Assume the 1+\sin x =t

\therefore \cos x dx = dt

\implies \int \frac{\cos x }{\sqrt{1+\sin x}}dx = \int \frac{dt}{\sqrt t}

= \frac{t^{\frac{1}{2}}}{\frac{1}{2}} +C

= 2\sqrt t +C

Now, back substituted the value of t.

= 2{\sqrt{1+\sin x}} +C , where C is any constant value.

Question:29 Integrate the functions \cot x \: log \sin x

Answer:

Given function \cot x \: log \sin x ,

Assume the \log \sin x =t

\therefore \frac{1}{\sin x }.\cos x dx =dt

\cot x dx =dt

\implies \int \cot x \log \sin x dx =\int t dt

= \frac{t^2}{2}+C

Now, back substituted the value of t.

= \frac{1}{2}(\log \sin x )^2+C , where C is any constant value.

Question:30 Integrate the functions \frac{\sin x }{1+ \cos x }

Answer:

Given function \frac{\sin x }{1+ \cos x } ,

Assume the 1+\cos x =t

\therefore -\sin x dx =dt

\implies \int \frac{\sin x}{1+\cos x}dx = \int -\frac{dt}{t}

= -\log|t| +C

Now, back substituted the value of t.

= -\log|1+\cos x | +C , where C is any constant value.

Question:31 Integrate the functions \frac{\sin x }{( 1+ \cos x )^2}

Answer:

Given function \frac{\sin x }{( 1+ \cos x )^2} ,

Assume the 1+\cos x =t

\therefore -\sin x dx =dt

\implies \int \frac{\sin x}{(1+\cos x)^2}dx = \int -\frac{dt}{t^2}

= -\int t^{-2}dt

= \frac{1}{t}+C

Now, back substituted the value of t.

= \frac{1}{1+\cos x} +C , where C is any constant value.

Question:32 Integrate the functions \frac{1}{1+ \cot x }

Answer:

Given function \frac{1}{1+ \cot x }

Assume that I = \int \frac{1}{1+ \cot x } dx

Now solving the assumed integral;

I = \int \frac{1}{1+ \frac{\cos x }{\sin x} } dx

= \int \frac{\sin x }{\sin x + \cos x } dx

= \frac{1}{2}\int \frac{2\sin x }{\sin x + \cos x } dx

= \frac{1}{2}\int \frac{(\sin x+ \cos x ) +(\sin x -\cos x ) }{(\sin x + \cos x) } dx

=\frac{1}{2}\int 1 dx + \frac{1}{2} \int \frac{\sin x -\cos x }{\sin x +\cos x } dx

=\frac{1}{2}(x) + \frac{1}{2} \int \frac{\sin x -\cos x }{\sin x +\cos x } dx

Now, to solve further we will assume \sin x + \cos x =t

Or, (\cos x -\sin x)dx =dt

\therefore I = \frac{x}{2}+ \frac{1}{2}\int \frac{-(dt)}{t}

= \frac{x}{2}- \frac{1}{2}\log|t| +C

Now, back substituting the value of t,

= \frac{x}{2}- \frac{1}{2}\log|\sin x + \cos x| +C

Question:33 Integrate the functions \frac{1}{1- \tan x }

Answer:

Given function \frac{1}{1- \tan x }

Assume that I = \int \frac{1}{1- \tan x } dx

Now solving the assumed integral;

I = \int \frac{1}{1-\frac{\sin x}{\cos x }} dx

= \int \frac{\cos x }{\cos x - \sin x } dx

= \frac{1}{2}\int \frac{2\cos x }{\cos x - \sin x } dx

= \frac{1}{2}\int \frac{(\cos x -\sin x ) +(\cos x +\sin x ) }{(\cos x - \sin x) } dx

=\frac{1}{2}\int 1 dx + \frac{1}{2} \int \frac{\cos x +\sin x }{\cos x -\sin x } dx

=\frac{1}{2}(x) + \frac{1}{2} \int \frac{\cos x +\sin x }{\cos x -\sin x } dx

Now, to solve further we will assume \cos x - \sin x =t

Or, (-\sin x-\cos x )dx =dt

\therefore I = \frac{x}{2}+ \frac{1}{2}\int \frac{-(dt)}{t}

= \frac{x}{2}- \frac{1}{2}\log|t| +C

Now, back substituting the value of t,

= \frac{x}{2}- \frac{1}{2}\log|\cos x - \sin x| +C

Question:34 Integrate the functions \frac{\sqrt { \tan x } }{\sin x \cos x }

Answer:

Given function \frac{\sqrt { \tan x } }{\sin x \cos x }

Assume that I = \int \frac{\sqrt { \tan x } }{\sin x \cos x }dx

Now solving the assumed integral;

Multiplying numerator and denominator by \cos x ;

I = \int \frac{\sqrt{\tan x }\times\cos x}{\sin x \cos x\times \cos x}dx

= \int \frac{\sqrt{\tan x }}{\tan x \cos^2 x } dx

= \int \frac{\sec^2 x }{\sqrt{\tan x }}dx

Now, to solve further we will assume \tan x =t

Or, \sec^2{x}dx =dt

\therefore I =\int \frac{dt}{\sqrt t}

=2\sqrt t +C

Now, back substituting the value of t,

= 2\sqrt{\tan x } +C

Question:35 Integrate the functions \frac{( 1+ \log x )^2}{x}

Answer:

Given function \frac{( 1+ \log x )^2}{x}

Assume that 1+\log x =t

\therefore \frac{1}{x}dx =dt

= \int \frac{(1+\log x )^2}{x}dx = \int t^2 dt

= \frac{t^3}{3}+C

Now, back substituting the value of t,

= \frac{(1+\log x )^3}{3}+C

Question:36 Integrate the functions \frac{( x+1)( x+ \log x )^2}{x }

Answer:

Given function \frac{( x+1)( x+ \log x )^2}{x }

Simplifying to solve easier;

\frac{( x+1)( x+ \log x )^2}{x } = \left ( \frac{x+1}{x} \right )\left ( x+\log x \right )^2

=\left ( 1+\frac{1}{x} \right )\left ( x+\log x \right )^2

Assume that x+\log x =t

\therefore \left ( 1+\frac{1}{x} \right )dx = dt

= \int \left ( 1+\frac{1}{x} \right )\left ( x+\log x \right )^2 dx = \int t^2 dt

= \frac{t^3}{3}+C

Now, back substituting the value of t,

= \frac{(x+\log x )^3}{3}+C

Question:37 Integrate the functions \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }

Answer:

Given function \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }

Assume that x^4 =t

\therefore 4x^3 dx =dt

\Rightarrow \int \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }dx = \frac{1}{4} \int \frac{\sin(\tan^{-1} t)}{1+t^2}dt ......................(1)

Now to solve further we take \tan ^{-1} t = u

\therefore \frac{1}{1+t^2} dt =du

So, from the equation (1), we will get

\Rightarrow \int \frac{x ^3 \sin ( \tan ^{-1} x ^ 4 )}{1 + x ^8 }dx =\frac{1}{4}\int \sin u\ du

= \frac{1}{4}(-\cos u) +C

Now back substitute the value of u,

= \frac{-1}{4}\cos (\tan^{-1} t) +C

and then back substituting the value of t,

= \frac{-1}{4}\cos (\tan^{-1} x^4) +C

Question:38 Choose the correct answer \int \frac{10 x^ 9 + 10 ^x \log _ e 10 dx }{x ^{10}+ 10 ^x }dx\: \: \: equals

(A) 10^x - x^{10} + C \\\\(B) 10^x + x^{10} + C\\\\ (C) (10^x - x^{10})^{-1} + C \\\\ (D) log (10^x + x^{10}) + C

Answer:

Given integral \int \frac{10 x^ 9 + 10 ^x \log _ e 10 dx }{x ^{10}+ 10 ^x }dx

Taking the denominator x^{10} +10^x = t

Now differentiating both sides we get

\therefore \left ( 10x^9+10^x\log_{e}10 \right )dx = dt

\implies \int \frac{10x^9+10^x\log_{e}10}{x^{10}+10^x} dx = \int \frac{dt}{t}

= \log t +C

Back substituting the value of t,

= \log (x^{10}+10^x) +C

Therefore the correct answer is D.

Question:39 Choose the correct answer \int \frac{dx }{\sin ^ 2 x \cos ^2 x }\: \: \: equals

(A) \tan x + \cot x + C \\\\ (B) \tan x - \cot x + C\\\\ (C) \tan x \cot x + C\\\\ (D) \tan x - \cot 2x + C

Answer:

Given integral \int \frac{dx }{\sin ^ 2 x \cos ^ 2x }

\int \frac{dx }{\sin ^ 2 x \cos ^ 2x } = \int \frac{1}{\sin ^2 x \cos ^2 x } dx

=\int \frac{\sin ^2 x +\cos^2 x }{\sin^2 x \cos^2 x}dx \left ( \because \sin ^2 x +\cos^2 x =1 \right )

=\int \frac{\sin^2 x }{\sin^2 x \cos^2 x}dx + \int \frac{\cos^2 x}{\sin^2 x \cos^2 x}dx

=\int \sec^2 x dx + \int cosec^2 x dx

=\tan x -\cot x +C

Therefore, the correct answer is B.


Class 12 Integrals NCERT Solutions Exercise: 7.3

Question:1 Find the integrals of the functions \sin ^ 2 ( 2x+ 5 )

Answer:

\sin ^ 2 ( 2x+ 5 )

using the trigonometric identity

sin^2x=\frac{1-cos2x}{2}

we can write the given question as

= \frac{1-\cos 2(2x+5)}{2} = \frac{1-\cos (4x+10)}{2}
\\=\int \frac{1-\cos (4x+10)}{2}dx\\ =\frac{1}{2}\int dx - \frac{1}{2}\int \cos(4x+10)dx\\ =\frac{x}{2}-\frac{1}{2}[\sin(4x+10)/4]\\ =\frac{x}{2}-\frac{\sin(4x+10)}{8}+C

Question:2 Find the integrals of the functions \sin 3x \cos 4x

Answer:

Using identity \sin A\cos B = 1/2[sin(A+B)+sin(A-B)]

, therefore the given integral can be written as

\int \sin 3x\cos 4x=\frac{1}{2}\int sin(3x+4x)+sin(3x-4x)\ dx

=\frac{1}{2}\int sin(7x)-sin(x)\ dx\\ =1/2[\int \sin (7x) dx-\int \sin x\ dx]\\ =\frac{1}{2}[(-1/7)\cos 7x+\cos x+ C]\\ = \frac{\cos x}{2}-\frac{\cos 7x}{14}+C

Question:3 Find the integrals of the functions \cos 2x \cos 4x \cos 6x

Answer:

Using identity
cosAcosB = \frac{1}{2}[cos(A+B)+cos(A-B)]

\int \cos 2x.\cos 4x.\cos 6x = \int \cos 2x. \frac{1}{2}[(\cos 10x)+\cos 2x]dx

Again use the same identity mentioned in the first line

\\= \frac{1}{2}\int (\cos 2x.\cos 10x+\cos 2x. \cos 2x)dx\\ =\frac{1}{2}\int\frac{1}{2}({\cos12x +\cos 8x})dx+\frac{1}{2}\int (\frac{1+\cos 4x}{2})dx\\ =\frac{\sin 12x}{48}+\frac{\sin 8x}{32}+\frac{\sin 4x}{16}+ x/4+C

Question:4 Find the integrals of the functions \sin ^ 3 ( 2x +1 )

Answer:

\int \sin^3(2x+1)dx = \int \sin^2(2x+1).\sin(2x+1)dx

The integral can be written as

= \int (1-\cos^2(2x+1)).\sin(2x+1)dx
Let
\\\cos (2x+1) =t\\ \sin (2x+1)dx = -dt/2

\\=\frac{-1}{2}\int (1-t^2)dt\\ =\frac{-1}{2}[t-t^3/3]\\ =\frac{t^3}{6}-\frac{t}{2}

Now, replace the value of t, we get;

=\frac{\cos^3(2x+1)}{6}-\frac{\cos(2x+1)}{2}+C

Question:5 Find the integrals of the functions \sin ^3 x \cos ^ 3 x

Answer:

I = \int \sin^3x.\cos^3x\ dx

rewrite the integral as follows

\\=\int cos^3x.sin^2x.\sin x\ dx\\ =\int cos^3x(1-\cos^2x)\sin x\ dx
Let \cos x = t \Rightarrow \sin x dx =-dt

\\=-\int t^3(1-t^2)dt\\ =-\int(t^3-t^5)dt\\ =-[\frac{t^4}{4}]+[\frac{t^6}{6}] +C\\ =\frac{\cos^6x}{6}-\frac{cos^4x}{4}+C ......(replace the value of t as cos\ x )

Question:6 Find the integrals of the functions \sin x \sin 2x \sin 3x

Answer:

Using the formula
sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))

we can write the integral as follows

\int \sin x.\sin 2x\sin 3x\ dx = \int \sin x\frac{1}{2}[\cos x-\cos 5x]dx
\\=\frac{1}{2} \int [\sin x.\cos x-\sin x.\cos 5x]dx\\ =\frac{1}{2}\int \frac{\sin 2x}{2}dx-\frac{1}{2}\int \sin x. \cos 5x\ dx\\ =-\frac{\cos 2x}{8}-\frac{1}{4}\int[\sin 6x -\sin 4x]\\ =-\frac{\cos 2x}{8}-\frac{1}{4}[\frac{-\cos 6x}{6}+\frac{\cos 4x}{4}]\\ =-\frac{\cos 2x}{8}+\frac{\cos 6x}{24}-\frac{\cos 4x}{16}+C

Question:7 Find the integrals of the functions \sin 4x \sin 8x

Answer:

Using identity

sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))

we can write the following integral as

\sin 4x \sin 8x =
\\=\frac{1}{2}\int(\cos 4x - \cos 12x) dx\\ =\frac{1}{2} [\int\cos 4x\ dx - \int \cos 12x\ dx]\\ =\frac{\sin 4x}{8}-\frac{\sin 12x}{24}+C

Question:8 Find the integrals of the functions \frac{1- \cos x }{1+ \cos x }

Answer:

We know the identities

\\1+\cos 2A = 2\cos^2A\\ 1-\cos 2A = 2\sin^2 A

Using the above relations we can write

\frac{1-\cos x}{1+\cos x}=\frac{\sin^2x/2}{\cos^2x/2} = \tan^2x/2

=\int \tan^2x/2 =\int (\sec^2x/2-1)dx
\\=\int (\sec^2x/2)dx-\int dx\\ = 2[\tan x/2]-{x}+C

Question:9 Find the integrals of the functions \frac {\cos x }{1 + \cos x }

Answer:

The integral is rewritten using trigonometric identities

\frac{\cos x}{1+ \cos x}= \frac{\cos^2x/2-\sin^2x/2}{2\cos^2x/2} =\frac{1}{2}[1-\tan^2x/2]
\\=\int \frac{1}{2}[1-\tan^2x/2] dx\\ =\frac{1}{2}\int 1-[sec^2\frac{x}{2}-1]=\frac{1}{2}\int 2-sec^2\frac{x}{2}\\=x-tan\frac{x}{2}+c

Question:10 Find the integrals of the functions \sin ^ 4 x

Answer:

\sin ^ 4 x can be written as follows using trigonometric identities
\\=\sin^2x.\sin^2x\\ =\frac{1}{4}(1-\cos 2x)^{2}\\ =\frac{1}{4}(1+\cos^22x-2\cos 2x)\\ =\frac{1}{4}(1+\frac{1}{2}(1+\cos 4x)-2\cos 2x)\\ =3/8+\frac{\cos 4x}{8}-\frac{\cos 2x}{2}

Therefore,
\Rightarrow \int \sin^4x\ dx = \int \frac{3}{8}dx+\frac{1}{8}\int\cos 4x\ dx -\frac{1}{2}\int\cos 2x\ dx
= \frac{3x}{8}+\frac{\sin 4x}{32} -\frac{\sin 2x}{4}+C

Question:11 Find the integrals of the functions \cos ^ 4 2x

Answer:

cos^42x=cos^32xcos2x

now using the identity

cos^3x=\frac{cos3x+3cosx}{4}

cos^32xcos2x=\frac{cos6x +3cos2x}{4}cos2x=\frac{cos6xcos2x+3cos^22x}{4}

now using the below two identities

cosacosb=\frac{cos(a+b)+cos(a-b)}{2}\\and\ cos^22x=\frac{1+cos4x}{2}\\

the value

cos^42x=cos^32xcos2x\\=\frac{cos6xcos2x+3cos^22x}{4}=\frac{cos 4x+cos8x}{8}+\frac{3}{4}\frac{1+cos4x}{2} .

the integral of the given function can be written as

\int cos^42x=\int \frac{cos 4x+cos8x}{8}+\int \frac{3}{4}\frac{1+cos4x}{2}\\ \\=\frac{3}{8}x+\frac{sin4x}{8}+\frac{sin8x}{64}+C

Question:12 Find the integrals of the functions \frac{\sin ^ 2x }{1+ \cos x }

Answer:

Using trigonometric identities we can write the given integral as follows.

\frac{\sin ^ 2x }{1+ \cos x }

\\=\frac{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}{2\cos^2\frac{x}{2}}\\ =2\sin^2\frac{x}{2}\\ =1-\cos x

\therefore \int \frac{sin^22x}{1+\cos x} = \int (1-\cos x)dx
\\= \int 1dx-\int\cos x\ dx\\ =x-\sin x+C

Question:13 Find the integrals of the functions \frac{\cos 2x - \cos 2 \alpha }{\cos x - \cos \alpha }

Answer:

We know that,

\cos A-\cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})

Using this identity we can rewrite the given integral as

\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos\alpha}=\frac{-2\sin\frac{2x+2\alpha}{2}\sin\frac{2x-2\alpha}{2}}{-2\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}

\\=\frac{\sin(x+\alpha)\sin(x-\alpha)}{\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}\\ =\frac{[2\sin\frac{x+\alpha}{2}\cos \frac{x+\alpha}{2}][2\sin\frac{x-\alpha}{2}\cos\frac{x-\alpha}{2}]}{\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}\\ =4\cos\frac{x+\alpha}{2}\cos\frac{x-\alpha}{2}\\ =2[\cos x+\cos \alpha]

\therefore \int\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos\alpha}=\int 2\cos x\ dx +\int 2\cos \alpha\ dx
=2[\sin x + x\cos \alpha]+C

Question:14 Find the integrals of the functions \frac{\cos x - \sin x }{1+ \sin 2x }

Answer:

\frac{\cos x-\sin x}{1+2\sin x}=\frac{\cos x-\sin x}{(sin^2x+cos^2x)+2 sin x.\cos x}
=\frac{\cos x-\sin x}{(\sin x+\cos x)^2}


\\sin x+\cos x =t\\ \therefore (\cos x-\sin x)dx = dt

Now,
=\int \frac{dt}{t^2}\\ =\int t^-2\ dt\\ =-t^{-1}+C\\ =-\frac{1}{(\sin x+\cos x)}+C

Question:15 Find the integrals of the functions \tan ^ 3 2x \sec 2x

Answer:

\tan^32x.\sec 2x = \tan^22x.\tan 2x.\sec 2x
\\= (\sec^22x-1).\tan 2x.\sec 2x\\ =\sec^22x.\tan 2x-\tan 2x.\sec 2x

Therefore integration of \tan ^ 3 2x \sec 2x =
\\=\int\sec^22x.\tan 2x\ dx-\int\tan 2x.\sec 2x\ dx\\ =\int\sec^22x.\tan 2x\ dx-\sec 2x/2+C\\ .....................(i)
Let assume

\sec 2x = t
So, that 2\sec 2x.\tan 2x\ dx =dt
Now, the equation (i) becomes,

\\\Rightarrow \frac{1}{2}\int t^2\ dt-\frac{\sec 2x}{2}+C\\ \Rightarrow \frac{t^3}{6}-\frac{\sec 2x}{2}+C\\ =\frac{(\sec 2x)^3}{6}-\frac{\sec 2x}{2}+C

Question:16 Find the integrals of the functions \tan ^ 4x

Answer:

the given question can be rearranged using trigonometric identities

tan^4x=(\sec^2x-1).\tan^2x\\ =\sec^2x.\tan^2x-\tan^2x\\ =\sec^2x.\tan^2x-\sec^2x+1

Therefore, the integration of \tan^4x = \\=\int \sec^2x.\tan^2x\ dx-\int\sec^2x\ dx+\int dx\\ =(\int \sec^2x.\tan^2x\ dx)-\tan x+x+C\\ ...................(i)
Considering only \int \sec^2x.\tan^2x\ dx
let \tan x =t\Rightarrow \sec^2x\ dx =dt

\int \sec^2x\tan^2x\ dx = \int t^2\ dt = t^3/3=\frac{\tan^3x}{3}

now the final solution is,

\int \tan^4x =\frac{\tan^3x}{3}-\tan x+x+C

Question:17 Find the integrals of the functions \frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }

Answer:

\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }

now splitting the terms we can write

\\=\frac{\sin^3x}{\sin^2x.\cos^2x}+\frac{\cos^3x}{\sin^2x.\cos^2x}\\ =\frac{\sin x}{cos^2x}+\frac{\cos x}{\sin^2x}\\ =\tan x.\sec x+\cot xcosec x

Therefore, the integration of
\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }

\\=\int (\tan x\sec x+\cot xcosec x)dx\\ =\sec x-cosec\ x+C

Question:18 Find the integrals of the functions \frac{\cos 2 x + 2 \sin ^ 2x }{\cos ^ 2 x }

Answer:

The integral of the above equation is

\\=\int (\frac{\cos 2x+2\sin^2x}{\cos^2x})dx\\ =\int (\frac{\cos 2x+(1-\cos 2x)}{\cos^2x}\\ =\int\frac{1}{\cos^2x}\\ =\int \sec^2x\ dx =\tan x+C

Thus after evaluation, the value of integral is tanx+ c

Question:19 Find the integrals of the functions \frac{1}{\sin x \cos ^3 x }

Answer:

Let
We can write 1 = \sin^2x +\cos^2x
Then, the equation can be written as
I =\frac{\sin^2x +\cos^2x}{\sin x\cos^3x}

I =\int (\tan x+\frac{1}{\tan x})\sec^2 x dx
put the value of tan x = t
So, that \sec^2xdx =dt

\\\Rightarrow I=\int (t+\frac{1}{t})dt\\ =\frac{t^2}{2}+\log\left | t \right |+C\\ =\log\left | \tan x \right |+\frac{1}{2}\tan^2x+C

Question:20 Find the integrals of the functions \frac{\cos 2x }{( \cos x + \sin x )^2}

Answer:

we know that cos2x= cos^2x-sin^2x
therefore,

\frac{\cos 2x }{( \cos x + \sin x )^2}
\frac{\cos 2x}{1+\sin 2x}\\ \Rightarrow \int \frac{\cos 2x}{1+\sin 2x}\\ let 1+sin2x =t \Rightarrow 2cos2x\ dx = dt
Now the given integral can be written as

\therefore \int \frac{\cos 2x}{(\cos x+\sin x)^2}=\frac{1}{2}\int \frac{1}{t}dt
\\\Rightarrow \frac{1}{2}\log\left | t \right |+C\\ \Rightarrow \frac{1}{2}\log\left | 1+\sin 2x \right |+C\\=log|sin^2x+cos^2x+2sinxcosx|+C\\=\frac{1}{2}log|(sinx+cosx)^2|+C=log|sinx+cosx|+C

Question:21 Find the integrals of the functions \sin ^ { -1} ( \cos x )

Answer:

using the trigonometric identities we can evaluate the following integral as follows

\dpi{100} \int \sin^{-1}(\cos x)dx = \int \sin^{-1}(sin(\frac{\pi}{2}-x))dx\\=\int(\frac{\pi}{2}-x)dx=\frac{\pi x}{2}-\frac{x^2}{2}+C

Question:22 Find the integrals of the functions \frac{1}{\cos ( x-a ) \cos ( x-b )}

Answer:

Using the trigonometric identities following integrals can be simplified as follows

\frac{1}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}[\frac{\sin(a-b)}{\cos(x-a)\cos(x-b)}]

=\frac{1}{\sin(a-b)}[\frac{\sin[(x-b)-(x-a)]}{\cos(x-a)\cos(x-b)}]

=\frac{1}{\sin(a-b)}[\frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)}]

=\frac{tan(x-b)-\tan (x-a)}{\sin(a-b)}

=\frac{1}{\sin(a-b)}\int tan(x-b)-\tan (x-a)dx
\\=\frac{1}{\sin(a-b)}[-\log\left | \cos(x-b) \right |+\log\left | \cos(x-a) \right |]\\ =\frac{1}{\sin(a-b)}(\log\left | \frac{\cos(x-a)}{\cos(x-b)} \right |)

Question:23 Choose the correct answer

\int \frac{\sin ^ 2 x - \cos ^ 2 x dx }{\sin ^ 2 x \cos ^ 2x } dx \: \:is \: \:equal \: \: to stgdrffd

Answer:

The correct option is (A)

On reducing the above integral becomes \sec^2x-csc^2x
\int\sec^2x-csc^2x\ dx = \tan x+ \cot x+C

Question:24 Choose the correct answer \int \frac{e ^x ( 1+x)}{\cos ^ 2 ( e ^xx )} dx \: \: equals

\\(A) -\cot (ex^x) + C \\\\ (B) \tan (xe^x) + C\\\\ (C) \tan (e^x) + C \\\\ (D) \cot (e^x) + C

Answer:

The correct option is (B)

Let e^xx = t .
So, (e^x.x+ 1.e^x)dx = dt
(1+ x ) e^x\ dx =dt

therefore,

\int \frac{e^x(1+x)}{\cos^2(e^x.x)}dx =\int\frac{dt}{\cos^2t}
\\=\int \sec^2t dt\\ =\tan t +C\\ =\tan(e^x.x)+C


NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.4

Question:1 Integrate the functions \frac{3x^ 2 }{x^6 + 1 }

Answer:

The given integral can be calculated as follows

Let x^3 = t
, therefore, 3x^2 dx =dt

\Rightarrow \int\frac{3x^2}{x^6+1}=\int \frac{dt}{t^2+1}

\\=\tan^{-1} t +C\\ =tan^{-1}(x^3)+C

Question:2 Integrate the functions \frac{1}{\sqrt { 1+ 4 x^2 }}

Answer:

\frac{1}{\sqrt { 1+ 4 x^2 }}
let suppose 2x = t
therefore 2dx = dt

\int \frac{1}{\sqrt{1+4x^2}} =\frac{1}{2}\int \frac{dt}{1+t^2}
\\=\frac{1}{2}[\log\left | t+\sqrt{1+t^2} \right |]+C\\ =\frac{1}{2}\log\left | 2x+\sqrt{4x^2+1} \right |+C .................using formula \int\frac{1}{\sqrt{x^2+a^2}}dt = \log\left | x+\sqrt{x^2+a^2} \right |

Question:3 Integrate the functions \frac{1}{\sqrt { ( 2- x)^2+ 1 }}

Answer:

\frac{1}{\sqrt { ( 2- x)^2+ 1 }}

let suppose 2-x =t
then, -dx =dt
\Rightarrow \int\frac{1}{\sqrt{(2-x)^2+1}}dx = -\int \frac{1}{\sqrt{t^2+1}}dt

using the identity

\int \frac{1}{\sqrt{x^2+1}}dt=log\left | x+\sqrt{x^2+1} \right |

\\= -\log\left | t+\sqrt{t^2+1} \right |+C\\ =-\log\left | 2-x+\sqrt{(2-x)^2+1} \right |+C\\ =\log \left | \frac{1}{(2-x)+\sqrt{x^2-4x+5}} \right |+C

Question:4 Integrate the functions \frac{1}{\sqrt {9 - 25 x^2 }}

Answer:

\frac{1}{\sqrt {9 - 25 x^2 }}
Let assume 5x =t,
then 5dx = dt

\Rightarrow \int \frac{1}{\sqrt{9-25x^2}}=\frac{1}{5}\int \frac{1}{\sqrt{9-t^2}}dt
\\=\frac{1}{5}\int \frac{1}{\sqrt{3^2-t^2}}dt\\ =\frac{1}{5}\sin^{-1}(\frac{t}{3})+C\\ =\frac{1}{5}\sin^{-1}(\frac{5x}{3})+C

The above result is obtained using the identity

\\\int \frac{1}{\sqrt{a^2-x^2}}dt\\ =\frac{1}{a}sin^{-1}\frac{x}{a}

Question:5 Integrate the functions \frac{3x }{1+ 2 x ^ 4 }

Answer:

\frac{3x }{1+ 2 x ^ 4 }


Let {\sqrt{2}}x^2 = t
\therefore 2\sqrt{2}xdx =dt

The integration can be done as follows

\Rightarrow \int \frac{3x}{1+2x^4}= \frac{3}{2\sqrt{2}}\int \frac{dt}{1+t^2}
\\= \frac{3}{2\sqrt{2}}[\tan^{-1}t]+C\\ =\frac{3}{2\sqrt{2}}[\tan^{-1}(\sqrt{2}x^2)]+C

Question:6 Integrate the functions \frac{x ^ 2 }{1- x ^ 6 }

Answer:

\frac{x ^ 2 }{1- x ^ 6 }

let x^3 =t
then 3x^2dx =dt

using the special identities we can simplify the integral as follows

\int \frac{x^2}{1-x^6}dx =\frac{1}{3}\int \frac{dt}{1-t^2}
=\frac{1}{3}[\frac{1}{2}\log\left | \frac{1+t}{1-t} \right |]+C\\ =\frac{1}{6}\log\left | \frac{1+x^3}{1-x^3} \right |+C

Question:7 Integrate the functions \frac{x-1 }{\sqrt { x^2 -1 }}

Answer:


We can write above eq as
\frac{x-1 }{\sqrt { x^2 -1 }} =\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx ............................................(i)

for \int \frac{x}{\sqrt{x^2-1}}dx let x^2-1 = t \Rightarrow 2xdx =dt

\therefore \int \frac{x}{\sqrt{x^2-1}}dx=\frac{1}{2}\int \frac{dt}{\sqrt{t}}
\\=\frac{1}{2}\int t^{1/2}dt\\ =\frac{1}{2}[2t^{1/2}]\\ =\sqrt{t}\\ =\sqrt{x^2-1}
Now, by using eq (i)
=\int \frac{x}{\sqrt{x^2-1}}dx-\int \frac{1}{\sqrt{x^2-1}}dx
\\=\sqrt{x^2-1}-\int \frac{1}{\sqrt{x^2}-1}dx\\ =\sqrt{x^2-1}-\log\left | x+\sqrt{x^2-1} \right |+C

Question:8 Integrate the functions \frac{x ^ 2 }{\sqrt { x^6 + a ^ 6 }}

Answer:

The integration can be down as follows

\frac{x ^ 2 }{\sqrt { x^6 + a ^ 6 }}
let x^3 = t \Rightarrow 3x^2dx =dt

\therefore \frac{x^2}{\sqrt{x^6+a^6}}=\frac{1}{3}\int \frac{dt}{\sqrt{t^2+(a^3)^2}}
\\=\frac{1}{3}\log\left | t+\sqrt{t^2+a^6} \right |+C\\ =\frac{1}{3}\log\left | x^3+\sqrt{x^6+a^6} \right |+C ........................using \int \frac{dx}{\sqrt{x^2+a^2}} = \log\left | x+\sqrt{x^2+a^2} \right |

Question:9 Integrate the functions \frac{\sec ^ 2 x }{\sqrt { \tan ^ 2 x+ 4 }}

Answer:

The integral can be evaluated as follows

\frac{\sec ^ 2 x }{\sqrt { \tan ^ 2 x + 4 }}
let \tan x =t \Rightarrow sec^2x dx =dt

\Rightarrow \int \frac{\sec^2x}{\sqrt{\tan^2x+4}}dx = \int \frac{dt}{\sqrt{t^2+2^2}}
\\= \log\left | t+\sqrt{t^2+4} \right |+C\\ =\log \left | \tan x+\sqrt{ tan^2x+4} \right |+C

Question:10 Integrate the functions \frac{1 }{ \sqrt { x ^ 2 + 2 x + 2 }}

Answer:

\frac{1 }{ \sqrt { x ^ 2 + 2 x + 2 }}
the above equation can be also written as,
=\int\frac{1}{\sqrt{(1+x)^2+1^2}}dx
let 1+x = t
then dx = dt
therefore,

\\=\int\frac{1}{\sqrt{t^2+1^2}}dx\\ =\log\left | t+\sqrt{t^2+1} \right |+C\\ =\log\left | (1+x)+\sqrt{(1+x)^2+1} \right |+C\\ =\log\left | (1+x)+\sqrt{(x^2+2x+2} \right |+C

Question:11 Integrate the functions \frac{1}{9 x ^2 + 6x + 5 }

Answer:

\frac{1}{9 x ^2 + 6x + 5 }
this denominator can be written as
9x^2+6x+5=9[x^2+\frac{2}{3}x+\frac{5}{9}]\\=9[(x+\frac{1}{3})^2+(\frac{2}{3})^2] Now,
\frac{1}{9}\int \frac{1}{(x+\frac{1}{3})^2+(\frac{2}{3})^2}dx =\frac{1}{9} [\frac{3}{2}\tan^{-1}(\frac{(x+1/3)}{2/3})] +C\\=\frac{1}{6} \tan^{-1}(\frac{3x+1}{2})] +C
......................................by using the form (\int \frac{1}{x^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{x}{a}))

Question:12 Integrate the functions \frac{1}{\sqrt{ 7-6x - x ^ 2 }}

Answer:

the denominator can be also written as,
7-6x-x^2=16-(x^2+6x+9)
=4^2-(x+3)^2

therefore

\int \frac{1}{\sqrt{7-6x-x^2}}dx=\int \frac{1}{\sqrt{4^2-(x+3)^2}}dx
Let x+3 = t
then dx =dt

\Rightarrow \int \frac{1}{\sqrt{4^2-(x+3)^2}}dx=\int \frac{1}{\sqrt{4^2-t^2}}dt ......................................using formula \int \frac{1}{\sqrt{a^2-x^2}}=\sin^{-1}(\frac{x}{a})
\\= sin^{-1}(\frac{t}{4})+C\\ =\sin^{-1}(\frac{x+3}{4})+C

Question:13 Integrate the functions \frac{1}{\sqrt { ( x-1)( x-2 )}}

Answer:

(x-1)(x-2) can be also written as
= x^2-3x+2
= (x-\frac{3}{2})^2-(\frac{1}{2})^2

therefore

\int \frac{1}{\sqrt{(x-1)(x-2)}}dx= \int \frac{1}{\sqrt{(x-\frac{3}{2})^2-(\frac{1}{2})^2}}dx
let suppose
x-3/2 = t \Rightarrow dx =dt
Now,

\Rightarrow \int \frac{1}{\sqrt{(x-\frac{3}{2})^2-(\frac{1}{2})^2}}dx = \int \frac{1}{\sqrt{t^2-(\frac{1}{2})^2}}dt .............by using formula \int \frac{1}{\sqrt{x^2-a^2}}=\log\left | x+\sqrt{x^2+a^2} \right |
\\= \log \left | t+\sqrt{t^2-(1/2)^2} \right |+C\\ = \log \left | (x-\frac{3}{2})+\sqrt{x^2-3x+2} \right |+C

Question:14 Integrate the functions \frac{1}{\sqrt { 8 + 3 x - x ^ 2 }}

Answer:

We can write denominator as
\\=8-(x^2-3x+\frac{9}{4}-\frac{9}{4})\\ =\frac{41}{4}-(x-\frac{3}{2})^2

therefore
\Rightarrow \int \frac{1}{\sqrt{8+3x-x^2}}dx= \int \frac{1}{\sqrt{\frac{41}{4}-(x-\frac{3}{2})^2}}
let x-3/2 = t \Rightarrow dx =dt

\therefore
\\=\int \frac{1}{\sqrt{(\frac{\sqrt{41}}{2})^2-t^2}}dt\\ =\sin^{-1}(\frac{t}{\frac{\sqrt{41}}{2}})+C\\ =\sin^{-1}(\frac{2x-3}{\sqrt{41}})+C

Question:15 Integrate the functions \frac{1}{\sqrt {(x-a)( x-b )}}

Answer:

(x-a)(x-b) can be written as x^2-(a+b)x+ab
\\x^2-(a+b)x+ab+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4}\\ (x-\frac{(a+b)}{2}^2)^2-\frac{(a-b)^2}{4}

\Rightarrow \int\frac{1}{\sqrt{(x-a)(x-b)}}dx=\int \frac{1}{\sqrt{(x-\frac{(a+b)}{2}^2)^2-\frac{(a-b)^2}{4}}}dx
let
x-\frac{(a+b)}{2}=t \Rightarrow dx =dt
So,
\\=\int \frac{1}{\sqrt{t^2-(\frac{a-b}{2})^2}}dt\\ =\log \left | t+\sqrt{t^2-(\frac{a-b}{2})^2} \right |+C\\ =\log \left | x-(\frac{a+b}{2})+\sqrt{(x-a)(x-b)} \right |+C

Question:16 Integrate the functions \frac{4x+1 }{\sqrt {2x ^ 2 + x -3 }}

Answer:

let
\\4x+1 = A\frac{d}{dx}(2x^2+x-3)+B\\ 4x+1=A(4x+1)+B\\ 4x+1=4Ax+A+B

By equating the coefficient of x and constant term on each side, we get
A = 1 and B=0

Let (2x^2+x-3) = t\Rightarrow (4x+1)dx =dt

\therefore \int \frac{4x+1}{\sqrt{2x^2+x-3}}dx= \int\frac{1}{\sqrt{t}}dt
\\= 2\sqrt{t}+C\\ =2\sqrt{2x^2+x-3}+C

Question:17 Integrate the functions \frac{x+ 2 }{\sqrt { x ^2 -1 }}

Answer:

let x+2 =A\frac{d}{dx}(x^2-1)+B=A(2x)+B
By comparing the coefficients and constant term on both sides, we get;

A=1/2 and B=2
then x+2 = \frac{1}{2}(2x)+2

\int \frac{x+2}{\sqrt{x^2-1}}dx =\int\frac{1/2(2x)+2}{x^2-1}dx
\\=\frac{1}{2}\int\frac{(2x)}{\sqrt{x^2-1}}dx+\int \frac{2}{\sqrt{x^2-1}}dx\\ =\frac{1}{2}[2\sqrt{x^2-1}]+2\log\left | x+\sqrt{x^2-1} \right |+C\\ =\sqrt{x^2-1}+2\log\left | x+\sqrt{x^2-1} \right |+C

Question:18 Integrate the functions \frac{5x -2 }{1+ 2x +3x^2 }

Answer:

let
\\5x+2 = A\frac{d}{dx}(1+2x+3x^2)+B\\ 5x+2= A(2+6x)+B = 2A+B+6Ax
By comparing the coefficients and constants we get the value of A and B

A = 5/6 and B = -11/3

NOW,
I = \frac{5}{6}\int \frac{6x+2}{3x^2+2x+1}dx-\frac{11}{3}\int \frac{dx}{3x^2+2x+1}
I = I_{1}-\frac{11}{3}I_{2} ...........................(i)

put 3x^2+2x+1 =t \Rightarrow (6x+2)dx =dt
Thus
I_{1}=\frac{5}{6}\int \frac{dt}{t} =\frac{5}{6}\log t =\frac{5}{6}\log (3x^2+2x+1)+c1
I_{2}= \int \frac{dx}{3x^2+2x+1} = \frac{1}{3}\int\frac{dx}{(x+1/3)^2+(\sqrt{2}/3)^2}
\\=\frac{1}{\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt{2}})+c2

\therefore I = I_1+I_2
I = \frac{5}{6}\log(3x^2+2x+1)-\frac{11}{3}\frac{1}{\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt2})+C

Question:19 Integrate the functions \frac{6x + 7 }{\sqrt {( x-5 )( x-4)}}

Answer:

let
6x+7 = A\frac{d}{dx}(x^2-9x+20)+B =A(2x-9)+B
By comparing the coefficients and constants on both sides, we get
A =3 and B =34

I =\int \frac{6x+7}{\sqrt{x^2-9x+20}}dx = \int \frac{3(2x+9)}{\sqrt{x^2-9x+20}}dx+34\int\frac{dx}{\sqrt{x^2-9x+20}} I = I_1+I_2 ....................................(i)

Considering I_1

I_1 =\int \frac{2x-9}{\sqrt{x^2-9x+20}}dx let x^2-9x+20 = t \Rightarrow (2x-9)dx =dt

I_1=\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}=2\sqrt{x^2-9x+20}

Now consider I_2

I_2=\int \frac{dx}{\sqrt{x^2-9x+20}}
here the denominator can be also written as
Dr = (x-\frac{9}{2})^2-(\frac{1}{2})^2

\therefore I_2 = \int \frac{dx}{\sqrt{(x-\frac{9}{2})^2-(\frac{1}{2})^2}}
\\= \log\left | (x-\frac{9}{2})^2+\sqrt{x^2-9x+20} \right |

Now put the values of I_1 and I_2 in eq (i)

\\I = 3I_1+34I_2\\ I=6\sqrt{x^2-9x+20}+34\log\left | (x-\frac{9}{2})+\sqrt{x^2-9x+20} \right |+C

Question:20 Integrate the functions \frac{x +2 }{\sqrt { 4x - x ^ 2 }}

Answer:

let
x+2 = A\frac{d}{dx}(4x-x^2)+B = A(4-2x)+B
By equating the coefficients and constant term on both sides we get

A = -1/2 and B = 4

(x+2) = -1/2(4-2x)+4

\\\therefore \int \frac{x+2}{\sqrt{4x-x^2}}dx = -\frac{1}{2}\int \frac{4-2x}{\sqrt{4x-x^2}}+4\int \frac{dx}{\sqrt{4x-x^2}}\\ \ I =\frac{-1}{2}I_1+4I_2 ....................(i)

Considering I_1
\int \frac{4-2x}{\sqrt{4x-x^2}}dx
let 4x-x^2 =t \Rightarrow (4-2x)dx =dt
I_1=\int \frac{dt}{\sqrt{t}} = 2\sqrt{t}=2\sqrt{4x-x^2}
now, I_2

I_2 =\int \frac{dx}{\sqrt{4x-x^2}} = \int \frac{dx}{\sqrt{2^2-(x-2)^2}}
=\sin^{-1}(\frac{x-2}{2})

put the value of I_1 and I_2

I =-\sqrt{4x-x^2}+4\sin^{-1}(\frac{x-2}{2})+C

Question:21 Integrate the functions \frac{x +2 }{\sqrt { x^ 2 + 2x +3 }}

Answer:

\frac{x +2 }{\sqrt { x^ 2 + 2x +3 }}
\int \frac{x+2}{\sqrt{x^2+2x+3}}dx = \frac{1}{2}\int \frac{2(x+2)}{\sqrt{x^2+2x+3}}dx
\\= \frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\frac{1}{2}\int \frac{2}{\sqrt{x^2+2x+3}}dx\\ =\frac{1}{2}\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx+\int \frac{1}{\sqrt{x^2+2x+3}}dx\\ I=\frac{1}{2}I_1+I_2 ...........(i)

take I_1

\int \frac{2x+2}{\sqrt{x^2+2x+3}}dx
let x^2+2x+3 = t \Rightarrow (2x+2)dx =dt

I_1=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^2+2x+3}
considering I_2

= \int \frac{dx}{\sqrt{x^2+2x+3}}= \int \frac{dx}{\sqrt{(x+1)^2+(\sqrt{2})^2}}
= \log \left | (x+1)+\sqrt{x^2+2x+3} \right |
putting the values in equation (i)

I=\sqrt{x^2+2x+3} +\log \left | (x+1)+\sqrt{x^2+2x+3} \right |+C

Question:22 Integrate the functions \frac{x + 3 }{x ^ 2 - 2x - 5 }

Answer:

Let (x+3) =A\frac{d}{dx}(x^2-2x+5)+B= A(2x-2)+B

By comparing the coefficients and constant term, we get;

A = 1/2 and B =4

\\\int \frac{x+3}{x^2-2x+5}dx = \frac{1}{2}\int \frac{2x-2}{x^2-2x+5}dx +4\int \frac{1}{x^2-2x+5}dx\\ I=I_1+I_2 ..............(i)

\\\Rightarrow I_1\\ =\int \frac{2x-2}{x^2-2x-5}dx
put x^2-2x-5 =t \Rightarrow (2x-2)dx =dt

=\int \frac{dt}{t} = \log t = \log (x^2-2x-5)

\\\Rightarrow I_2\\ = \int \frac{1}{x^2-2x-5}dx\\ =\int \frac{1}{(x-1)^2+(\sqrt{6})^2}dx\\ =\frac{1}{2\sqrt{6}}\log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})

I=I_1+I_2

=\frac{1}{2}\log\left | x^2-2x-5 \right |+\frac{2}{\sqrt{6}}\log(\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}})+C

Question:23 Integrate the functions \frac{5x + 3 }{\sqrt { x^2 + 4x +10 }}

Answer:

let
5x+3 = A\frac{d}{dx}(x^2+4x+10)+B = A(2x+4)+B
On comparing, we get

A =5/2 and B = -7

\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx = \frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}dx-7\int \frac{dx}{\sqrt{x^2+4x+10}}dx I = 5/2I_1-7I_2 ...........................................(i)

\\\Rightarrow I_1\\ \int \frac{2x+4}{\sqrt{x^2+4x+10}}dx
put
x^2+4x+10= t \Rightarrow (2x+4)dx = dt

=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}=2\sqrt{x^2+4x+10}

\\\Rightarrow I_2\\ =\int \frac{1}{\sqrt{x^2+4x+10}}dx \\ =\int \frac{1}{\sqrt{(x+2)^2+(\sqrt{6})^2}}dx\\ =\log \left | (x+2)+\sqrt{x^2+4x+10} \right |

I = 5\sqrt{x^2+4x+10}-7\log\left | (x+2)+\sqrt{x^2+4x+10} \right |+C

Question:24 Choose the correct answer

\int \frac{dx }{x^2 + 2x +2 }\: \: equals

(A) x \tan^{-1} (x + 1) + C\\\\ (B) \tan^{-1} (x + 1) + C\\\\ (C) (x + 1) \tan^{-1}x + C \\\\ (D) \tan^{-1}x + C

Answer:

The correct option is (B)

\int \frac{dx }{x^2 + 2x +2 }\: \: equals
the denominator can be written as (x+1)^2+1
now, \int \frac{dx}{(x+1)^2+1} = tan^{-1}(x+1)+C

Question:25 Choose the correct answer \int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals

A) \frac{1}{9} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C \\\\B ) \frac{1}{2} \sin ^{-1}\left ( \frac{8x-9}{9} \right )+ C \\\\ C) \frac{1}{3} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C \\\\ D ) \frac{1}{2} \sin ^{-1}\left ( \frac{9x-8}{8} \right )+ C

Answer:

The following integration can be done as

\int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals
\int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x)}}= \int \frac{1}{\sqrt{-4(x^2-\frac{9}{4}x+81/64-81/64)}}dx
\\= \int \frac{1}{\sqrt{-4[(x-9/8)^2-(9/8)^2]}}dx\\ =\frac{1}{2}\int \frac{1}{\sqrt{-(x-9/8)^2+(9/8)^2}}dx\\ =\frac{1}{2}[\sin^{-1}(\frac{x-9/8}{9/8})]+C\\ =\frac{1}{2}\sin^{-1}(\frac{8x-9}{9})+C

The correct option is (B)


NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.5

Question:1 Integrate the rational functions \frac{x }{( x +1)( x+2)}

Answer:

Given function \frac{x }{( x +1)( x+2)}

Partial function of this function:

\frac{x }{( x +1)( x+2)} = \frac{A}{(x+1)}+\frac{B}{(x+2)}

\implies x = A(x+2)+B(x+1)

Now, equating the coefficients of x and constant term, we obtain

A+B =1

2A+B =0

On solving, we get

A=-1\ and\ B =2

\therefore \frac{x}{(x+1)(x+2)} = \frac{-1}{(x+1)}+\frac{2}{(x+2)}

\implies \int \frac{x}{(x+1)(x+2)} dx =\int \frac{-1}{(x+1)}+\frac{2}{(x+2)} dx

=-\log|x+1| +2\log|x+2| +C

=\log(x+2)^2-\log|x+1|+C

=\log\frac{(x+2)^2}{(x+1)}+C

Question:2 Integrate the rational functions \frac{1}{x^2 -9 }

Answer:

Given function \frac{1}{x^2 -9 }

The partial function of this function:

\frac{1}{(x+3)(x-3)}= \frac{A}{(x+3)}+\frac{B}{(x-3)}

1 = A(x-3)+B(x+3)

Now, equating the coefficients of x and constant term, we obtain

A+B =1

-3A+3B =1

On solving, we get

A=-\frac{1}{6}\ and\ B =\frac{1}{6}

\frac{1}{(x+3)(x-3)}= \frac{-1}{6(x+3)} +\frac{1}{6(x-3)}

\int \frac{1}{(x^2-9)}dx = \int \left ( \frac{-1}{6(x+3)}+\frac{1}{6(x-3)} \right )dx

=-\frac{1}{6}\log|x+3| +\frac{1}{6}\log|x-3| +C

= \frac{1}{6}\log\left | \frac{x-3}{x+3} \right |+C

Question:3 Integrate the rational functions \frac{3x -1}{( x-1)(x-2)(x-3)}

Answer:

Given function \frac{3x -1}{( x-1)(x-2)(x-3)}

Partial function of this function:

\frac{3x -1}{( x-1)(x-2)(x-3)}= \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}

3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) .(1)

Now, substituting x=1,2,\ and\ 3 respectively in equation (1), we get

A =1,\ B=-5,\ and\ C=4

\therefore \frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{1}{(x-1)} -\frac{5}{(x-2)}+\frac{4}{(x-3)}

That implies \int \frac{3x-1}{(x-1)(x-2)(x-3)} dx = \int \left \{ \frac{1}{(x-1)}-\frac{5}{(x-2)}+\frac{4}{(x-3)} \right \}dx

= \log|x-1|-5\log|x-2|+4\log|x-3|+C

Question:4 Integrate the rational functions \frac{x }{( x-1)(x-2)(x-3)}

Answer:

Given function \frac{x }{( x-1)(x-2)(x-3)}

Partial function of this function:

\frac{x }{( x-1)(x-2)(x-3)}= \frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}

x = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) .....(1)

Now, substituting x=1,2,\ and\ 3 respectively in equation (1), we get

A =\frac{1}{2},\ B=-2,\ and\ C=\frac{3}{2}

\therefore \frac{x}{(x-1)(x-2)(x-3)} = \frac{1}{2(x-1)} -\frac{2}{(x-2)}+\frac{3}{2(x-3)}

That implies \int \frac{x}{(x-1)(x-2)(x-3)} dx = \int \left \{ \frac{1}{2(x-1)}-\frac{2}{(x-2)}+\frac{3}{2(x-3)} \right \}dx

= \frac{1}{2}\log|x-1|-2\log|x-2|+\frac{3}{2}\log|x-3|+C

Question:5 Integrate the rational functions \frac{2x}{x^2 + 3x +2 }

Answer:

Given function \frac{2x}{x^2 + 3x +2 }

Partial function of this function:

\frac{2x}{x^2 + 3x +2 }= \frac{A}{(x+1)}+\frac{B}{(x+2)}

2x = A(x+2)+B(x+1) ...........(1)

Now, substituting x=-1\ and\ -2 respectively in equation (1), we get

A ={-2},\ B=4

\frac{2x}{x^2 + 3x +2 }= \frac{-2}{(x+1)}+\frac{4}{(x+2)}

That implies \int \frac{2x}{x^2 + 3x +2 }dx= \int \left \{ \frac{-2}{(x+1)}+\frac{4}{(x+2)} \right \}dx

=4\log|x+2| -2\log|x+1| +C

Question:6 Integrate the rational functions \frac{1- x^2 }{ x ( 1- 2x )}

Answer:

Given function \frac{1- x^2 }{ x ( 1- 2x )}

Integral is not a proper fraction so,

Therefore, on dividing (1-x^2) by x(1-2x) , we get

\frac{1- x^2 }{ x ( 1- 2x )} = \frac{1}{2} +\frac{1}{2}\left ( \frac{2-x}{x(1-2x)} \right )

Partial function of this function:

\frac{2-x}{x(1-2x)} =\frac{A}{x}+\frac{B}{(1-2x)}

(2-x) =A(1-2x)+Bx ...........(1)

Now, substituting x=0\ and\ \frac{1}{2} respectively in equation (1), we get

A =2,\ B=3

\therefore \frac{2-x}{x(1-2x)} = \frac{2}{x}+\frac{3}{1-2x}

No, substituting in equation (1) we get

\frac{1-x^2}{(1-2x)} = \frac{1}{2}+\frac{1}{2}\left \{ \frac{2}{3}+\frac{3}{(1-2x)} \right \}

\implies \int \frac{1-x^2}{x(1-2x)}dx =\int \left \{ \frac{1}{2}+\frac{1}{2}\left ( \frac{2}{x}+\frac{3}{1-2x} \right ) \right \}dx

=\frac{x}{2}+\log|x| +\frac{3}{2(-2)}\log|1-2x| +C

=\frac{x}{2}+\log|x| -\frac{3}{4}\log|1-2x| +C

Question:7 Integrate the rational functions \frac{x }{( x^2+1 )( x-1)}

Answer:

Given function \frac{x }{( x^2+1 )( x-1)}

Partial function of this function:

\frac{x }{( x^2+1 )( x-1)} = \frac{Ax+b}{(x^2+1)} +\frac{C}{(x-1)}

x = (Ax+B)(x-1)+C(x^2+1)

x=Ax^-Ax+Bc-B+Cx^2+C

Now, equating the coefficients of x^2, x and the constant term, we get

A+C = 0

-A+B =1 and -B+C = 0

On solving these equations, we get

A = -\frac{1}{2}, B= \frac{1}{2},\ and\ C=\frac{1}{2}

From equation (1), we get

\therefore \frac{x}{(x^2+1)(x-1)} = \frac{\left ( -\frac{1}{2}x+\frac{1}{2} \right )}{x^2+1}+\frac{\frac{1}{2}}{(x-1)}

\implies \int \frac{x}{(x^2+1)(x-1)}

=-\frac{1}{2}\int \frac{x}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2} \int \frac{1}{x-1}dx

=- \frac{1}{4} \int \frac{2x}{x^2+1} dx +\frac{1}{2} \tan^{-1}x + \frac{1}{2} \log|x-1| +C

Now, consider \int \frac{2x}{x^2+1} dx ,

and we will assume (x^2+1) = t \Rightarrow 2xdx =dt

So, \int \frac{2x}{x^2+1}dx = \int \frac{dt}{t} =\log|t| = \log|x^2+1|

\therefore \int \frac{x}{(x^2+1)(x-1)} =-\frac{1}{4}\log|x^2+1| +\frac{1}{2}\tan^{-1}x +\frac{1}{2}\log|x-1| +C or

\frac{1}{2}\log|x-1| - \frac{1}{4}\log|x^2+1|+\frac{1}{2}\tan^{-1}x +C

Question:8 Integrate the rational functions \frac{x }{( x+1)^2 ( x+2)}

Answer:

Given function \frac{x }{( x+1)^2 ( x+2)}

Partial function of this function:

\frac{x }{( x+1)^2 ( x+2)} = \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+2)}

x = A(x-1)(x+2)+B(x+2)+C(x-1)^2

Now, putting x=1 in the above equation, we get

B =\frac{1}{3}

By equating the coefficients of x^2 and constant term, we get

A+C=0

-2A+2B+C = 0

then after solving, we get

A= \frac{2}{9}\ and\ C=\frac{-2}{9}

Therefore,

\frac{x}{(x-1)^2(x+2)} = \frac{2}{9(x-1)}+\frac{1}{3(x-1)^2}-\frac{2}{9(x+2)}

\int \frac{x}{(x-1)^2(x+2)}dx= \frac{2}{9}\int \frac{1}{(x-1)}dx+\frac{1}{3}\int \frac{1}{(x-1)^2}dx-\frac{2}{9}\int \frac{1}{(x+2)}dx

= \frac{2}{9}\log|x-1|+\frac{1}{3}\left ( \frac{-1}{x-1} \right )-\frac{2}{9}\log|x+2|+C

\frac{2}{9}\log\left | \frac{x-1}{x+2} \right | -\frac{1}{3(x-1)}+C

Question:9 Integrate the rational functions \frac{3x+ 5 }{x^3 - x^2 - x +1 }

Answer:

Given function \frac{3x+ 5 }{x^3 - x^2 - x +1 }

can be rewritten as \frac{3x+ 5 }{x^3 - x^2 - x +1 } = \frac{3x+5}{(x-1)^2(x+1)}

Partial function of this function:

\frac{3x+5}{(x-1)^2(x+1)}= \frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)}

3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)^2

3x+5 = A(x^2-1)+B(x+1)+C(x^2+1-2x) ................(1)

Now, putting x=1 in the above equation, we get

B =4

By equating the coefficients of x^2 and x , we get

A+C=0

B-2C =3

then after solving, we get

A= -\frac{1}{2}\ and\ C=\frac{1}{2}

Therefore,

\frac{3x+5}{(x-1)^2(x+1)}= \frac{-1}{2(x-1)}+\frac{4}{(x-1)^2}+\frac{1}{2(x+1)}

\int \frac{3x+5}{(x-1)^2(x+1)}dx= \frac{-1}{2}\int \frac{1}{(x-1)}dx+4\int \frac{1}{(x-1)^2} dx+\frac{1}{2}\int \frac{1}{(x+1)}dx

= -\frac{1}{2}\log|x-1| +4\left ( \frac{-1}{x-1} \right ) +\frac{1}{2}\log|x+1| +C

=\frac{1}{2}\log|\frac{x+1}{x-1}| - \frac{4}{(x-1)} + +C

Question:10 Integrate the rational functions \frac{2x -3 }{(x^2 -1 )( 2x+3)}

Answer:

Given function \frac{2x -3 }{(x^2 -1 )( 2x+3)}

can be rewritten as \frac{2x -3 }{(x^2 -1 )( 2x+3)} = \frac{2x-3}{(x+1)(x-1)(2x+3)}

The partial function of this function:

\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{A}{(x+1)} +\frac{B}{(x-1)}+\frac{C}{(2x+3)}

\Rightarrow (2x-3) =A(x-1)(2x+3)+B(x+1)(2x+3)+C(x+1)(x-1) \Rightarrow (2x-3) =A(2x^2+x-3)+B(2x^2+5x+3)+C(x^2-1) \Rightarrow (2x-3) =(2A+2B+C)x^2+(A+5B)x+(-3A+3B-C)

Equating the coefficients of x^2\ and\ x , we get

B=-\frac{1}{10},\ A =\frac{5}{2},\ and\ C= -\frac{24}{5}

Therefore,

\frac{2x -3 }{(x^2 -1 )( 2x-3)} = \frac{5}{2(x+1)} -\frac{1}{10(x-1)}-\frac{24}{5(2x+3)}

\implies \int \frac{2x-3}{(x^2-1)(2x+3)}dx = \frac{5}{2}\int \frac{1}{(x+1)}dx -\frac{1}{10}\int \frac{1}{x-1}dx -\frac{24}{5}\int \frac{1}{(2x+3)}dx = \frac{5}{2}\log|x+1| -\frac{1}{10}\log|x-1| -\frac{24}{10}\log|2x+3|

= \frac{5}{2}\log|x+1| -\frac{1}{10}\log|x-1| -\frac{12}{5}\log|2x+3|+C

= -\frac{1}{2}\log|x-1| +4\left ( \frac{-1}{x-1} \right ) +\frac{1}{2}\log|x+1| +C

=\frac{1}{2}\log|\frac{x+1}{x-1}| - \frac{4}{(x-1)} + +C

= \frac{2}{9}\log|x-1|+\frac{1}{3}\left ( \frac{-1}{x-1} \right )-\frac{2}{9}\log|x+2|+C

\frac{2}{9}\log\left | \frac{x-1}{x+2} \right | -\frac{1}{3(x-1)}+C

Question:11 Integrate the rational functions \frac{5x}{(x+1)(x^2-4)}

Answer:

Given function \frac{5x}{(x+1)(x^2-4)}

can be rewritten as \frac{5x}{(x+1)(x^2-4)} = \frac{5x}{(x+1)(x+2)(x-2)}

The partial function of this function:

\frac{5x }{(x+1)( x+2)(x-2)} = \frac{A}{(x+1)} +\frac{B}{(x+2)}+\frac{C}{(x-2)}

\Rightarrow (5x) =A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2)

Now, substituting the value of x =-1,-2,\ and\ 2 respectively in the equation above, we get

A=\frac{5}{3},\ B =\frac{-5}{2},\ and\ C= \frac{5}{6}

Therefore,

\frac{5x }{(x+1)( x+2)(x-2)} = \frac{5}{3(x+1)} -\frac{5}{2(x+2)}+\frac{5}{6(x-2)}

\implies \int \frac{5x}{(x+1)(x^2-4)}dx = \frac{5}{3}\int \frac{1}{(x+1)}dx -\frac{5}{2}\int \frac{1}{x+2}dx+\frac{5}{6}\int \frac{1}{(x-2)}dx = \frac{5}{3}\log|x+1| -\frac{5}{2}\log|x+2| +\frac{5}{6}\log|x-2|+C

Question:12 Integrate the rational functions \frac{x^3 + x +1}{ x^2-1}

Answer:

Given function \frac{x^3 + x +1}{ x^2-1}

As the given integral is not a proper fraction.

So, we divide (x^3+x+1) by x^2-1 , we get

\frac{x^3 + x +1}{ x^2-1} = x+\frac{2x+1}{x^2-1}

can be rewritten as \frac{2x+1}{x^2-1} =\frac{A}{(x+1)} +\frac{B}{(x-1)}

2x+1 ={A}{(x-1)} +{B}{(x+1)} ....................(1)

Now, substituting x =1\ and\ x=-1 in equation (1), we get

A =\frac{1}{2}\ and\ B=\frac{3}{2}

Therefore,

\frac{x^3+x+1 }{(x^2-1)} =x+\frac{1}{2(x+1)}+\frac{3}{2(x-1)}

\implies \int \frac{x^3+x+1 }{(x^2-1)}dx =\int xdx +\frac{1}{2}\int \frac{1}{(x+1)} dx+\frac{3}{2} \int \frac{1}{(x-1)}dx

= \frac{x^2}{2}+\frac{1}{2}\log|x+1| +\frac{3}{2}\log|x-1| +C

Question:13 Integrate the rational functions \frac{2}{(1-x)(1+ x^2)}

Answer:

Given function \frac{2}{(1-x)(1+ x^2)}

can be rewritten as \frac{2}{(1-x)(1+ x^2)} = \frac{A}{(1-x)}+\frac{Bx+C}{1+x^2}

2 =A(1+x^2)+(Bx+C)(1-x) ....................(1)

2 =A +Ax^2 +Bx-Bx^2+C-Cx

Now, equating the coefficient of x^2, x, and constant term, we get

A-B= 0 , B-C = 0 , and A+C =2

Solving these equations, we get

A=1, B=1,\ and\ C=1

Therefore,

\therefore \frac{2}{(1-x)(1+ x^2)} = \frac{1}{(1-x)}+\frac{x+1}{1+x^2}

\implies \int \frac{2}{(1-x)(1+ x^2)}dx =\int \frac{1}{(1-x)} dx+ \int \frac{x}{1+x^2}dx +\int \frac{1}{1+x^2}dx = -\int \frac{1}{x-1}dx +\frac{1}{2}\int \frac{2x}{1+x^2}dx +\int\frac{1}{1+x^2}dx

=-\log|x-1| +\frac{1}{2}\log|1+x^2| +\tan^{-1}x+C

Question:14 Integrate the rational functions \frac{3x-1}{(x+2)^2}

Answer:

Given function \frac{3x-1}{(x+2)^2}

can be rewritten as \frac{3x-1}{(x+2)^2} = \frac{A}{(x+2)}+\frac{B}{(x+2)^2}

3x-1 = A(x+2)+B

Now, equating the coefficient of x and constant term, we get

A=3 and 2A+B = -1 ,

Solving these equations, we get

B=-7

Therefore,

\frac{3x-1}{(x+2)^2} = \frac{3}{(x+2)}-\frac{7}{(x+2)^2}

\implies \int\frac{3x-1}{(x+2)^2}dx = 3 \int \frac{1}{(x+2)}dx-7\int \frac{x}{(x+2)^2}dx

\implies 3\log|x+2| -7\left ( \frac{-1}{(x+2)}\right )+C

\implies 3\log|x+2| + \frac{7}{(x+2)} +C

Question:15 Integrate the rational functions \frac{1}{x^4 -1 }

Answer:

Given function \frac{1}{x^4 -1 }

can be rewritten as \frac{1}{x^4 -1 } = \frac{1}{(x^2-1)(x^2+1)} =\frac{1}{(x+1)(x-1)(1+x^2)}

The partial fraction of above equation,

\frac{1}{(x+1)(x-1)(1+x^2)} = \frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{Cx+D}{(x^2+1)}

1 = A(x-1)(x^2+1) +B(x+1)(x^2+1)+(Cx+D)(x^2-1)

1 = A(x^3+x-x^2-1)+B(x^3+x+x^2+1)+Cx^3+Dx^2-Cx-D 1 = (A+B+C)x^3 +(-A+B+D)x^2+(A+B-C)x+(-A+B-D)

Now, equating the coefficient of x^3,x^2,x and constant term, we get

A+B+C = 0 and -A+B+D = 0

A+B-C = 0 and -A+B-D = 1

Solving these equations, we get

A= -\frac{1}{4}, B=\frac{1}{4},C=0,\ and\ D = -\frac{1}{2}

Therefore,

\frac{1}{x^4-1} = \frac{-1}{4(x+1)}+\frac{1}{4(x-1)}-\frac{1}{2(x^2+1)}

\implies \int \frac{1}{x^4-1}dx = -\frac{1}{4}\log|x-1| +\frac{1}{4}\log|x-1| -\frac{1}{2}\tan^{-1}x +C

= \frac{1}{4}\log|\frac{x-1}{x+1}| -\frac{1}{2}\tan^{-1}x +C

Question:16 Integrate the rational functions \frac{1}{x ( x^n+1)}

[Hint: multiply numerator and denominator by x ^{n-1} and put x ^n = t ]

Answer:

Given function \frac{1}{x ( x^n+1)}

Applying Hint multiplying numerator and denominator by x^{n-1} and putting x^n =t

\frac{1}{x ( x^n+1)} = \frac{x^{n-1}}{x^{n-1}x(x^n+1)} = \frac{x^{n-1}}{x^n(x^n+1)}

Putting x^n =t

\therefore x^{n-1}dx =dt

can be rewritten as \int \frac{1}{x ( x^n+1)}dx =\int \frac{x^{n-1}}{x^n(x^n+1)}dx = \frac{1}{n} \int \frac{1}{t(t+1)}dt

Partial fraction of above equation,

\frac{1}{t(t+1)} =\frac{A}{t}+\frac{B}{(t+1)}

1 = A(1+t)+Bt ................(1)

Now, substituting t = 0,-1 in equation (1), we get

A=1\ and\ B=-1

\therefore \frac{1}{t(t+1)} = \frac{1}{t}- \frac{1}{(1+t)}

\implies \int \frac{1}{x(x^n+1)}dx = \frac{1}{n} \int \left \{ \frac{1}{t}-\frac{1}{(t+1)} \right \}dx

= \frac{1}{n} \left [ \log|t| -\log|t+1| \right ] +C

= -\frac{1}{n} \left [ \log|x^n| -\log|x^n+1| \right ] +C

= \frac{1}{n} \log|\frac{x^n}{x^n+1}| +C

Question:17 Integrate the rational functions \frac{\cos x }{(1- \sin x )( 2- \sin x )}

[Hint : Put \sin x = t ]

Answer:

Given function \frac{\cos x }{(1- \sin x )( 2- \sin x )}

Applying the given hint: putting \sin x =t

We get, \cos x dx =dt

\therefore \int \frac{\cos x }{(1- \sin x )( 2- \sin x )}dx = \int \frac{dt}{(1-t)(2-t)}

Partial fraction of above equation,

\frac{1}{(1-t)(2-t)} =\frac{A}{(1-t)}+\frac{B}{(2-t)}

1 = A(2-t)+B(1-t) ................(1)

Now, substituting t = 2\ and\ 1 in equation (1), we get

A=1\ and\ B=-1

\therefore \frac{1}{(1-t)(2-t)} = \frac{1}{(1-t)} - \frac{1}{(2-t)}

\implies \int \frac{\cos x }{(1-\sin x)(2-\sin x )}dx = \int \left \{ \frac{1}{1-t}-\frac{1}{(2-t)} \right \}dt

= -\log|1-t| +\log|2-t| +C

= \log\left | \frac{2-t}{1-t} \right |+C

Back substituting the value of t in the above equation, we get

= \log\left | \frac{2-\sin x}{1- \sin x} \right |+C

Question:18 Integrate the rational functions \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}

Answer:

Given function \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}

We can rewrite it as: \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \frac{(4x^2+10)}{(x^2+3)(x^2+4)}

Partial fraction of above equation,

\frac{(4x^2+10)}{(x^2+3)(x^2+4)} =\frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+4)}

4x^2+10 = (Ax+B)(x^2+4)+(Cx+D)(x^2+3)

4x^2+10 = Ax^3+4Ax+Bx^2+4B+Cx^3+3Cx+Dx^2+3D

4x^2+10 = (A+C)x^3+(B+D)x^2+(4A+3C)x+(3D+4B)

Now, equating the coefficients of x^3, x^2, x and constant term, we get

A+C=0 , B+D = 4 , 4A+3C = 0 , 4B+3D =10

After solving these equations, we get

A= 0, B =-2, C=0,\and\ D=6

\therefore \frac{4x^2+10}{(x^2+3)(x^2+4)} = \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)}

\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \left ( \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)} \right )

\implies \int \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} dx= \int \left \{ 1+ \frac{2}{(x^2+3)} - \frac{6}{(x^2+4)} \right \}dx

= \int \left \{ 1+ \frac{2}{(x^2+(\sqrt3)^2)} - \frac{6}{(x^2+2^2)} \right \}dx

= x+2\left ( \frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt 3} \right ) - 6\left ( \frac{1}{2}\tan^{-1}\frac{x}{2} \right )+C

= x+\frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3} -3\tan^{-1}\frac{x}{2}+C

Question:19 Integrate the rational functions \frac{2x }{( x^2 +1)( x^2 +3)}

Answer:

Given function \frac{2x }{( x^2 +1)( x^2 +3)}

Taking x^2 = t \Rightarrow 2xdx=dt

\therefore \int \frac{2x }{( x^2 +1)( x^2 +3)}dx = \int \frac{dt}{(t+1)(t+3)}

The partial fraction of above equation,

\frac{1}{(t+3)(t+3)} = \frac{A}{(t+1)}+\frac{B}{(t+3)}

1= A(t+3)+B(t+1) ..............(1)

Now, substituting t = -3\ and\ t = -1 in equation (1), we get

A =\frac{1}{2}\ and\ B = -\frac{1}{2}

\therefore\frac{1}{(t+3)(t+3)} = \frac{1}{2(t+1)}-\frac{1}{2(t+3)}

\implies \int \frac{2x }{( x^2 +1)( x^2 +3)}dx = \int \left \{ \frac{1}{2(t+1)}-\frac{1}{2(t+3)} \right \}dt

= \frac{1}{2}\log|t+1|- \frac{1}{2}\log|t+3| +C

= \frac{1}{2}\log\left | \frac{t+1}{t+3} \right | +C

= \frac{1}{2}\log\left | \frac{x^2+1}{x^2+3} \right | +C

Question:20 Integrate the rational functions \frac{1}{x (x^4 -1)}

Answer:

Given function \frac{1}{x (x^4 -1)}

So, we multiply numerator and denominator by x^3 , to obtain

\frac{1}{x (x^4 -1)} = \frac{x^3}{x^4(x^4-1)}

\therefore \int \frac{1}{x(x^4-1)}dx =\int\frac{x^3}{x^4(x^4-1)}dx

Now, putting x^4 = t

we get, 4x^3dx =dt

Taking x^2 = t \Rightarrow 2xdx=dt

\therefore \int \frac{1}{x(x^4-1)}dx =\frac{1}{4}\int \frac{dt}{t(t-1)}

Partial fraction of above equation,

\frac{1}{t(t-1)} = \frac{A}{t}+\frac{B}{(t-1)}

1= A(t-1)+Bt ..............(1)

Now, substituting t = 0\ and\ t = 1 in equation (1), we get

A = -1\ and\ B=1

\Rightarrow \frac{1}{t(t+1)} = -\frac{1}{t}+\frac{1}{t-1}

\Rightarrow \int \frac{1}{x(x^4+1)}dx =\frac{1}{4}\int \left \{ \frac{-1}{t}+\frac{1}{t-1} \right \}dt

= \frac{1}{4} \left [ -\log|t|+\log|t-1| \right ]+C

= \frac{1}{4}\log\left | \frac{t-1}{t} \right |+C

Back substituting the value of t,

=\frac{1}{4}\log \left | \frac{x^4-1}{x^4} \right | +C

Question:21 Integrate the rational functions \frac{1}{( e ^x-1)} [Hint : Put e ^x= t ]

Answer:

Given function \frac{1}{( e ^x-1)}

So, applying the hint: Putting e^x = t

Then e^x dx= dt

\int \frac{1}{( e ^x-1)}dx = \int\frac{1}{t-1}\times\frac{dt}{t} = \int \frac{1}{t(t-1)}dt


Partial fraction of above equation,

\frac{1}{t(t-1)} = \frac{A}{t}+\frac{B}{(t-1)}

1= A(t-1)+Bt ..............(1)

Now, substituting t = 0\ and\ t = 1 in equation (1), we get

A = -1\ and\ B=1

\Rightarrow \frac{1}{t(t+1)} = -\frac{1}{t}+\frac{1}{t-1}

\implies \int \frac{1}{t(t-1)}dt = \log \left | \frac{t-1}{t} \right |+C

Now, back substituting the value of t,

= \log \left | \frac{e^x-1}{e^x} \right |+C

Question:22 Choose the correct answer \int \frac{x dx }{( x-1)(x-2) } \: \: equals

A ) \log |\frac{(x-1)^2}{x-2}| + C \\\\ B) \log |\frac{(x-2)^2}{x-1}| + C \\\\ C ) \log |(\frac{x-1}{x-2})^2| + C \\\\ D ) \log |{(x-1)^2}({x-2})| + C

Answer:

Given integral \int \frac{x dx }{( x-1)(x-2) }

Partial fraction of above equation,

\frac{x}{(x-1)(x-2)} = \frac{A}{(x-1)}+\frac{B}{(x-2)}

x= A(x+2)+B(x-1) ..............(1)

Now, substituting x = 1\ and\ x = 2 in equation (1), we get

A = -1\ and\ B=2

\therefore \frac{x}{(x-1)(x-2)} = -\frac{1}{(x-1)}+\frac{2}{(x-2)}

\implies \int \frac{x}{(x-1)(x-2)}dx = \int \left \{ \frac{-1}{(x-1)}+\frac{2}{(x-2)} \right \}dx

= -\log|x-1| +2log|x-2| +C

=\log \left | \frac{(x-2)^2}{x-1} \right | +C

Therefore, the correct answer is B.

Question:23 Choose the correct answer \int \frac{dx}{x ( x ^2+1)} \: \: equals

A ) \log |x| - \frac{1}{2} \log ( x^2 +1 ) + C \\\\ B ) \log |x|+ \frac{1}{2} \log ( x^2 +1 ) + C \\\\ C )- \log |x| + \frac{1}{2} \log ( x^2 +1 ) + C \\\\ D ) \frac{1}{2}\log |x| +\log ( x^2 +1 ) + C

Answer:

Given integral \int \frac{dx}{x ( x ^2+1)}

Partial fraction of above equation,

\frac{1}{x ( x ^2+1)} = \frac{A}{x}+\frac{Bx+c}{x^2+1}

1= A(x^2+1)+(Bx+C)x

Now, equating the coefficients of x^2,x, and the constant term, we get

A+B = 0 , C=0 , A=1

We have the values, A = 1\ and\ B=-1,\ and\ C=0

\therefore \frac{1}{x ( x ^2+1)} = \frac{1}{x}+\frac{-x}{x^2+1}

\implies \int \frac{1}{x ( x ^2+1)}dx =\int \left \{ \frac{1}{x}+\frac{-x}{x^2+1}\right \}dx

= \log|x| -\frac{1}{2}\log|x^2+1| +C

Therefore, the correct answer is A.


NCERT solutions for maths chapter 7 class 12 Integrals Exercise: 7.6

Question:1 Integrate the functions x \sin x

Answer:

Given function is
f(x)=x \sin x
We will use integrate by parts method
\int x\sin x = x.\int \sin xdx - \int(\frac{d(x)}{dx}.\int sin x dx)dx\\ \\ \int x\sin x = x.(-\cos x)- \int (1.(-\cos x))dx\\ \\ \int x\sin x= -x\cos x+\sin x + C
Therefore, the answer is -x\cos x+\sin x + C

Question:2 Integrate the functions x \sin 3x

Answer:

Given function is
f(x)=x \sin 3x
We will use integration by parts method
\int x\sin 3x = x.\int \sin 3xdx - \int(\frac{d(x)}{dx}.\int sin 3x dx)dx\\ \\ \int x\sin 3x = x.(\frac{-\cos 3x}{3})- \int (1.(\frac{-\cos 3x}{3}))dx\\ \\ \int x\sin 3x= -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C

Therefore, the answer is -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C

Question:3 Integrate the functions x ^ 2 e ^x

Answer:

Given function is
f(x)=x^2e^x
We will use integration by parts method
\int x^2e^x= x^2.\int e^xdx - \int(\frac{d(x^2)}{dx}.\int e^x dx)dx\\ \\ \int x^2e^x = x^2.e^x- \int (2x.e^x)dx\\
Again use integration by parts in \int (2x.e^x)dx\\
\int (2x.e^x)dx = 2x.\int e^x dx - \int (\frac{d(2x)}{dx}.\int e^xdx)dx\\ \int 2x.e^x dx = 2xe^x- \int 2.e^xdx\\ \int 2x.e^x dx = 2xe^x- 2e^x
Put this value in our equation
we will get,
\int x^2.e^x dx =x^2e^x -2xe^x+ 2e^x + C\\ \int x^2.e^x dx = e^x(x^2-2x+2)+ C

Therefore, answer is e^x(x^2-2x+2)+ C

Question:4 Integrate the functions x \log x

Answer:

Given function is
f(x)=x.\log x
We will use integration by parts method
\int x.\log xdx= \log x.\int xdx - \int(\frac{d(\log x)}{dx}.\int x dx)dx\\ \\ \int x\log xdx = \log x.\frac{x^2}{2}- \int (\frac{1}{x}.\frac{x^2}{2})dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \frac{x^2}{4}+ C

Therefore, the answer is \frac{x^2}{2}\log x- \frac{x^2}{4}+ C

Question:5 Integrate the functions x \log 2x

Answer:

Given function is
f(x)=x.\log 2 x
We will use integration by parts method
\int x.\log 2xdx= \log 2x.\int xdx - \int(\frac{d(\log 2x)}{dx}.\int x dx)dx\\ \\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \int (\frac{2}{2x}.\frac{x^2}{2})dx\\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C

Therefore, the answer is \log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C

Question:6 Integrate the functions x^ 2 \log x

Answer:

Given function is
f(x)=x^2.\log x
We will use integration by parts method
\int x^2.\log xdx= \log x.\int x^2dx - \int(\frac{d(\log x)}{dx}.\int x^2 dx)dx\\ \\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \int (\frac{1}{x}.\frac{x^3}{3})dx\\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \int \frac{x^2}{3}dx\\ \int x^2\log xdx = \log x.\frac{x^3}{3}- \frac{x^3}{9}+ C

Therefore, the answer is \log x.\frac{x^3}{3}- \frac{x^3}{9}+ C

Question:7 Integrate the functions x \sin ^{ -1} x

Answer:

Given function is
f(x)=x.\sin^{-1} x
We will use integration by parts method
\int x.\sin^{-1} xdx= \sin^{-1} x.\int xdx - \int(\frac{d(\sin^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\
Now, we need to integrate \int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\
\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\left ( \int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{1}{2}\left ( \frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x-\sin^{-1}x \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{x\sqrt{1-x^2}}{4} -\frac{\sin^{-1}x}{4}+C
Put this value in our equation

Therefore, the answer is \int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x\sqrt{1-x^2}}{4}

Question:8 Integrate the functions x \tan ^{-1} x

Answer:

Given function is
f(x)=x.\tan^{-1} x
We will use integration by parts method
\int x.\tan^{-1} xdx= \tan^{-1} x.\int xdx - \int(\frac{d(\tan^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}- \int (\frac{1}{1+x^2}.\frac{x^2}{2})dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\int \left ( \frac{x^2+1}{1+x^2}-\frac{1}{1+x^2} \right )dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\int \left ( 1-\frac{1}{1+x^2} \right )dx\\ \\ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\left ( x- \tan^{-1}x \right )+C\\ \\ \int x\tan^{-1}xdx = \frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C

Put this value in our equation
\int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \frac{x}{4\sqrt{1-x^2}}-\frac{\sin^{-1}x}{4}+C\\ \int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x}{4\sqrt{1-x^2}}

Therefore, the answer is \frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C

Question:9 Integrate the functions x\cos ^{ -1} x

Answer:

Given function is
f(x)=x.\cos^{-1} x
We will use integration by parts method
\int x.\cos^{-1} xdx= \cos^{-1} x.\int xdx - \int(\frac{d(\cos^{-1} x)}{dx}.\int x dx)dx\\ \\ \int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}- \int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\
Now, we need to integrate \int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\
\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} \right )dx\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\left ( \int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{1}{2}\left ( \frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}\cos^{-1}x+\cos^{-1}x \right )\\ \\ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C
Put this value in our equation
\int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}-\left ( \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C \right )\\ \\ \int x\cos^{-1} xdx =\frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}

Therefore, the answer is \frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}

Question:10 Integrate the functions ( \sin ^{-1}x ) ^ 2

Answer:

Given function is
f(x)=( \sin ^{-1}x ) ^ 2
we will use integration by parts method
\int (\sin^{-1}x)^2= (\sin^{-1}x)^2.\int 1dx-\int \left ( \frac{d( (\sin^{-1}x)^2)}{dx} .\int 1dx\right )dx\\ \\ \int (\sin^{-1}x)^2 = (\sin^{-1}x)^2.x-\int \left ( \sin^{-1}.\frac{2x}{\sqrt{1-x^2}} \right )dx\\ \int (\sin^{-1}x)^2 = (\sin^{-1}x)^2.x + \left [ \sin^{-1}x.\int \frac{-2x}{\sqrt{1-x^2}}dx-\int \left ( \frac{d(\sin^{-1}x)}{dx}. \int \frac{-2x}{\sqrt{1-x^2}}dx\right ) \right ]\\ \\ . \ \ \ \ \ = (\sin^{-1}x)^2.x + \left [ \sin^{-1}x.2\sqrt{1-x^2}- \int \frac{1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]\\ \\ . \ \ \ \ \ = (\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C
Therefore, answer is (\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C

Question:11 Integrate the functions \frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}

Answer:

Consider \int \frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}dx =I

So, we have then: I = \frac{-1}{2}\int \frac{-2x}{\sqrt{1-x^2}}. \cos^{-1}x dx

After taking \cos ^{-1}x as a first function and \left ( \frac{-2x}{\sqrt{1-x^2}} \right ) as second function and integrating by parts, we get

I =-\frac{1}{2}\left [ \cos^{-1}x\int\frac{-2x}{\sqrt{1-x^2}}dx - \int\left \{ \left ( \frac{d}{dx}\cos^{-1}x \right )\int \frac{-2x}{\sqrt{1-x^2}}dx \right \}dx \right ] =-\frac{1}{2}\left [ \cos^{-1}x.2{\sqrt{1-x^2}} + \int \frac{-1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]

=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-\int2dx \right ]

=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-2x \right ]+C

Or, -\left \sqrt{1-x^2}\cos^{-1}x +x\right +C

Question:12 Integrate the functions x \sec ^2 x

Answer:

Consider x \sec ^2 x

So, we have then: I =\int x\sec^2 x dx

After taking x as a first function and \sec^2x as second function and integrating by parts, we get

I =x\int \sec^2 x dx -\int \left \{ \left ( \frac{d}{dx}x \right )\int \sec^2 x dx \right \}dx

= x\tan x -\int1.\tan x dx

= x\tan x +\log|\cos x | +C

Question:13 Integrate the functions \tan ^{-1} x

Answer:

Consider \tan ^{-1} x

So, we have then: I =\int 1.\tan^{-1}x dx

After taking \tan^{-1}x as a first function and 1 as second function and integrating by parts, we get

I = \tan^{-1}x \int 1dx -\int \left \{ \left ( \frac{d}{dx}\tan^{-1}x \right )\int1.dx \right \}dx

= \tan^{-1}x.x -\int \frac{1}{1+x^2}.xdx

= x\tan^{-1}x -\frac{1}{2}\int \frac{2x}{1+x^2}dx

= x\tan^{-1}x -\frac{1}{2}\log|1+x^2|+C

= x\tan^{-1}x -\frac{1}{2}\log(1+x^2)+C

Question:14 Integrate the functions x ( \log x )^ 2

Answer:

Consider x ( \log x )^ 2

So, we have then: I = \int x(\log x)^2 dx

After taking (\log x )^2 as a first function and x as second function and integrating by parts, we get

I = (\log x )^2 \int xdx -\int \left \{ \left ( \frac{d}{dx} (\log x)^2 \right )\int x.dx \right \}dx

= (\log x)^2 .\frac{x^2}{2} - \int \frac{2\log x }{x}.\frac{x^2}{2} dx

= (\log x)^2 .\frac{x^2}{2} - \int x\log x dx

= (\log x)^2 .\frac{x^2}{2} - \left ( \frac{x^2 \log x }{2} -\frac{x^2}{4} \right )+C

Question:15 Integrate the functions ( x^2 + 1 ) \log x

Answer:

Consider ( x^2 + 1 ) \log x

So, we have then: I = \int (x^2+1) \log x dx = \int x^2 \log x dx +\int \log x dx

Let us take I = I_{1} +I_{2} ....................(1)

Where, I_{1} = \int x^2\log x dx and I_{2} = \int \log x dx

So, I_{1} = \int x^2\log x dx

After taking \log x as a first function and x^2 as second function and integrating by parts, we get

I = \log x \int x^2dx -\int \left \{ \left ( \frac{d}{dx} \log x \right )\int x^2.dx \right \}dx

= \log x .\frac{x^3}{3} - \int \frac{1}{x}.\frac{x^3}{3} dx

= \log x .\frac{x^3}{3} - \frac{x^3}{9} +C_{1} ....................(2)

I_{2} = \int \log x dx

After taking \log x as a first function and 1 as second function and integrating by parts, we get

I_{2} = \log x \int 1.dx - \int \left \{ \left ( \frac{d}{dx}\log x \right ) \int 1.dx \right \}dx

= \log x .x -\int \frac{1}{x}. xdx

= x\log x -\int 1 dx

= x\log x -x +C_{2} ................(3)

Now, using the two equations (2) and (3) in (1) we get,

I = \frac{x^3}{3}\log x -\frac{x^3}{9} +C_{1} +x\log x - x +C_{2}

= \frac{x^3}{3}\log x -\frac{x^3}{9} +x\log x - x +(C_{1}+C_{2})

=\left ( \frac{x^3}{3}+x \right ) \log x -\frac{x^3}{9} -x+C

Question:16 Integrate the functions e ^ x ( \sin x + \cos x )

Answer:

Let suppose
I = e ^ x ( \sin x + \cos x )
f(x) = \sin x \Rightarrow f'(x) = \cos x
we know that,
I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C
Thus, the solution of the given integral is given by

\therefore I = e^x\sin x +C

Question:17 Integrate the functions \frac{x e ^x }{( 1+ x )^2}

Answer:

\frac{x e ^x }{( 1+ x )^2}
Let suppose
I = \int \frac{e^x(x)}{(1+x)^2}dx
by rearranging the equation, we get
\Rightarrow \int e^x[\frac{1}{1+x}-\frac{1}{(1+x)^2}]dx
let
f(x)=\frac{1}{1+x} \Rightarrow f'(x)= -\frac{1}{(1+x)^2}
It is known that \int e^x[f(x)+f'(x)]=e^x[f(x)]+C
therefore the solution of the given integral is

I = \frac{e^x}{1+x}+C

Question:18 Integrate the functions e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )

Answer:

Let
I =e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )
substitute 1 =\sin ^2\frac{x}{2}+\cos^2\frac{x}{2} and \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}

\\\Rightarrow e^x(\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}})\\ =e^x(\frac{1}{2}\sec^2\frac{x}{2}+\tan\frac{x}{2})\\
let
f(x) =\tan\frac{x}{2} \Rightarrow f'(x)=\frac{1}{2}\sec^2\frac{x}{2}
It is known that \int e^x[f(x)+f'(x)]=e^x[f(x)]+C
Therefore the solution of the given integral is

I = e^x\tan\frac{x}{2} +C

Question:19 Integrate the functions e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )

Answer:

e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )
It is known that
\int e^x[f(x)+f'(x)]=e^x[f(x)]+C

let
f(x)=\frac{1}{x}\Rightarrow f'(x)=-\frac{1}{x^2}
Therefore the required solution of the given above integral is
I = e^x.\frac{1}{x}+C

Question:20 Integrate the functions \frac{( x-3)e ^x }{( x-1)^3}

Answer:

\frac{( x-3)e ^x }{( x-1)^3}
It is known that \int e^x[f(x)+f'(x)]=e^x[f(x)]+C

So, By adjusting the given equation, we get
\int\frac{( x-3)e ^x }{( x-1)^3} =\int e^x(\frac{x-1-2}{(x-1)^3}) =\int e^x({\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3})}dx

to let
f(x)=\frac{1}{(x-1)^2}\Rightarrow f'(x)=-\frac{2}{(x-1)^3}
Therefore the required solution of the given I=\frac{e^x}{(x-1)^2}+C integral is

Question:21 Integrate the functions e ^{ 2x } \sin x

Answer:

Let
I =e ^{ 2x } \sin x
By using integrating by parts, we get

\\=\sin x\int e ^{ 2x }dx-\int(\frac{d}{dx}\sin x.\int e^{2x}dx)\ dx\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}\int e^{2x}.\cos x\ dx\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}[\cos x\int e^{2x}dx-\int (\frac{d}{dx}\cos x.\int e^{2x}dx)\ dx]\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{2}[\cos x.\frac{e^x}{2}+\frac{1}{2}\int e^{2x}\sin x dx]\\ =\frac{\sin x.e^{2x}}{2}-\frac{1}{4}\cos x.e^{2x}-\frac{1}{4}I\\ \Rightarrow \frac{5}{4}I =\frac{\sin x.e^{2x}}{2}-\frac{1}{4}\cos x.e^{2x}\\ I = \frac{e^{2x}}{5}[2\sin x-\cos x]+C

Question:22 Integrate the functions \sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )

Answer:

\sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )

\int \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx
let
x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta

\\=\int\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int2\theta \sec^2\theta d\theta\\
Taking \theta as a first function and \sec^2\theta as a second function, by using by parts method

\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]+C\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]+C\\ =2x\tan^{-1}x+2\log (1+x^2)^{-1/2}\\ =2x\tan^{-1}x-\log(1+x^2)+C

Question:23 Choose the correct answer

\int x ^ 2 e ^{x ^3 } dx \: \: equals

A ) \frac{1}{3} e ^{x^3} + C \\\\ B) \frac{1}{3} e ^{x^2} + C \\\\ C ) \frac{1}{2} e ^{x^3} + C \\\\ D ) \frac{1}{2} e ^{x^2} + C

Answer:

the integration can be done ass follows

let x^3 =t\Rightarrow 3x^2dx=dt
\Rightarrow I =\frac{1}{3}\int e^tdt =\frac{1}{3}e^t+C=\frac{1}{3}e^x^3+C

Question:24 Choose the correct answer

\int e ^ x \sec ( 1+ \tan x ) dx \: \: \: equals

A ) e ^ x \cos x + C \\\\ B) e ^ x \sec x + C \\\\ C ) e ^ x \sin x + C\\\\D ) e ^ x \tan x + C

Answer:

we know that,
I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C
from above integral
let
f(x)=\sec x\Rightarrow f'(x)= \sec x.\tan x
thus, the solution of the above integral is
I=e^x\sec x+C

NCERT class 12 maths ch 7 question answer Exercise: 7.7

Question:1 Integrate the functions in Exercises 1 to 9.

\sqrt{4 - x^2}

Answer:

Given function \sqrt{4 - x^2} ,

So, let us consider the function to be;

I = \int \sqrt{4-x^2}dx

= \int \sqrt{(2)^2-x^2}dx

Then it is known that, = \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C

Therefore, I = \frac{x}{2}\sqrt{4-x^2} +\frac{4}{2}\sin^{-1}{\frac{x}{2}}+C

= \frac{x}{2}\sqrt{4-x^2} +2\sin^{-1}{\frac{x}{2}}+C


Question:2 Integrate the functions in Exercises 1 to 9.

\sqrt{1 - 4x^2}

Answer:

Given function to integrate \sqrt{1 - 4x^2}

Now we can rewrite as

= \int \sqrt{1 - (2x)^2}dx

As we know the integration of this form is \left [ \because \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a} \right ]

= \frac{(\frac{2x}{2})\sqrt{1^2-(2x)^2}+\frac{1^2}{2}\sin^{-1}\frac{2x}{1}}{2\rightarrow Coefficient\ of\ x\ in\ 2x} +C

= \frac{1}{2}\left [ x\sqrt{1-4x^2}+\frac{1}{2}\sin^{-1}2x \right ]+C

= \frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\sin^{-1}2x+C


Question:3 Integrate the functions in Exercises 1 to 9.

\sqrt{x^2 + 4x + 6}

Answer:

Given function \sqrt{x^2 + 4x + 6} ,

So, let us consider the function to be;

I = \int\sqrt{x^2 + 4x + 6}dx

= \int\sqrt{(x^2 + 4x + 4)+2}dx = \int\sqrt{(x + 2)^2 +(\sqrt2)^2}dx

And we know that, \int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C

\Rightarrow I = \frac{x+2}{2}\sqrt{x^2+4x+6}+\frac{2}{2}\log\left | (x+2)+\sqrt{x^2+4x+6} \right |+C

= \frac{x+2}{2}\sqrt{x^2+4x+6}+\log\left | (x+2)+\sqrt{x^2+4x+6} \right |+C


Question:4 Integrate the functions in Exercises 1 to 9.

\sqrt{x^2 + 4x +1}

Answer:

Given function \sqrt{x^2 + 4x +1} ,

So, let us consider the function to be;

I = \int\sqrt{x^2 + 4x + 1}dx

= \int\sqrt{(x^2 + 4x + 4)-3}dx = \int\sqrt{(x + 2)^2 -(\sqrt3)^2}dx

And we know that, \int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C

\therefore I = \frac{x+2}{2}\sqrt{x^2+4x+1}-\frac{3}{2}\log\left | (x+2)+\sqrt{x^2+4x+1} \right |+C


Question:5 Integrate the functions in Exercises 1 to 9.

\sqrt{1-4x-x^2}

Answer:

Given function \sqrt{1-4x-x^2} ,

So, let us consider the function to be;

I = \int\sqrt{1-4x-x^2}dx

= \int\sqrt{1-(x^2+4x+4-4)}dx = \int\sqrt{1+4 -(x+2)^2}dx

= \int\sqrt{(\sqrt5)^2 -(x+2)^2}dx

And we know that, \int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C

\therefore I = \frac{x+2}{2}\sqrt{1-4x-x^2}+\frac{5}{2}\sin^{-1}\left ( \frac{x+2}{\sqrt5} \right )+C


Question:6 Integrate the functions in Exercises 1 to 9.

\sqrt{x^2 + 4x - 5}

Answer:

Given function \sqrt{x^2 + 4x - 5} ,

So, let us consider the function to be;

I = \int\sqrt{x^2+4x-5}dx

a = \int\sqrt{(x^2+4x+4)-9}dx = \int\sqrt{(x+2)^2 -(3)^2}dx

And we know that, \int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}|+C

\therefore I = \frac{x+2}{2}\sqrt{x^2+4x-5}-\frac{9}{2}\log\left | (x+2)+ \sqrt{x^2+4x-5} \right |+C


Question:7 Integrate the functions in Exercises 1 to 9.

\sqrt{1 + 3x - x^2}

Answer:

Given function \sqrt{1 + 3x - x^2} ,

So, let us consider the function to be;

I = \int\sqrt{1+3x-x^2}dx

= \int\sqrt{(1-\left ( x^2-3x+\frac{9}{4}-\frac{9}{4} \right )}dx = \int \sqrt{\left ( 1+\frac{9}{4} \right )-\left ( x-\frac{3}{2} \right )^2}dx = \int \sqrt{\left ( \frac{\sqrt{13}}{2} \right )^2-\left ( x-\frac{3}{2} \right )^2}dx

And we know that, \int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C

\therefore I = \frac{x-\frac{3}{2}}{2}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\left ( \frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}} \right )+C

= \frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\left ( \frac{2x-3}{\sqrt{13}} \right )+C


Question:8 Integrate the functions in Exercises 1 to 9.

\sqrt{x^2 + 3x}

Answer:

Given function \sqrt{x^2 + 3x} ,

So, let us consider the function to be;

I = \int\sqrt{x^2+3x}dx

= \int\sqrt{x^2+3x+\frac{9}{4}-\frac{9}{4}}dx

= \int\sqrt{\left ( x+\frac{3}{2} \right )^2-\left ( \frac{3}{2} \right )^2 }dx

And we know that, \int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C

\therefore I = \frac{x+\frac{3}{2}}{2}\sqrt{x^2+3x}-\frac{\frac{9}{4}}{2}\log \left | \left ( x+\frac{3}{2} \right )+\sqrt{x^2+3x} \right |+C

= \frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}\log\left | \left ( x+\frac{3}{2} \right )+\sqrt{x^2+3x} \right |+C


Question:9 Integrate the functions in Exercises 1 to 9.

\sqrt{1 + \frac{x^2}{9}}

Answer:

Given function \sqrt{1 + \frac{x^2}{9}} ,

So, let us consider the function to be;

I = \int\sqrt{1+\frac{x^2}{9}}dx = \frac{1}{3}\int \sqrt{9+x^2}dx

= \frac{1}{3}\int \sqrt{3^2+x^2}dx

And we know that, \int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C

\therefore I = \frac{1}{3}\left [ \frac{x}{2}\sqrt{x^2+9} +\frac{9}{2}\log|x+\sqrt{x^2+9}| \right ]+C

= \frac{x}{6}\sqrt{x^2+9} +\frac{3}{2}\log\left | x+\sqrt{x^2+9} \right |+C


Question:10 Choose the correct answer in Exercises 10 to 11.

\int \sqrt{1+x^2}dx is equal to

(A) \frac{x}{2}\sqrt{1+x^2} + \frac{1}{2}\log\left |\left(x + \sqrt{1+x^2} \right )\right| +C

(B) \frac{2}{3}(1+x^2)^{\frac{3}{2}} + C

(C) \frac{2}{3}x(1+x^2)^{\frac{3}{2}} + C

(D) \frac{x^2}{2}\sqrt{1+x^2} + \frac{1}{2}x^2\log\left |x + \sqrt{1+x^2} \right| +C

Answer:

As we know that, \int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C

So, \int \sqrt{1+x^2}dx = \frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\log|x+\sqrt{x^2+1}| +C

Therefore the correct answer is A.


Question:11 Choose the correct answer in Exercises 10 to 11.

\int \sqrt{x^2 - 8x+7}dx is equal to

(A) \frac{1}{2}(x-4)\sqrt{x^2-8x+7} + 9\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C

(B) \frac{1}{2}(x+4)\sqrt{x^2-8x+7} + 9\log\left|x+4+\sqrt{x^2 -8x+7}\right| +C

(C) \frac{1}{2}(x-4)\sqrt{x^2-8x+7} -3\sqrt2\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C

(D) \frac{1}{2}(x-4)\sqrt{x^2-8x+7} -\frac{9}{2}\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C

Answer:

Given integral \int \sqrt{x^2 - 8x+7}dx

So, let us consider the function to be;

I = \int\sqrt{x^2-8x+7}dx =\int\sqrt{(x^2-8x+16)-(9)}dx

=\int\sqrt{(x-4)^2-(3)^2}dx

And we know that, \int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C

I = \frac{(x-4)}{2}\sqrt{x^2-8x+7}-\frac{9}{2}\log|(x-4)+\sqrt{x^2-8x+7}| +C

Therefore the correct answer is D.



NCERT class 12 maths ch 7 question answer - Exercise:7.8

Question:1 Evaluate the following definite integrals as a limit of sums.

\int_a^b x dx

Answer:

We know that,
\int_{a}^{b}f(x)dx = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+...+f(a+(n-1)h)]
\therefore \int_{a}^{b}xdx = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[a+(a+h)...(a+2h)..a+(n-1)h]
\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[(a+a...a+a)_{n}+(h+2h+3h....(n-1)h)]\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[na+h(1+2+3..+n-1)]\\ = (b-a)\lim_{x\rightarrow \infty }\frac{1}{n}[na+h(\frac{n(n-1)}{2})]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{n-1}{2}h]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{(n-1)(b-a)}{2n}]\\ = (b-a)\lim_{x\rightarrow \infty }[a+\frac{(1-\frac{1}{n})(b-a)}{2}]\\ = (b-a)[a+\frac{(b-a)}{2}]\\ =(b-a)(b+a)/2\\ =\frac{(b^2-a^2)}{2}

This is how the integral is evaluated using limit of a sum

Question:2 Evaluate the following definite integrals as limit of sums.

\int_0^5 (x + 1)dx

Answer:

We know that
let I =\int_{0}^{5}(x+1)dx
\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}
Here a = 0, b = 5 and f(x)=(x+1)
therefore h=\frac{5}{n}


\int_{0}^{5}(x+1)dx=5\lim_{x\rightarrow \infty }\frac{1}{n}[f(0)+f(5/n)+.....+f((n-1)5/n)]

=5\lim_{x\rightarrow \infty }\frac{1}{n}[1+(5/n+1)+....+(1+\frac{5(n-1)}{n})]\\ =5\lim_{x\rightarrow \infty }\frac{1}{n}[(1+1..+1)_{n}+\frac{5}{n}(1+2+3+...+n-1)]\\ =5\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{5}{n}\frac{n(n-1)}{2}]\\ =5\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{5(n-1)}{2}]\\ =5\lim_{x\rightarrow \infty }[1+\frac{5(1-\frac{1}{n})}{2}]\\ =5[1+\frac{5}{2}]\\ =\frac{35}{2}

Question:3 Evaluate the following definite integrals as limit of sums.

\int_2^3 x^2dx

Answer:

We know that

\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}
here a = 2 and b = 3 , so h = 1/n


\int_{2}^{3}x^2dx=(3-2)\lim_{x\rightarrow \infty }\frac{1}{n}[f(2)+f(2+\frac{1}{n})+f(2+\frac{2}{n})+....+f(2+\frac{(n-1)}{n})]

\\=(1)\lim_{x\rightarrow \infty }\frac{1}{n}[2^2+(2+\frac{1}{n})^2+......+(2+\frac{(n-1)}{n})^2]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[(2^2+2^2+...2^2)_{n}+(\frac{1}{n})^2+(\frac{2}{n})^2+....(\frac{n-1}{n})^2+4(\frac{1}{n}+\frac{2}{n}+.....+\frac{n-1}{n})\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[4n+\frac{n(n-1)(2n-1)}{6n^2}+\frac{4}{n}.\frac{n(n-1)}{2}]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}[4n+(1-\frac{(1-\frac{1}{n})(2n-1)}{6})+\frac{4(n-1)}{2}]
\\=\lim_{x\rightarrow \infty }\frac{1}{n}[4n+(1-\frac{n(1-\frac{1}{n})(2-\frac{1}{n})}{6})+\frac{4(n-1)}{2}]\\ =\lim_{x\rightarrow \infty }\frac{1}{n}.n[4+(1-\frac{(1-\frac{1}{n})(2-\frac{1}{n})}{6})+2-\frac{2}{n}]\\ =4+\frac{2}{6}+2 =\frac{19}{3}

Question:4 Evaluate the following definite integrals as limit of sums.

\int_{1}^4(x^2-x)dx

Answer:

Let
\\I = \int_{1}^{4}(x^2-x)dx =\int_{1}^{4}x^2dx-\int_{1}^{4}xdx\\ I = I_1-I_2

\int_{1}^{4}x^2dx=(4-1)\lim_{x\rightarrow \infty }\frac{1}{n}[f(1)+f(1+h)+f(1+2h)+.....+f(1+(n-1)h)]

=(4-1)\lim_{x\rightarrow \infty }\frac{1}{n}[f(1)+f(1+h)+f(1+2h)+.....+f(1+(n-1)h)]\\ =3\lim_{x\rightarrow \infty }\frac{1}{n}[1^2+(1+\frac{3}{n})^2+(1+2.\frac{3}{n})^2+......+(1+(n-1).\frac{3}{n})^2]\\ =3\lim_{x\rightarrow \infty }\frac{1}{n}[(1^2+..1^2)_{n}+(\frac{3}{n})^2(1^2+2^2+3^2+....+(n-1)^2)+2.\frac{3}{n}(1+2+3..+n-1)]\\ =3\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{9}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{6}{n}(\frac{n(n-1)}{2})]

=3\lim_{x\rightarrow \infty }\frac{1}{n}[n+\frac{9}{n^2}(\frac{n(n-1)(2n-1)}{6})+\frac{6}{n}(\frac{n(n-1)}{2})]\\ =3\lim_{x\rightarrow \infty }[1+\frac{9}{6}(1-\frac{1}{n})(2-\frac{1}{n})+3(1-\frac{1}{n})]\\ =3[1+\frac{9}{6}.2+3]\\ = 21

for the second part, we already know the general solution of \int_{a}^{b}xdx = \frac{(b^2-a^2)}{2}
So, here a = 1 and b = 4
therefore \int_{1}^{4}xdx = \frac{(4^2-1^2)}{2}=\frac{15}{2}

So, I = 21-\frac{15}{2} = \frac{27}{2}

Question:5 Evaluate the following definite integrals as limit of sums.

. \int_{-1}^1 e^xdx

Answer:

let I = \int_{-1}^{1}e^xdx
We know that
\\\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)+...+f(a+(n-1)h)]\\ h = \frac{b-a}{n}
Here a =-1, b = 1 and f(x) = e^x
therefore h = 2/n
I = 2.\lim_{x\rightarrow \infty }\frac{1}{n}[f(-1)+f(-1+\frac{2}{n})+.....+f(-1+(n-1).\frac{2}{n})]
\\ =2.\lim_{x\rightarrow \infty }\frac{1}{n}[e^{-1}+e^{-1+\frac{2}{n}}+e^{-1+2.\frac{2}{n}}+...+e^{-1+(n-1).\frac{2}{n}}]\\ = 2.\lim_{x\rightarrow \infty }\frac{1}{n}[e^{-1}(1+e^{2/n}+e^{4/n}+...+e^{(n-1).\frac{2}{n}})]\\ =
By using sum of n terms of GP S =\frac{a(r^n-1)}{r-1} ....where a = 1st term and r = ratio

\\=2\lim_{n\rightarrow \infty }\frac{e^{-1}}{n}[\frac{1.(e^{\frac{2}{n}.n}-1)}{e^\frac{2}{n}-1}]\\ =2\lim_{n\rightarrow \infty }\frac{e^{-1}}{n}(\frac{e^2-1}{e^{2/n}-1})\\ =\frac{e^{-1}(e^2-1)}{\lim_{\frac{2}{n}\rightarrow \infty }\frac{e^{2/n}-1}{2/n}}\\ =\frac{e^2-1}{e} .........using [\lim_{x\rightarrow \infty }(\frac{e^x-1}{x})=1]

Question:6 Evaluate the following definite integrals as limit of sums.

\int_0^4(x + e^{2x})dx

Answer:

It is known that,
\int_{0}^{4}(x+e^{2x})dx = 4\lim_{x\rightarrow \infty }\frac{1}{n}[f(0)+f(h)+f(2h)+....+f(n-1)h]
\\=4\lim_{x\rightarrow \infty }\frac{1}{n}[(0+e^0)+(h+e^2h)+(2h+e^4h)+......+((n-1)h+e^{2(n-1)h})]\\ = 4\lim_{x\rightarrow \infty }\frac{1}{n}[h(1+2+3+.....+n-1)+(\frac{e^{2nh}-1}{e^{2h}-1})]\\ = 4\lim_{x\rightarrow \infty }\frac{1}{n}[\frac{4}{n}(\frac{n(n-1)}{2})+(\frac{e^8-1}{e^{8/n}-1})]
\\=4\lim_{x\rightarrow \infty }[4.\frac{1-\frac{1}{n}}{2}+\frac{\frac{e^8-1}{8}}{\frac{e^{8/n}-1}{\frac{8}{n}}}]\\ =4(2)+4[(\frac{e^8-1}{8})]\\ ==8+e^8/2-1/2\\ =\frac{15+e^8}{2} ..........................( \lim_{x\rightarrow 0}\frac{e^x-1}{x}=1 )

NCERT class 12 maths ch 7 question answer - Exercise:7.9

Question:1 Evaluate the definite integrals in Exercises 1 to 20.

\int_{-1}^{1} (x+1)dx

Answer:

Given integral: I = \int_{-1}^{1} (x+1)dx

Consider the integral \int (x+1)dx

\int (x+1)dx = \frac{x^2}{2}+x

So, we have the function of x , f(x) = \frac{x^2}{2}+x

Now, by Second fundamental theorem of calculus, we have

I = f(1)-f(-1)

= \left ( \frac{1}{2}+1\right ) - \left (\frac{1}{2}-1 \right ) = \frac{1}{2}+1-\frac{1}{2}+1 = 2

Question:2 Evaluate the definite integrals in Exercises 1 to 20.

\int_2^3\frac{1}{x}dx

Answer:

Given integral: I = \int_2^3\frac{1}{x}dx

Consider the integral \int_2^3\frac{1}{x}dx

\int \frac{1}{x}dx = \log|x|

So, we have the function of x , f(x) = \log|x|

Now, by Second fundamental theorem of calculus, we have

I = f(3)-f(2)

=\log|3|-\log|2| = \log \frac{3}{2}

Question:3 Evaluate the definite integrals in Exercises 1 to 20.

\int_1^2(4x^3-5x^2 + 6x +9)dx

Answer:

Given integral: I = \int_1^2(4x^3-5x^2 + 6x +9)dx

Consider the integral I = \int (4x^3-5x^2 + 6x +9)dx

\int (4x^3-5x^2 + 6x +9)dx = 4\frac{x^4}{4} -5\frac{x^3}{3}+6\frac{x^2}{2}+9x

= x^4 -\frac{5x^3}{3}+3x^2+9x

So, we have the function of x , f(x) = x^4 -\frac{5x^3}{3}+3x^2+9x

Now, by Second fundamental theorem of calculus, we have

I = f(2)-f(1)

=\left \{ 2^4-\frac{5(2)^3}{3}+3(2)^2+9(2)\right \} - \left \{ 1^4-\frac{5(1)^3}{3}+3(1)^2+9(1) \right \}

=\left \{ 16-\frac{40}{3}+12+18\right \} - \left \{ 1-\frac{5}{3}+3+9 \right \}

=\left \{ 46-\frac{40}{3}\right \} - \left \{ 13-\frac{5}{3}\right \}

=\left \{ 33-\frac{35}{3} \right \} = \left \{ \frac{99-35}{3} \right \}

= \frac{64}{3}

Question:4 Evaluate the definite integrals in Exercises 1 to 20.

\int_0^\frac{\pi}{4}\sin 2x dx

Answer:

Given integral: \int_0^\frac{\pi}{4}\sin 2x dx

Consider the integral \int \sin 2x dx

\int \sin 2x dx = \frac{-\cos 2x }{2}

So, we have the function of x , f(x) = \frac{-\cos 2x }{2}

Now, by Second fundamental theorem of calculus, we have

I = f(\frac{\pi}{4})-f(0)

= \frac{-\cos 2(\frac{\pi}{4})}{2} + \frac{\cos 0}{2}

=\frac{1}{2} - 0

= \frac{1}{2}

Question:5 Evaluate the definite integrals in Exercises 1 to 20.

\int_0^\frac{\pi}{2}\cos 2x dx

Answer:

Given integral: \int_0^\frac{\pi}{2}\cos 2x dx

Consider the integral \int \cos 2x dx

\int \cos 2x dx = \frac{\sin 2x }{2}

So, we have the function of x , f(x) = \frac{\sin 2x }{2}

Now, by Second fundamental theorem of calculus, we have

I = f(\frac{\pi}{2})-f(0)

= \frac{1}{2}\left \{ \sin 2(\frac{\pi}{2}) - \sin 0 \right \}

= \frac{1}{2}\left \{ 0 - 0 \right \} = 0

Question:6 Evaluate the definite integrals in Exercises 1 to 20.

\int_4^5 e^x dx

Answer:

Given integral: \int_4^5 e^x dx

Consider the integral \int e^x dx

\int e^x dx = e^x

So, we have the function of x , f(x) = e^x

Now, by Second fundamental theorem of calculus, we have

I = f(5) -f(4)

= e^5 -e^4

= e^4(e-1)

Question:7 Evaluate the definite integrals in Exercises 1 to 20.

\int^\frac{\pi}{4}_0 \tan x dx

Answer:

Given integral: \int^\frac{\pi}{4}_0 \tan x dx

Consider the integral \int \tan x dx

\int \tan x dx = -\log|\cos x |

So, we have the function of x , f(x) = -\log|\cos x |

Now, by Second fundamental theorem of calculus, we have

I = f(\frac{\pi}{4}) -f(0)

= -\log\left | \cos \frac{\pi}{4} \right | +\log|\cos 0|

= -\log\left | \cos \frac{1}{\sqrt2} \right | +\log|1|

= -\log\left | \frac{1}{\sqrt2} \right | + 0 = -\log (2)^{-\frac{1}{2}}

= \frac{1}{2}\log (2)

Question:8 Evaluate the definite integrals in Exercises 1 to 20.

\int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec}xdx

Answer:

Given integral: \int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec}xdx

Consider the integral \int\textup{cosec}xdx

\int\textup{cosec}xdx = \log|cosec x -\cot x |

So, we have the function of x , f(x) =\log|cosec x -\cot x |

Now, by Second fundamental theorem of calculus, we have

I = f(\frac{\pi}{4}) -f(\frac{\pi}{6})

= \log|cosec \frac{\pi}{4} -\cot \frac{\pi}{4} | - \log|cosec \frac{\pi}{6} -\cot \frac{\pi}{6} |

= \log|\sqrt2 -1 | - \log|2 -\sqrt3 |

= \log \left ( \frac{\sqrt2 -1}{2-\sqrt3} \right )

Question:9 Evaluate the definite integrals in Exercises 1 to 20.

\int_0^1\frac{dx}{\sqrt{1-x^2}}

Answer:

Given integral: \int_0^1\frac{dx}{\sqrt{1-x^2}}

Consider the integral \int \frac{dx}{\sqrt{1-x^2}}

\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x

So, we have the function of x , f(x) = \sin^{-1}x

Now, by Second fundamental theorem of calculus, we have

I = f(1) -f(0)

= \sin^{-1}(1) -\sin^{-1}(0)

= \frac{\pi}{2} - 0

= \frac{\pi}{2}

Question:10 Evaluate the definite integrals in Exercises 1 to 20.

\int_0^1\frac{dx}{1 + x^2}

Answer:

Given integral: \int_0^1\frac{dx}{1 + x^2}

Consider the integral \int\frac{dx}{1 + x^2}

\int\frac{dx}{1 + x^2} = \tan^{-1}x

So, we have the function of x , f(x) =\tan^{-1}x

Now, by Second fundamental theorem of calculus, we have

I = f(1) -f(0)

= \tan^{-1}(1) -\tan^{-1}(0)

= \frac{\pi}{4} - 0

= \frac{\pi}{4}

Question:11 Evaluate the definite integrals in Exercises 1 to 20.

\int_2^3 \frac{dx}{x^2 -1 }

Answer:

Given integral: \int_2^3 \frac{dx}{x^2 -1 }

Consider the integral \int \frac{dx}{x^2 -1 }

\int \frac{dx}{x^2 -1 } = \frac{1}{2}\log\left | \frac{x-1}{x+1} \right |

So, we have the function of x , f(x) =\frac{1}{2}\log\left | \frac{x-1}{x+1} \right |

Now, by Second fundamental theorem of calculus, we have

I = f(3) -f(2)

= \frac{1}{2}\left \{ \log\left | \frac{3-1}{3+1} \right | - \log\left | \frac{2-1}{2+1} \right | \right \}

= \frac{1}{2}\left \{ \log\left | \frac{2}{4} \right | -\log\left | \frac{1}{3} \right | \right \}

= \frac{1}{2}\left \{ \log \frac{1}{2} -\log \frac{1}{3} \right \} = \frac{1}{2}\log\frac{3}{2}

Question:12 Evaluate the definite integrals in Exercises 1 to 20.

\int_0^\frac{\pi}{2}\cos^2 x dx

Answer:

Given integral: \int_0^\frac{\pi}{2}\cos^2 x dx

Consider the integral \int \cos^2 x dx

\int \cos^2 x dx = \int \frac{1+\cos 2x}{2} dx = \frac{x}{2}+\frac{\sin 2x }{4}

= \frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )

So, we have the function of x , f(x) =\frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )

Now, by Second fundamental theorem of calculus, we have

I = f(\frac{\pi}{2}) -f(0)

= \frac{1}{2}\left \{ \left ( \frac{\pi}{2}-\frac{\sin \pi}{2} \right ) -\left ( 0+\frac{\sin 0}{2} \right ) \right \}

= \frac{1}{2}\left \{ \frac{\pi}{2}+0-0-0 \right \}

= \frac{\pi}{4}

Question:13 Evaluate the definite integrals in Exercises 1 to 20.

\int_2^3\frac{xdx}{x^2+1}

Answer:

Given integral: \int_2^3\frac{xdx}{x^2+1}

Consider the integral \int \frac{xdx}{x^2+1}

\int \frac{xdx}{x^2+1} = \frac{1}{2}\int \frac{2x}{x^2+1}dx =\frac{1}{2}\log(1+x^2)

So, we have the function of x , f(x) =\frac{1}{2}\log(1+x^2)

Now, by Second fundamental theorem of calculus, we have

I = f(3) -f(2)

= \frac{1}{2}\left \{ \log(1+(3)^2)-\log(1+(2)^2) \right \}

= \frac{1}{2}\left \{ \log(10)-\log(5) \right \} = \frac{1}{2}\log\left ( \frac{10}{5} \right ) = \frac{1}{2}\log2

Question:14 Evaluate the definite integrals in Exercises 1 to 20.

\int_0^1\frac{2x+3}{5x^2+1}dx

Answer:

Given integral: \int_0^1\frac{2x+3}{5x^2+1}dx

Consider the integral \int \frac{2x+3}{5x^2+1}dx

Multiplying by 5 both in numerator and denominator:

\int \frac{2x+3}{5x^2+1}dx = \frac{1}{5}\int \frac{5(2x+3)}{5x^2+1}dx

=\frac{1}{5}\int \frac{10x+15}{5x^2+1}dx

= \frac{1}{5} \int \frac{10x}{5x^2+1} dx +3\int \frac{1}{5x^2+1} dx

= \frac{1}{5}\int \frac{10x}{5x^2+1}+3\int \frac{1}{5\left ( x^2+\frac{1}{5} \right )}dx

= \frac{1}{5}\log(5x^2+1) +\frac{3}{5}\times \frac{1}{\frac{1}{\sqrt5}} \tan^{-1}\frac{x}{\frac{1}{\sqrt5}}

= \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )

So, we have the function of x , f(x) = \frac{1}{5}\log(5x^2+1) +\frac{3}{\sqrt5}\tan^{-1}(\sqrt5 x )

Now, by Second fundamental theorem of calculus, we have

I = f(1) -f(0)

= \left \{ \frac{1}{5}\log(1+5)+\frac{3}{\sqrt5}\tan^{-1}(\sqrt5) \right \} - \left \{ \frac{1}{5}\log(1)+\frac{3}{\sqrt5}\tan^{-1}(0) \right \}

= \frac{1}{5}\log 6 +\frac{3}{\sqrt 5}\tan^{-1}{\sqrt5}

Question:15 Evaluate the definite integrals in Exercises 1 to 20.

\int_0^1xe^{x^2}dx

Answer:

Given integral: \int_0^1xe^{x^2}dx

Consider the integral \int xe^{x^2}dx

Putting x^2 = t which gives, 2xdx =dt

As, x\rightarrow0 ,t \rightarrow0 and as x\rightarrow1 ,t \rightarrow1 .

So, we have now:

\therefore I = \frac{1}{2}\int_0^1 e^t dt

= \frac{1}{2}\int e^t dt = \frac{1}{2} e^t

So, we have the function of x , f(x) = \frac{1}{2} e^t

Now, by Second fundamental theorem of calculus, we have

I = f(1) -f(0)

= \frac{1}{2}e^1 -\frac{1}{2}e^0 = \frac{1}{2}(e-1)

Question:16 Evaluate the definite integrals in Exercises 1 to 20.

\int_1^2\frac{5x^2}{x^2 + 4x +3}

Answer:

Given integral: I = \int_1^2\frac{5x^2}{x^2 + 4x +3}

So, we can rewrite the integral as;

I = \int_1^2 \frac{5x^2}{x^2 + 4x +3}= \int_1^2 \left ( 5 - \frac{20x+15}{x^2 + 4x +3} \right ) dx

= \int_1^2 5 dx - \int_1^2 \frac{20x+15}{x^2+4x+3}dx

= [5x]_1^2 - \int_1^2 \frac{20x+15}{x^2+4x+3}dx

I = 5-I_1 where I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx . ................(1)

Now, consider I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx

Take numerator 20x+15 = A \frac{d}{dx}\left ( x^2+4x+3 \right )+B

= 2A x+(4A+B)

We now equate the coefficients of x and constant term, we get

A= 10 \and\ B =-25

\Rightarrow I_1 = 10\int_1^2 \frac{2x+4}{x^2+4x+3}dx -25\int_1^2 \frac{dx}{x^2+4x+3}

Now take denominator x^2+4x+3 = t

Then we have (2x+4)dx =dt

\Rightarrow I_{1} =10\int \frac{dt}{t} -25\int \frac{dx}{(x+2)^2-1^2}

= 10\log t -25\left [ \frac{1}{2}\log\left ( \frac{x+2-1}{x+2+1} \right ) \right ]

=[10\log(x^2+4x+3)]_1^2 -25 \left [ \frac{1}{2}\log\left ( \frac{x+1}{x+3} \right ) \right ]_1^2

= \left [ 10\log15 -10\log 8 \right ] -25 \left [ \frac{1}{2}\log\frac{3}{5} -\frac{1}{2}\log\frac{2}{4} \right ]

= \left [ 10\log5 +10\log3 -10\log4-10\log2 \right ] -\frac{25}{2}\left [ \log3 -\log5-\log2+\log4 \right ] = \left ( 10+\frac{25}{2} \right )\log5 + \left ( -10-\frac{25}{2} \right )\log 4 + \left ( 10-\frac{25}{2} \right )\log 3 + \left ( -10+\frac{25}{2} \right )\log 2 = \frac{45}{2}\log5 -\frac{45}{2}\log4 - \frac{5}{2}\log3 +\frac{5}{2}\log2

= \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log \frac{3}{2}

Then substituting the value of I_{1} in equation (1), we get

I= 5 -\left ( \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log\frac{3}{2} \right )

= 5 -\frac{5}{2}\left ( 9\log\frac{5}{4}-\log\frac{3}{2} \right )

Question:17 Evaluate the definite integrals in Exercises 1 to 20.

\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx

Answer:

Given integral: \int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx

Consider the integral \int (2\sec^2x + x^3 + 2)dx

\int (2\sec^2x + x^3 + 2)dx = 2\tan x +\frac{x^4}{4}+2x

So, we have the function of x , f(x) = 2\tan x +\frac{x^4}{4}+2x

Now, by Second fundamental theorem of calculus, we have

I = f(\frac{\pi}{4}) -f(0)

= \left \{ \left ( 2\tan\frac{\pi}{4}+\frac{1}{4}\left ( \frac{\pi}{4} \right )^4+2\frac{\pi}{4} \right ) - \left ( 2\tan 0 +0 +0 \right ) \right \}

=2\tan\frac{\pi}{4} +\frac{\pi^4}{4^5} +\frac{\pi}{2}

2+\frac{\pi}{2}+\frac{\pi^4}{1024}

Question:18 Evaluate the definite integrals in Exercises 1 to 20.

\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx

Answer:

Given integral: \int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx

Consider the integral \int (\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx

can be rewritten as: -\int (\cos^2\frac{x}{2} - \sin^2\frac{x}{2})dx = -\int_0^{\pi} \cos x dx

= \sin x

So, we have the function of x , f(x) =\sin x

Now, by Second fundamental theorem of calculus, we have

I = f(\pi) - f(0)

\Rightarrow \sin \pi - \sin 0 = 0-0 =0

Question:19 Evaluate the definite integrals in Exercises 1 to 20.

\int_0^2\frac{6x+3}{x^2+ 4}

Answer:

Given integral: \int_0^2\frac{6x+3}{x^2+ 4}

Consider the integral \int \frac{6x+3}{x^2+ 4}

can be rewritten as: \int \frac{6x+3}{x^2+ 4} = 3\int \frac{2x+1}{x^2+4}dx

= 3\int \frac{2x}{x^2+4}dx +3\int \frac{1}{x^2+4}dx

= 3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}

So, we have the function of x , f(x) =3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}

Now, by Second fundamental theorem of calculus, we have

I = f(2) - f(0)

= \left \{ 3\log(2^2+4)+\frac{3}{2}\tan^{-1}\left ( \frac{2}{2} \right ) \right \}- \left \{ 3\log(0+4)+\frac{3}{2}\tan^{-1}\left ( \frac{0}{2} \right ) \right \} =3\log 8 +\frac{3}{2}\tan^{-1}1 -3\log 4 -\frac{3}{2}\tan^{-1} 0

=3\log 8 +\frac{3}{2}\times\frac{\pi}{4} -3\log 4 -0

=3\log \frac{8}{4} +\frac{3\pi}{8}

or we have =3\log 2 +\frac{3\pi}{8}

Question:20 Evaluate the definite integrals in Exercises 1 to 20.

\int_0^1(xe^x + sin\frac{\pi x}{4})dx

Answer:

Given integral: \int_0^1(xe^x + sin\frac{\pi x}{4})dx

Consider the integral \int (xe^x + sin\frac{\pi x}{4})dx

can be rewritten as: x\int e^x dx - \int \left \{ \left ( \frac{d}{dx}x \right )\int e^x dx \right \}dx +\left \{ \frac{-\cos \frac{\pi x}{4}}{\frac{\pi}{4}} \right \}

= xe^x -\int e^x dx -\frac{4\pi}{\pi} \cos \frac{x}{4}

= xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}

So, we have the function of x , f(x) = xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}

Now, by Second fundamental theorem of calculus, we have

I = f(1) - f(0)

= \left (1.e^t-e^t - \frac{4}{\pi}\cos \frac{\pi}{4} \right ) - \left ( 0.e^0 -e^0 -\frac{4}{\pi}\cos 0 \right )

= e-e -\frac{4}{\pi}\left ( \frac{1}{\sqrt2} \right )+1+\frac{4}{\pi}

Question:21 Choose the correct answer in Exercises 20 and 21.

\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}

(A) \frac{\pi}{3}

(B) \frac{2\pi}{3}

(C) \frac{\pi}{6}

(D) \frac{\pi}{12}

Answer:

Given definite integral \int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}

Consider \int \frac{dx}{1 +x^2} = \tan^{-1}x

we have then the function of x, as f(x) = \tan^{-1}x

By applying the second fundamental theorem of calculus, we will get

\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2} = f(\sqrt3) - f(1)

= \tan^{-1}\sqrt{3} - \tan^{-1}1

=\frac{\pi}{3} - \frac{\pi}{4}

= \frac{\pi}{12}

Therefore the correct answer is D.

Question:22 Choose the correct answer in Exercises 21 and 22.

\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2} equals

(A) \frac{\pi}{6}

(B) \frac{\pi}{12}

(C) \frac{\pi}{24}

(D) \frac{\pi}{4}

Answer:

Given definite integral \int_0^\frac{2}{3}\frac{dx}{4+ 9x^2}

Consider \int \frac{dx}{4+ 9x^2} = \int \frac{dx}{2^2+(3x)^2}

Now, putting 3x = t

we get, 3dx=dt

Therefore we have, \int \frac{dx}{2^2+(3x)^2} = \frac{1}{3}\int \frac{dt}{2^2+t^2}

= \frac{1}{3}\left ( \frac{1}{2}\tan^{-1}\frac{t}{2} \right ) = \frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )

we have the function of x , as f(x) =\frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )

So, by applying the second fundamental theorem of calculus, we get

\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2} = f(\frac{2}{3}) - f(0)

= \frac{1}{6}\tan^{-1}\left ( \frac{3}{2}.\frac{2}{3} \right ) -\frac{1}{6}\tan^{-1}0

= \frac{1}{6}\tan^{-1}1 - 0

= \frac{1}{6}\times \frac{\pi}{4} = \frac{\pi}{24}

Therefore the correct answer is C.

NCERT solutions for class 12 maths chapter 7 Integrals - Exercise:7.10

Question:1 Evaluate the integrals in Exercises 1 to 8 using substitution.

\int_0^1\frac{x}{x^2 +1}dx

Answer:

\int_0^1\frac{x}{x^2 +1}dx
let x^2+1 = t \Rightarrow xdx =dt/2
when x = 0 then t = 1 and when x =1 then t = 2
\therefore \int_{o}^{1}\frac{x}{x^2+1}dx=\frac{1}{2}\int_{1}^{2}\frac{dt}{t}
\\=\frac{1}{2}[\log\left | t \right |]_{1}^{2}\\ =\frac{1}{2}\log 2

Question:2 Evaluate the integrals in Exercises 1 to 8 using substitution.

\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi

Answer:

\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi
let \sin \phi = t \Rightarrow \cos \phi d\phi = dt
when \phi =0,t\rightarrow 0 and \phi =\pi/2,t\rightarrow 1

using the above substitution we can evaluate the integral as

\\\therefore \int_{0}^{1}\sqrt{t}(1-t^2)dt\\ =\int_{0}^{1} t^\frac{1}{2}(1+t^4-2t^2)dt\\ =\int_{0}^{1}t^\frac{1}{2}dt+\int_{0}^{1}t^{9/2}dt-2\int_{0}^{1}t^{5/2}dt\\ =[2t^{3/2}/3+2t^{11/2}/11+4t^{7/2}/7]^1_0\\ =\frac{64}{231}

Question:3 Evaluate the integrals in Exercises 1 to 8 using substitution.

\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx

Answer:

\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx
let
x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta
when x = 0 then \theta= 0 and when x = 1 then \theta= \pi/4

\\=\int_{0}^{\pi/4}\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int_{0}^{\pi/4}2\theta \sec^2\theta d\theta\\
Taking \theta as a first function and \sec^2\theta as a second function, by using by parts method

\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]^{\pi/4}_0\\ =2[\pi/4+\log(1/\sqrt{2})]\\ =\pi/4-\log 2

Question:4 Evaluate the integrals in Exercises 1 to 8 using substitution.

\int_0^2x\sqrt{x+2} . (Put {x+2} = t^2 )

Answer:

Let x+2 = t^2\Rightarrow dx =2tdt
when x = 0 then t = \sqrt{2} and when x=2 then t = 2

I=\int_{0}^{2}x\sqrt{x+2}dx

\\=2\int_{\sqrt{2}}^{2}(t^2-2)t^2dt\\ =2\int_{\sqrt{2}}^{2}(t^4-2t^2)dt\\ =2[t^5/5-\frac{2}{3}t^3]^2_{\sqrt{2}}\\ =2[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}]\\ =\frac{16\sqrt{2}(\sqrt{2}+1)}{15}

Question:5 Evaluate the integrals in Exercises 1 to 8 using substitution.

\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx

Answer:

\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx =I
let \cos x =t\Rightarrow -\sin x dx = dt
when x=0 then t = 1 and when x= \pi/2 then t = 0

\\I=\int_{1}^{0}\frac{dt}{1+t^2}\\ =[\tan ^{-1}t]^0_1\\ =\pi/4

Question:6 Evaluate the integrals in Exercises 1 to 8 using substitution.

\int_0^2\frac{dx}{x + 4 - x^2}

Answer:

By adjusting, the denominator can also be written as (\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2 =x+4-x^2
Now,
\Rightarrow \int_{0}^{2}\frac{dx}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}
let x-1/2 = t\Rightarrow dx=dt
when x= 0 then t =-1/2 and when x =2 then t = 3/2

\\\Rightarrow\int_{-1/2}^{3/2}\frac{dt}{(\frac{\sqrt{17}}{2})^2-t^2}\\ =\frac{1}{2.\frac{\sqrt{17}}{2}}\log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}\\ =\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}/2+3/2}{\sqrt{17}/2-3/2}-\log\frac{\sqrt{17}/2-1/2}{\sqrt{17}/2+1/2}]\\ =\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}+3}{\sqrt{17}-3/}.\frac{\sqrt{17}+1}{\sqrt{17}+1}]
\\ =\frac{1}{\sqrt{17}}[\log (\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}})]\\ =\frac{1}{\sqrt{17}}[\log (\frac{5+\sqrt{17}}{5-\sqrt{17}})]
On rationalisation, we get

=\frac{1}{\sqrt{17}}\log \frac{21+5\sqrt{17}}{4}

Question:7 Evaluate the integrals in Exercises 1 to 8 using substitution.

\int_{-1}^1\frac{dx}{x^2 +2x + 5}

Answer:

\int_{-1}^1\frac{dx}{x^2 +2x + 5}
the Dr can be written as x^2+2x+5 = (x+1)^2+2^2
and put x+1 = t then dx =dt

when x= -1 then t = 0 and when x = 1 then t = 2

\\\Rightarrow \int_{0}^{2}\frac{dt}{t^2+2^2}\\ =\frac{1}{2}[\tan^{-1}\frac{t}{2}]^2_0\\ =\frac{1}{2}( \pi/4)\\ =\frac{\pi}{8}

Question:8 Evaluate the integrals in Exercises 1 to 8 using substitution.

\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx

Answer:

\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx
let 2x =t \Rightarrow 2dx =dt
when x = 1 then t = 2 and when x = 2 then t= 4

\\=\frac{1}{2}\int_{2}^{4}(\frac{2}{t}-\frac{2}{t^2})e^tdt\\
let
\frac{1}{t} = f(t)\Rightarrow f'(t)=-\frac{1}{t^2}
\Rightarrow \int_{2}^{4}(\frac{1}{t}-\frac{1}{t^2})e^tdt =\int_{2}^{}4e^t[f(t)+f'(t)]dt
\\=[e^tf(t)]^4_2\\ =[e^t.\frac{1}{t}]^4_2\\ =\frac{e^4}{4}-\frac{e^2}{2}\\ =\frac{e^2(e^2-2)}{4}

Question:9 Choose the correct answer in Exercises 9 and 10.

The value of the integral \int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx is

(A) 6

(B) 0

(C) 3

(D) 4

Answer:

The value of integral is (A) = 6

\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx
\int_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1)^\frac{1}{3}}{x^3}dx\\
let
\frac{1}{x^2}-1 = t\Rightarrow \frac{dx}{x^3}=-dt/2
now, when x = 1/3, t = 8 and when x = 1 , t = 0

therefore

\\=-\frac{1}{2}\int_{8}^{0}t^{1/3}dt\\ =-\frac{1}{2}.\frac{3}{4}[t^4/3]^0_8\\ =-\frac{3}{8}[-2^4]\\ =6

Question:10 Choose the correct answer in Exercises 9 and 10.

If f(x) = \int_0^x t \sin t dt , then f'(x) is

(A) \cos x + x\sin x

(B) x\sin x

(C) x\cos x

(D) \sin x + x\cos x

Answer:

The correct answer is (B) = x\sin x

f(x) = \int_0^x t \sin t dt
by using by parts method,
\\=t\int \sin t dt - \int (\frac{d}{dt}t\int \sin t dt)dt\\ =[t(-\cos t )+\sin t]^x_0

f(x)= -x\cos x+sinx
So, f'(x)= -\cos x+x\sin x+\cos x\\ =x\sin x

NCERT solutions for class 12 maths chapter 7 Integrals - Exercise:7.11

Question:1 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\frac{\pi}{2}\cos^2 x dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{2}\cos^2 x dx ............................................................. (i)

By using

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get :-

I\ =\ \int_0^\frac{\pi}{2}\cos^2 x dx\ =\ \int_0^\frac{\pi}{2}\cos^2\ (\frac{\pi}{2}- x) dx

or

I\ =\ \int_0^\frac{\pi}{2}\sin^2 x dx ................................................................ (ii)

Adding both (i) and (ii), we get :-

\int_0^\frac{\pi}{2}\cos^2 x dx +\ \int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2I

or \int_0^\frac{\pi}{2}\ (cos^2 x\ +\ sin^2 x) dx\ =\ 2I

or \int_0^\frac{\pi}{2}1. dx\ =\ 2I

or 2I\ =\ \left [ x \right ] ^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}

or I\ =\ \frac{\Pi }{4}

Question:2 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx .......................................................................... (i)

By using ,

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin (\frac{\pi}{2}-x)}}{\sqrt{\sin (\frac{\pi}{2}-x)}+ \sqrt{\cos (\frac{\pi}{2}-x)}}dx


or I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx .......................................................(ii)

Adding (i) and (ii), we get,

2I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}\ +\ \sqrt{\cos x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx

or 2I\ =\ \int_0^\frac{\pi}{2}1.dx


or 2I\ =\ \left [ x \right ]^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}

Thus I\ =\ \frac{\Pi }{4}

Question:€‹3 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}

Answer:

We have I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x} ..................................................................(i)

By using :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}(\frac{\pi}{2}-x)dx}{\sin^\frac{3}{2}(\frac{\pi}{2}-x) + \cos^{\frac{3}{2}}(\frac{\pi}{2}-x)}


or I\ =\ \int^{\frac{\pi}{2}}_0\frac{\cos^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x} . ............................................................(ii)

Adding (i) and (ii), we get :

2I\ =\ \int^{\frac{\pi}{2}}_0\frac{\ (sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x)dx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}

or 2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx

or 2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}

Thus I\ =\ {\frac{\pi}{4}}

Question:4 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}

Answer:

We have I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x} ..................................................................(i)

By using :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 (\frac{\pi}{2}-x)dx}{\sin^5(\frac{\pi}{2}-x) + \cos^5(\frac{\pi}{2}-x)}

or I\ =\ \int_0^\frac{\pi}{2} \frac{\sin^5 xdx}{\sin^5x + \cos^5x} . ............................................................(ii)

Adding (i) and (ii), we get :

2I\ =\ \int_0^\frac{\pi}{2} \frac{\ ( sin^5x \ +\ cos^5 x)dx}{\sin^5x + \cos^5x}

or 2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx

or 2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}

Thus I\ =\ {\frac{\pi}{4}}

Question:5 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_{-5}^5|x+2|dx

Answer:

We have, I\ =\ \int_{-5}^5|x+2|dx

For opening the modulas we need to define the bracket :

If (x + 2) < 0 then x belongs to (-5, -2). And if (x + 2) > 0 then x belongs to (-2, 5).

So the integral becomes :-

I\ =\ \int_{-5}^{-2} -(x+2)dx\ +\ \int_{-2}^{5} (x+2)dx

or I\ =\ -\left [ \frac{x^2}{2}\ +\ 2x \right ]^{-2} _{-5}\ +\ \left [ \frac{x^2}{2}\ +\ 2x \right ]^{5} _{-2}

This gives I\ =\ 29

Question:6€‹ By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_2^8|x-5|dx

Answer:

We have, I\ =\ \int_{2}^8|x-5|dx

For opening the modulas we need to define the bracket :

If (x - 5) < 0 then x belongs to (2, 5). And if (x - 5) > 0 then x belongs to (5, 8).

So the integral becomes:-

I\ =\ \int_{2}^{5} -(x-5)dx\ +\ \int_{5}^{8} (x-5)dx

or I\ =\ -\left [ \frac{x^2}{2}\ -\ 5x \right ]^{5} _{2}\ +\ \left [ \frac{x^2}{2}\ -\ 5x \right ]^{8} _{5}

This gives I\ =\ 9

Question:7 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int^1_0x(1-x)^ndx

Answer:

We have I\ =\ \int^1_0x(1-x)^ndx

U sing the property : -

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get : -

I\ =\ \int^1_0x(1-x)^ndx\ =\ \int^1_0(1-x)(1-(1-x))^ndx

or I\ =\ \int^1_0(1-x)x^n\ dx

or I\ =\ \int^1_0(x^n\ -\ x^{n+1}) \ dx

or =\ \left [ \frac{x^{n+1}}{n+1}\ -\ \frac{x^{n+2}}{n+2} \right ]^1_0

or =\ \left [ \frac{1}{n+1}\ -\ \frac{1}{n+2} \right ]

or I\ =\ \frac{1}{(n+1)(n+2)}

Question:8 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\frac{\pi}{4}\log(1+\tan x)dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx

By using the identity

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx\ =\ \int_0^\frac{\pi}{4}\log(1+\tan (\frac{\pi}{4}-x))dx

or I\ =\ \int_0^\frac{\pi}{4}\log(1+\frac{1-\tan x}{1+\tan x})dx

or I\ =\ \int_0^\frac{\pi}{4}\log(\frac{2}{1+\tan x})dx

or I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ \int_0^\frac{\pi}{4}\log(1+ \tan x)dx

or I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ I

or 2I\ =\ \left [ x\log2 \right ]^{\frac{\Pi }{4}}_0

or I\ =\ \frac{\Pi }{8}\log2

Question:9 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^2x\sqrt{2-x}dx

Answer:

We have I\ =\ \int_0^2x\sqrt{2-x}dx

By using the identity

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get :

I\ =\ \int_0^2x\sqrt{2-x}dx\ =\ \int_0^2(2-x)\sqrt{2-(2-x)}dx

or I\ =\ \int_0^2(2-x)\sqrt{x}dx

or I\ =\ \int_0^2(2\sqrt{x}\ -\ x^\frac{3}{2} dx

or =\ \left [ \frac{4}{3}x^\frac{3}{2}\ -\ \frac{2}{5}x^\frac{5}{2} \right ]^2_0

or =\ \frac{4}{3}(2)^\frac{3}{2}\ -\ \frac{2}{5}(2)^\frac{5}{2}

or I\ =\ \frac{16\sqrt{2}}{15}

Question:10 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx

or I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log(2\sin x\cos x))dx

or I\ =\ \int_0^\frac{\pi}{2} (\log\sin x- \log\cos x\ -\ \log2)dx ..............................................................(i)

By using the identity :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get :

I\ =\ \int_0^\frac{\pi}{2} (\log\sin (\frac{\pi}{2}-x)- \log\cos (\frac{\pi}{2}-x)\ -\ \log2)dx

or I\ =\ \int_0^\frac{\pi}{2} (\log\cos x- \log\sin x\ -\ \log2)dx ....................................................................(ii)

Adding (i) and (ii) we get :-

2I\ =\ \int_0^\frac{\pi}{2} (- \log 2 -\ \log 2)dx

or I\ =\ -\log 2\left [ \frac{\Pi }{2} \right ]

or I\ =\ \frac{\Pi }{2}\log\frac{1}{2}

Question:11 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx

Answer:

We have I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx

We know that sin 2 x is an even function. i.e., sin 2 (-x) = (-sinx) 2 = sin 2 x.

Also,

I\ =\ \int_{-a}^af(x) dx\ =\ 2\int_{0}^af(x) dx

So,

I\ =\ 2\int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2\int_0^\frac{\pi}{2}\frac{(1-\cos2x)}{2} dx

or =\ \left [ x\ -\ \frac{\sin2x}{2} \right ]^{\frac{\Pi }{2}}_0

or I\ =\ \frac{\Pi }{2}

Question:12 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\pi\frac{xdx}{1+\sin x}

Answer:

We have I\ =\ \int_0^\pi\frac{xdx}{1+\sin x} ..........................................................................(i)

By using the identity :-

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin (\Pi -x)}

or I\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin x} ............................................................................(ii)


Adding both (i) and (ii) we get,

2I\ =\ \int_0^\pi\frac{\Pi}{1+\sin x} dx

or 2I\ =\ \Pi \int_0^\pi\frac{1-\sin x}{(1+\sin x)(1-\sin x)} dx\ =\ \Pi \int_0^\pi\frac{1-\sin x}{\cos^2 x} dx

or 2I\ =\ \Pi \int_0^\pi (\sec^2\ -\ \tan x \sec x) x dx

or I\ =\ \Pi

Question:13 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx

Answer:

We have I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx

We know that \sin^7x is an odd function.

So the following property holds here:-

\int_{-a}^{a}f(x)dx\ =\ 0

Hence

I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx\ =\ 0

Question:14 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^{2\pi}\cos^5xdx

Answer:

We have I\ =\ \int_0^{2\pi}\cos^5xdx

I t is known that :-

\int_0^{2a}f(x)dx\ =\ 2\int_0^{a}f(x)dx If f (2a - x) = f(x)

=\ 0 If f (2a - x) = - f(x)

Now, using the above property

\cos^5(\Pi - x)\ =\ - \cos^5x

Therefore, I\ =\ 0

Question:15 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx

Answer:

We have I\ =\ \int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx ................................................................(i)

By using the property :-

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get ,

I\ =\ \int^\frac{\pi}{2} _0\frac{\sin (\frac{\pi}{2}-x) - \cos (\frac{\pi}{2}-x) }{1+\sin (\frac{\pi}{2}-x)\cos (\frac{\pi}{2}-x)}dx

or I\ =\ \int^\frac{\pi}{2} _0\frac{\cos x - \sin x }{1+\sin x\cos x}dx ......................................................................(ii)

Adding both (i) and (ii), we get


2I\ =\ \int^\frac{\pi}{2} _0\frac{0 }{1+\sin x\cos x}dx

Thus I = 0

Question:16 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\pi\log(1 +\cos x)dx

Answer:

We have I\ =\ \int_0^\pi\log(1 +\tan x)dx .....................................................................................(i)

By using the property:-

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

or

I\ =\ \int_0^\pi\log(1 +\cos (\Pi -x))dx

I\ =\ \int_0^\pi\log(1 -\cos x)dx ....................................................................(ii)


Adding both (i) and (ii) we get,

2I\ =\ \int_0^\pi\log(1 +\cos x)dx\ +\ \int_0^\pi\log(1 -\cos x)dx

or 2I\ =\ \int_0^\pi\log(1 -\cos^2 x)dx\ =\ \int_0^\pi\log \sin^2 xdx

or 2I\ =\ 2\int_0^\pi\log \sin xdx

or I\ =\ \int_0^\pi\log \sin xdx ........................................................................(iii)

or I\ =\ 2\int_0^ \frac{\pi}{2} \log \sin xdx ........................................................................(iv)

or I\ =\ 2\int_0^ \frac{\pi}{2} \log \cos xdx .....................................................................(v)

Adding (iv) and (v) we get,

I\ =\ -\pi \log2

Question:17 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx

Answer:

We have I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx ................................................................................(i)

By using, we get


\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx\ =\ \int_0^a \frac{\sqrt {(a-x)}}{\sqrt {(a-x)} + \sqrt{x}}dx .................................................................(ii)


Adding (i) and (ii) we get :

2I\ =\ \int_0^a \frac{\sqrt x\ +\ \sqrt{a-x}}{\sqrt x + \sqrt{a-x}}dx

or 2I\ =\ \left [ x \right ]^a_0 = a

or I\ =\ \frac{a}{2}

Question:18 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^4 |x-1|dx

Answer:

We have, I\ =\ \int_{0}^4|x-1|dx

For opening the modulas we need to define the bracket :

If (x - 1) < 0 then x belongs to (0, 1). And if (x - 1) > 0 then x belongs to (1, 4).

So the integral becomes:-

I\ =\ \int_{0}^{1} -(x-1)dx\ +\ \int_{1}^{4} (x-1)dx

or I\ =\ \left [ x\ -\ \frac{x^2}{2}\ \right ]^{1} _{0}\ +\ \left [ \frac{x^2}{2}\ -\ x \right ]^{4} _{1}

This gives I\ =\ 5

Question:19 Show that \int_0^a f(x)g(x)dx = 2\int_0^af(x)dx if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4

Answer:

Let I\ =\ \int_0^a f(x)g(x)dx ........................................................(i)

This can also be written as :

I\ =\ \int_0^a f(a-x)g(a-x)dx

or I\ =\ \int_0^a f(x)g(a-x)dx ................................................................(ii)

Adding (i) and (ii), we get,

2I\ =\ \int_0^a f(x)g(a-x)dx +\ \int_0^a f(x)g(x)dx

2I\ =\ \int_0^a f(x)4dx

or I\ =\ 2\int_0^a f(x)dx

Question:20 Choose the correct answer in Exercises 20 and 21.

The value of is \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx is

(A) 0

(B) 2

(C) \pi

(D) 1

Answer:

We have

I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx

This can be written as :

I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}x^3dx +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} x\cos x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} \tan^5 x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} 1dx

Also if a function is even function then \int_{-a}^{a}f(x)\ dx\ =\ 2\int_{0}^{a}f(x)\ dx

And if the function is an odd function then : \int_{-a}^{a}f(x)\ dx\ =\ 0

Using the above property I become:-

I\ =\ 0+0+0+ 2\int_{0}^{\frac{\Pi }{2}}1.dx

or I\ =\ 2\left [ x \right ]^{\frac{\Pi }{2}}_0

or I\ =\ \Pi

Question:€‹21 Choose the correct answer in Exercises 20 and 21.

The value of \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx is

Answer:

We have

I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx .................................................................................(i)


By using :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin (\frac{\pi}{2}-x)}{4+3\cos (\frac{\pi}{2}-x)} \right )dx

or I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx .............................................................................(ii)

Adding (i) and (ii), we get:

2I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ +\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx

or 2I\ =\ \int_0^\frac{\pi}{2}\log1.dx

Thus I\ =\ 0


NCERT solutions for class 12 maths chapter 7 Integrals-Miscellaneous Exercise

Question:1 Integrate the functions in Exercises 1 to 24.

\frac{1}{x - x^3}

Answer:

Firstly we will simplify the given equation :-

\frac{1}{x - x^3}\ =\ \frac{1}{(x)(1-x)(1+x)}

Let

\frac{1}{(x)(1-x)(1+x)} =\ \frac{A}{x}\ +\ \frac{B}{1-x}\ +\ \frac{C}{1+x}

By solving the equation and equating the coefficients of x 2 , x and the constant term, we get

A\ =\ 1,\ B\ =\ \frac{1}{2},\ C\ =\ \frac{-1}{2}

Thus the integral can be written as :

\int \frac{1}{(x)(1-x)(1+x)}dx =\ \int \frac{1}{x}dx\ +\ \frac{1}{2}\int \frac{1}{1-x}dx\ +\ \frac{-1}{2}\int \frac{1}{1+x}dx

=\ \log x\ -\ \frac{1}{2}\log(1-x)\ +\ \frac{-1}{2}\log (1+x)

or =\ \frac{1}{2} \log \frac{x^2}{1-x^2}\ +\ C


Question:2 Integrate the functions in Exercises 1 to 24.

\frac{1}{\sqrt{x+a} + \sqrt{x+b}}

Answer:

At first we will simplify the given expression,

\frac{1}{\sqrt{x+a} + \sqrt{x+b}}\ =\ \frac{1}{\sqrt{x+a} + \sqrt{x+b}}\times\frac{\sqrt{x+a} - \sqrt{x+b}}{\sqrt{x+a} - \sqrt{x+b}}

or =\ \frac{\sqrt{x+a} - \sqrt{x+b}}{a-b}

Now taking its integral we get,

\int \frac{1}{\sqrt{x+a} + \sqrt{x+b}}\ =\ \frac{1}{a-b}\int (\sqrt{x+a} -\sqrt{x+b})dx

or =\ \frac{1}{a-b}\left [ \frac{(x+a)^{\frac{3}{2}}} {\frac{3}{2}}\ -\ \frac{(x+b)^{\frac{3}{2}}} {\frac{3}{2}} \right ]

or =\ \frac{2}{3(a-b)}\left [ (x+a)^{\frac{3}{2}}\ -\ (x+b)^{\frac{3}{2}} \right ]\ +\ C


Question:3€‹ Integrate the functions in Exercises 1 to 24.

\frac{1}{x\sqrt{ax-x^2}} [Hint: Put x = \frac{a}{t} ]

Answer:

Let

x = \frac{a}{t}\ dx\ \Rightarrow \ dx\ =\ \frac{-a}{t^2}dh

Using the above substitution we can write the integral is

\int \frac{1}{x\sqrt{ax-x^2}}\ =\ \int \frac{1}{\frac{a}{t}\sqrt{a.\frac{a}{t}\ -\ (\frac{a}{t})^2}} \frac{-a}{t^2}dt

or

=\ \frac{-1}{a}\int \frac{1}{\sqrt{(t-1)}}dt

or

=\ \frac{-1}{a}\ (2\sqrt{t-1})\ +\ C

or =\ \frac{-1}{a}\ (2\sqrt{\frac{a}{x}\ -\ 1})\ +\ C

or =\ \frac{-2}{a}\ \sqrt{\frac{a-x}{x}}\ +\ C

Question:4 Integrate the functions in Exercises 1 to 24.

\frac{1}{x^2(x^4 + 1)^\frac{3}{4}}

Answer:

For the simplifying the expression, we will multiply and dividing it by x -3 .

We then have,

\frac{x^{-3}}{x^2 x^{-3}(x^4 + 1)^\frac{3}{4}}\ =\ \frac{1}{x^5}\left [ \frac{x^4\ +\ 1}{x^4} \right ]^{\frac{-3}{4}}

Now, let

\frac{1}{x^4}\ =\ t\ \Rightarrow \ \frac{1}{x^5}dx\ =\ \frac{-dt}{4}

Thus,

\int \frac{1}{x^2(x^4 + 1)^\frac{3}{4}}\ =\ \int \frac{1}{x^5}\left ( 1+\ \frac{1}{x^4}^{\frac{-3}{4}}\ \right )dx

or =\ \frac{-1}{4} \int (1+t)^{\frac{-3}{4}}dt

=\ \frac{-1}{4} \frac{(1+\frac{1}{x^4})^{\frac{1}{4}}}{\frac{1}{4}}\ +\ C

=\ - \left [ 1+\frac{1}{x^4} \right ]^{\frac{1}{4}}\ +\ C

Question:5 Integrate the functions in Exercises 1 to 24.

\frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}} [Hint: \frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}} = \frac{1}{x^\frac{1}{3}(1 + \ x^\frac{1}{6})} , put x = t^6 ]

Answer:

Put x = t^6\ \Rightarrow \ dx = 6t^5dt

We get,

\int \frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}}dx\ =\ \int \frac{6t^5}{t^3+t^2}dt

or =\ 6\int \frac{t^3}{1+t}dt

or =\ 6\int \left \{ (t^2-t+1)-\frac{1}{1+t} \right \}dt

or =\ 6 \left [ \left ( \frac{t^3}{3} \right ) -\left ( \frac{t^2}{2} \right )+t - \log(1+t) \right ]

Now put x = t^6 in the above result :

=\ 2\sqrt{x} -3x^{\frac{1}{3}}+ 6x^{\frac{1}{6}} - 6 \log \left ( 1-x^\frac{1}{6} \right )\ +\ C

Question:6 Integrate the functions in Exercises 1 to 24.

\frac{5x}{(x+1)(x^2 + 9)}

Answer:

Let us assume that :

\frac{5x}{(x+1)(x^2 + 9)}\ =\ \frac{A}{(x+1)}\ +\ \frac{Bx + c}{x^2 + 9}

Solving the equation and comparing coefficients of x 2 , x and the constant term.

We get,

A\ =\ \frac{-1}{2}\ ;\ B\ =\ \frac{1}{2}\ ;\ C\ =\ \frac{9}{2}

Thus the equation becomes :

\frac{5x}{(x+1)(x^2 + 9)}\ =\ \frac{-1}{2(x+1)}\ +\ \frac{\frac{x}{2}+\frac{9}{2}}{x^2 + 9}

or

\int \frac{5x}{(x+1)(x^2 + 9)}\ =\ \int \left [ \frac{-1}{2(x+1)}\ +\ \frac{x+9}{2(x^2 + 9}) \right ]dx

or =\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{2} \int \frac{x}{x^2 +9}dx +\frac{9}{2} \int \frac{1}{x^2+9}dx

or =\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{4} \int \frac{2x}{x^2 +9}dx +\frac{9}{2} \int \frac{1}{x^2+9}dx

or =\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{4} \log {(x^2 +9)} +\frac{3}{2} \tan^{-1}\frac{x}{3}\ +\ C

Question:7 Integrate the functions in Exercises 1 to 24.

\frac{\sin x}{\sin (x-a)}

Answer:

We have,

I\ =\ \frac{\sin x}{\sin (x-a)}

Assume :- (x-a)\ =\ t \Rightarrow \ dx=dt

Putting this in above integral :

\int \frac{\sin x}{\sin (x-a)}dx\ =\ \int \frac{\sin (t+a)}{\sin t}dt

or =\ \int \frac{\sin t \cos a\ +\ \cos t \sin a }{\sin t}dt

or =\ \int (\cos a\ +\ \cot t \sin a)dt

or =\ t\cos a\ +\ \sin a \log |\sin t|\ +\ C

or =\ \sin a \log \left | \sin(x-a) \right | + x\cos a\ +\ C

Question:€‹9 Integrate the functions in Exercises 1 to 24.

\frac{\cos x}{\sqrt{4 - \sin^2 x}}

Answer:

We have the given integral

I\ =\ \frac{\cos x}{\sqrt{4 - \sin^2 x}}

Assume \sin x = t\ \Rightarrow \cos x dx = dt

So, this substitution gives,

\int \frac{\cos x}{\sqrt{4 - \sin^2 x}}\ =\ \int \frac{dt}{\sqrt{(2)^2 - (t)^2}}

=\ \sin^{-1}\frac{t}{2}\ +\ C

or =\ \sin^{-1}\left ( \frac{\sin x}{2} \right )\ +\ C

Question:10 Integrate the functions in Exercises 1 to 24.

\frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}

Answer:

We have

I\ =\ \int \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}

Simplifying the given expression, we get :

\frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}\ =\ \frac{(\sin^4x + \cos^4x)(\sin^4x - \cos^4x) }{1- 2\sin^ x\cos^2 x}

or =\ \frac{(\sin^4x + \cos^4x)(\sin^2x - \cos^2x)(\sin^2x + \cos^2x) }{1- 2\sin^ x\cos^2 x}

or =\ -\frac{(\sin^4x + \cos^4x)(\cos^2x - \sin^2x) }{1- 2\sin^ x\cos^2 x}

or =\ -\cos^2x - \sin^2x\ =\ -\cos 2x

Thus,

I\ =\ \int \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}\ =\ -\int \cos 2x\ dx

and =\ - \frac{\sin 2x}{2}\ +\ C

Question:11 Integrate the functions in Exercises 1 to 24.

\frac{1}{\cos(x+a)\cos(x+b)}

Answer:

For simplifying the given equation, we need to multiply and divide the expression by \sin (a-b) .

Thus we obtain :

\frac{1}{\cos(x+a)\cos(x+b)}\ =\ \frac{1}{\sin(a-b)}\times\frac{\sin (a-b)}{\cos(x+a)\cos(x+b)}

or = \frac{1}{ \sin (a-b)}\times \frac{\sin{\left [ (x+a) - (x+b) \right ]}}{\cos (x+a) \cos (x+b)}

or = \frac{1}{ \sin (a-b)}\times \left ( \frac{\sin (x+a) }{\cos (x+a) } - \frac{\sin(x+b)}{\cos (x+b)} \right )

or = \frac{1}{ \sin (a-b)}\times \left ( \tan(x+a)\ -\ \tan(x+b) \right )

Thus integral becomes :

\int \frac{1}{\cos(x+a)\cos(x+b)}\ =\ \frac{1}{\sin (a-b)} \times \int \left ( \tan(x+a)\ -\ \tan(x+b) \right )dx

or =\ \frac{1}{\sin (a-b)} \times \left [ -\log \left | \cos (x+a) \right | + \log \left | \cos(x+b) \right | \right ]\ +\ C

or =\ \frac{1}{\sin (a-b)} \times \log \left [ \frac{\cos(x+b) }{cos(x+a)} \right ]\ +\ C

Question:12€‹ Integrate the functions in Exercises 1 to 24.

\frac{x^3}{\sqrt{1-x^8}}

Answer:

Given that to integrate

\frac{x^3}{\sqrt{1-x^8}}

Let x^4 = t \implies 4x^3dx = dt

\therefore \int \frac{x^3}{\sqrt{1-x^8}}dx = \frac{1}{4}\int\frac{1}{\sqrt {1-t^2}}dt

= \frac{1}{4}sin^{-1}t + C= \frac{1}{4}sin^{-1}{x^4} + C

the required solution is \frac{1}{4}sin^{-1}{(x^4)} + C

Question:13 Integrate the functions in Exercises 1 to 24.

\frac{e^x}{(1 + e^x)(2 + e^x)}

Answer:

we have to integrate the following function

\frac{e^x}{(1 + e^x)(2 + e^x)}

Let 1+e^x = t \implies e^xdx = dt

using this we can write the integral as

\therefore \int\frac{e^x}{(1 + e^x)(2 + e^x)}dx = \int\frac{1}{t(1+t)}dt = \int\frac{(1+t)-t}{t(1+t)}dt

\\ = \int\left ( \frac{1}{t}-\frac{1}{t+1} \right )dt

\\ = \int\frac{1}{t}dt - \int\frac{1}{t+1}dt

\\ = \log t - \log (1+t) + C \\ = \log (1+e^x) - \log (2+e^x) + C \\ = \log\left ( \frac{e^x + 1}{e^x + 2} \right ) + C

Question:14 Integrate the functions in Exercises 1 to 24.

\frac{1}{(x^2 + 1)(x^2 +4)}

Answer:

Given,

\frac{1}{(x^2 + 1)(x^2 +4)}

Let I = \int\frac{1}{(x^2 + 1)(x^2 +4)}

Now, Using partial differentiation,

\frac{1}{(x^2 + 1)(x^2 +4)} = \frac{Ax + B}{(x^2 + 1)} + \frac{Cx +D}{(x^2 +4)}

\implies \frac{1}{(x^2 + 1)(x^2 +4)} = \frac{(Ax + B)(x^2 +4) + (Cx +D)(x^2 + 1)}{(x^2 + 1)(x^2 +4)}
\\ \implies1 = (Ax + B)(x^2 + 4)+(Cx + D)(x^2 + 1) \\ \implies 1 = Ax^3 +4Ax+ Bx^2 + 4B+ Cx^3 + Cx + Dx^2 + D \\ \implies (A+C)x^3 +(B+D)x^2 +(4A+C)x + (4B+D) = 1

Equating the coefficients of x, x^2, x^3 and constant value,

A + C = 0 \implies C = -A

B + D = 0 \implies B = -D

4A + C =0 \implies 4A = -C \implies 4A = A \implies A = 0 = C

4B + D = 1 \implies 4B – B = 1 \implies B = 1/3 = -D

Putting these values in equation, we have

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155444707593450

1554447076674147

1554447078146237

\implies I = \frac{1}{3}tan^{-1}x - \frac{1}{6}tan^{-1}\frac{x}{2} + C

Question:15 Integrate the functions in Exercises 1 to 24.

\cos^3 x \;e^{\log\sin x}

Answer:

Given,

\cos^3 x \;e^{\log\sin x}

I = \int \cos^3 x \;e^{\log\sin x} (let)

Let cos x = t \implies -sin x dx = dt \implies sin x dx = -dt

using the above substitution the integral is written as

\therefore \int cos^3xe^{\log sinx}dx = \int cos^3x.sinx dx

155444708344321

1554447084197794

1554447084982414

155444708573581

I = -\frac{cos^4x}{4} + C

Question:16Integrate the functions in Exercises 1 to 24.

e^{3\log x} (x^4 + 1)^{-1}

Answer:

Given the function to be integrated as

e^{3\log x} (x^4 + 1)^{-1}
= e^{\log x^3}(x^4 + 1)^{-1} = \frac{x^3}{x^4 + 1}

Let I = \int e^{3\log x} (x^4 + 1)^{-1}

Let x^4 = t \implies 4x^3 dx = dt

I = \int e^{3\log x} (x^4 + 1)^{-1} = \int \frac{x^3}{x^4 + 1}

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1554447092538842

\implies I = \frac{1}{4}\log(x^4 +1) + C

Question:€‹17 Integrate the functions in Exercises 1 to 24.

f'(ax +b)[f(ax +b)]^n

Answer:

Given,

f'(ax +b)[f(ax +b)]^n

Let I = \int f'(ax +b)[f(ax +b)]^n

Let f(ax +b) = t ⇒ a .f ' (ax + b)dx = dt

Now we can write the ntegral as

\int f'(ax +b)[f(ax +b)]^n = \frac{1}{a}\int t^ndt

\\ = \frac{1}{a}.\frac{t^{n+1}}{n+1} + C \\ = \frac{1}{a}.\frac{(f(ax+b))^{n+1}}{n+1} + C

\implies I = \frac{(f(ax+b))^{n+1}}{a(n+1)} + C


Question:18 Integrate the functions in Exercises 1 to 24.

\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}

Answer:

Given,

\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}

Let I = \int \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}

We know the identity that

sin (A+B) = sin A cos B + cos A sin B

\therefore \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}} = \frac{1}{\sqrt{\sin^3 x (\sin x \cos \alpha + \cos x \sin \alpha)}}

= \frac{1}{\sqrt{\sin^3 x . \sin x(\cos \alpha + \cot x \sin \alpha)}} = \frac{1}{\sqrt{\sin^4 x (\cos \alpha + \cot x \sin \alpha)}}

\frac{cosec^2 x}{\sqrt{(\cos \alpha + \cot x \sin \alpha)}}

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1554447106158144


1554447106897484


1554447107634823


1554447108399148


155444710916564


1554447109907513


1554447110653730


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1554447112159711


1554447112971688


Question:19 Integrate the functions in Exercises 1 to 24.

. \frac{\sin^{-1}\sqrt x - \cos^{-1}\sqrt x}{\sin^{-1}\sqrt x + \cos^{-1}\sqrt x}, \;\;\; x\in [0,1]

Answer:

We have

I\ =\ \int \frac{\sin^{-1}\sqrt x - \cos^{-1}\sqrt x}{\sin^{-1}\sqrt x + \cos^{-1}\sqrt x}\ dx

or =\ \int \frac{\sin^{-1}\sqrt x - \left ( \frac{\Pi }{2} - \sin^{-1}\sqrt x \right )}{\frac{\Pi }{2}}\ dx

or =\ \frac{2}{\Pi } \int \left ( \ 2sin^{-1}\sqrt x - \frac{\Pi }{2} \right )\ dx

or =\ \int \left (\frac{4}{\Pi } \sin^{-1}\sqrt x - 1 \right )\ dx

or =\ \frac{4}{\Pi }\int \sin^{-1}\sqrt x - 1 \ dx\ -\ \int 1 \ dx\ +\ C

or =\ \frac{4}{\Pi }\int \sin^{-1}\sqrt x \ dx\ -\ x +\ C

Thus I\ =\ \frac{4}{\Pi }I'\ -\ x +\ C


Now we will solve I'.

I'\ =\ \int \sin^{-1}\sqrt x \ dx

Put x = t 2 .

Differentiating the equation wrt x, we get

dx\ =\ 2t\ dt

Thus \int \sin^{-1}\sqrt x \ dx\ =\ \int \sin^{-1} t\ 2t \ dt

or =\ 2 \int t\ \sin^{-1} t\ \ dt

Using integration by parts, we get :

=\ 2 \left [ \sin^{-1}t \int t\ dt\ -\ \int \left ( \left ( \frac{d}{dt} \sin^{-1} t \right ) \int t\ dt \right ) \right ]\ dt

or =\ t^2 \sin^{-1}t\ -\ \int \frac{t^2}{\sqrt{1-t^2}}\ dt\ +\ C'

We know that

\int \frac{- t^2}{\sqrt{1-t^2}}\ dt\ =\ \frac{t}{2}\sqrt{1-t^2}\ -\ \frac{1}{2}\ \sin^{-1}t

Thus it becomes :

I'\ =\ t^2\sin^{-1} t\ +\ \frac{t}{2}\sqrt{1-t^2}\ -\ \frac{1}{2}\ \sin^{-1}t

So I come to be :-

I\ =\ \frac{4}{\Pi }I'\ -\ x +\ C

I\ =\ \sin^{-1}\sqrt{x} \left [ \frac{2(2x-1)}{\Pi } \right ]\ +\ \frac{2\sqrt{x-x^2}}{\Pi }\ -\ x\ +\ C

Question:€‹20 Integrate the functions in Exercises 1 to 24.

\sqrt{\frac{1-\sqrt x}{1 +\sqrt x}}

Answer:

Given,

\sqrt{\frac{1-\sqrt x}{1 +\sqrt x}} = I (let)

Let x= cos^2\theta \implies dx = -2sin\theta cos\theta d\theta

And \sqrt x= cos\theta \implies \theta = \cos^{-1}\sqrt x

using the above substitution we can write the integral as

\\ I = \int \sqrt{\frac{1-\sqrt {cos^2\theta}}{1 +\sqrt {cos^2\theta}}}(-2\sin\theta\cos\theta)d\theta \\ = -\int \sqrt{\frac{1-cos\theta}{1 +cos\theta}}(2\sin\theta\cos\theta)d\theta

\\ = -\int \sqrt{tan^2\frac{\theta}{2}}(2\sin\theta\cos\theta)d\theta \\ = -\int \sqrt{tan^2\frac{\theta}{2}}(2. 2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}\cos\theta)d\theta \\ = -4\int \sin^2\frac{\theta}{2}\cos\theta d\theta

\\ = -4\int \sin^2\frac{\theta}{2}(2cos^2\frac{\theta}{2} -1) d\theta

1554447146338657

1554447147081166

1554447147826729

1554447148562798

1554447149316945

1554447150058843

1554447150798562

1554447151538561

1554447152324930

1554447153875620

Question:21 Integrate the functions in Exercises 1 to 24.

\frac{2 + \sin 2x}{1 + \cos 2x}e^x

Answer:

Given to evaluate

\frac{2 + \sin 2x}{1 + \cos 2x}e^x

\frac{2 + \sin 2x}{1 + \cos 2x}e^x

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1554447156855423

1554447157709943

1554447158484136

1554447159245239

now the integral becomes

1554447160001344

Let tan x = f(x)

\implies f'(x) = sec^2x dx

1554447160771541

1554447161556589

1554447162309720

Question:22 Integrate the functions in Exercises 1 to 24.

\frac{x^2 + x + 1}{(x+1)^2 (x+2)}

Answer:

Given,

\frac{x^2 + x + 1}{(x+1)^2 (x+2)}

using partial fraction we can simplify the integral as

Let \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+2}

\\ \implies \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A(x+1)(x+2) + B(x+2) + C(x+1)^2}{(x+1)^2 (x+2)} \\ \implies \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{A(x^2 + 3x+2) + B(x+2) + C(x^2 + 2x+1)}{(x+1)^2 (x+2)}

\\ \implies x^2 + x + 1 = A(x^2 + 3x+2) + B(x+2) + C(x^2 + 2x+1) \\ = (A+C)x^2 + (3A+B+2C)x + (2A+2B+C)

Equating the coefficients of x, x 2 and constant value, we get:

A + C = 1

3A + B + 2C = 1

2A+2B+C =1

Solving these:

A= -2, B=1 and C=3

\implies \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{-2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}

\\ \implies \int \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \int\frac{-2}{x+1}dx+\int\frac{1}{(x+1)^2}dx+\int\frac{3}{x+2}dx \\ = -2\log(x+1) - \frac{1}{(x+1)} + 3\log (x+2) + C

Question:23 Integrate the functions in Exercises 1 to 24.

\tan^{-1}\sqrt{\frac{1-x}{1+x}}

Answer:

We have

I\ =\ \int \tan^{-1}\sqrt{\frac{1-x}{1+x}}

Let us assume that : x\ =\ \cos 2\Theta

Differentiating wrt x,

dx\ =\ -2 \sin 2\Theta\ d\Theta

Substituting this in the original equation, we get

\int \tan^{-1}\sqrt{\frac{1-x}{1+x}}\ =\ \int \tan^{-1}\sqrt{\frac{1-cos2\Theta }{1+cos2\Theta }}\times -2\sin 2\Theta \ d\Theta

or =\ -2\int \tan^{-1} (\frac{sin\Theta }{cos\Theta })\times \sin 2\Theta \ d\Theta

or =\ -2\int \Theta \sin 2\Theta \ d\Theta

Using integration by parts , we get

=\ -2\left ( \Theta \int \sin 2\Theta \ d\Theta\ - \int \frac{d\Theta }{d\Theta } \int \sin 2\Theta \ d\Theta\ \right )

or =\ -2\left ( \Theta \left ( \frac{-\cos 2\Theta }{2} \right ) - \int 1.\frac{-\cos 2\Theta }{2} \ d\Theta\ \right )

or =\ -2\left ( \frac{-\Theta \cos 2\Theta }{2}+ \frac{\sin 2\Theta }{4} \right )

Putting all the assumed values back in the expression,

=\ -2\left ( -\frac{1}{2}\left ( \frac{1}{2} \cos^{-1} x \right )+ \frac{\sqrt{1-x^2} }{4} \right )

or =\ \frac{1}{2}\left ( x \cos^{-1} x\ -\ \sqrt{1-x^2} \right )\ +\ C

Question:24 Integrate the functions in Exercises 1 to 24.

\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}

Answer:

\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}

Here let's first reduce the log function.

=\frac{\sqrt{x^2+1}}{x^4}\left [ \log (x^2+1)-\log x^2 \right ]dx

=\frac{\sqrt{x^2\left ( 1+\frac{1}{x^2} \right )}}{x^4}\left [ \log\frac{ (x^2+1)}{x^2} \right ]dx

=\int\frac{\sqrt{\left ( 1+\frac{1}{x^2} \right )}}{x^3}\left [ \log\left ( 1+\frac{1}{x^2} \right ) \right ]dx

Now, let

t=1+\frac{1}{x^2}

dt=\frac{-2}{x^3}dx

So our function in terms if new variable t is :

I=\frac{-1}{2}\int \left [\log t \right ]\cdot t^{\frac{1}{2}}dt

now let's solve this By using integration by parts

I=\frac{-1}{2}\int \left [(\log t)\frac{t^\frac{3}{2}}{\frac{3}{2}} -\int \frac{1}{t}\cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}dt\right ]

I=\frac{-1}{3}t^\frac{3}{2}\log t+\frac{1}{3}\int t^{\frac{1}{2}}dt

I=\frac{-1}{3}t^\frac{3}{2}\log t+\frac{1}{3}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}

I=\frac{2}{9}t^{\frac{3}{2}}-\frac{1}{3}t^{\frac{3}{2}}logt+c

I=\frac{1}{3}t^{\frac{3}{2}}\left [ \frac{2}{3}-\log t \right ]+c

I=\frac{1}{3}\left ( 1+\frac{1}{x^2} \right )^{\frac{3}{2}}\left [ \frac{2}{3}-\log \left ( 1+\frac{1}{x^2} \right ) \right ]+c

Question:25 Evaluate the definite integrals in Exercises 25 to 33.

\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx

Answer:

\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx

Since, we have e^x multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,

\int e^x(f(x)+f'(x))dx=e^xf(x)

So,

\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx

=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-2\sin \frac{x}{2}cos\frac{x}{2}}{2sin^2\frac{x}{2}} \right )dx

=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1}{2sin^2\frac{x}{2}} -\frac{2\sin \frac{x}{2}cos\frac{x}{2}}{2sin^2\frac{x}{2}}\right )dx

=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1}{2}cosec^2\frac{x}{2}-cot\frac{x}{2}\right )dx

=\int_\frac{\pi}{2}^\pi e^x \left(-cot\frac{x}{2}+\frac{1}{2}cosec^2\frac{x}{2}\right )dx

Here let's use the property

\int e^x(f(x)+f'(x))dx=e^xf(x)

so,

=\int_\frac{\pi}{2}^\pi e^x \left(-cot\frac{x}{2}+\frac{1}{2}cosec^2\frac{x}{2}\right )dx

=\left [ -e^xcot\frac{x}{2} \right ]_\frac{\pi}{2}^\pi

=\left [ -e^\pi cot\frac{\pi}{2} \right ]-\left [ -e^{\frac{\pi}{2}} cot\frac{\pi}{4} \right ]

=e^{\frac{\pi}{2}}

Question:26 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}

Answer:

\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}

First, let's convert sin and cos into tan and sec. (because we have a good relation in tan and square of sec,)

Let' divide both numerator and denominator by cos^4x

=\int_0^\frac{\pi}{4}\frac{\frac{\sin x\cos x}{cosxcosxsos^2x} }{1+\frac{\sin^4 x}{\cos^4x}}

=\int_0^\frac{\pi}{4}\frac{tanxsec^2x}{1+tan^4x}

Now lets change the variable

\\t=tan^2x \\dt=2tanxsec^2xdx

the limits will also change since the variable is changing

when\:x=0,t=tan^20=0

when\:x=\frac{\pi}{4},t=tan^2\frac{\pi}{4}=1

So, the integration becomes:

I=\frac{1}{2}\int_{0}^{1}\frac{dt}{1+t^2}

I=\frac{1}{2}\left [ tan^{-1}t \right ]_0^1

I=\frac{1}{2}\left [ tan^{-1}1 \right ]-\frac{1}{2}\left [ tan^{-1}0\right ]

I=\frac{1}{2}\left [ \frac{\pi}{4} \right ]-0

I=\frac{\pi}{8}

Question:27 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}

Answer:

Lets first simplify the function.

\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}=\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4(1-\cos^2 x)}=\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{4-3\cos^2 x}

\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{4-3\cos^2 x-4\:\: }{4-3\cos^2 x }dx=\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{4-3\cos^2 x\:\: }{4-3\cos^2 x }dx-\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{-4\:\: }{4-3\cos^2 x }dx

\\=\frac{-1}{3}\int_0^\frac{\pi}{2}1dx-\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{-4\:\: }{4-3\cos^2 x }dx \\ \\ \\=\frac{-1}{3} \left [ x \right ]_0^{\frac{\pi}{2}}-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4\:\: }{4-3\cos^2 x }dx

As we have a good relation in between squares of the tan and square of sec lets try to take our equation there,

=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{4sec^2x-3 }dx

AS we can write square of sec in term of tan,


=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{4(1+tan^2x)-3 }dx

=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{1+4tan^2x }dx

Now let's calculate the integral of the second function, (we already have calculated the first function)

=-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{1+4tan^2x }dx

let

\\t=2tanx, \\dt=2sec^2xdx

here we are changing the variable so we have to calculate the limits of the new variable

when x = 0, t = 2tanx = 2tan(0)=0

when x=\pi/2,t=2tan{\pi/2}=\infty

our function in terms of t is

=-\frac{2}{3}\int_0^\infty\frac{1 }{1+t^2 }dt

=\left [ tan^{-1} t\right ]_0^\infty=\left [ tan^{-1} \infty-tan^{-1} 0\right ]

=\frac{\pi}{2}

Hence our total solution of the function is

\\=-\frac{\pi}{6}+\frac{2}{3}*\frac{\pi}{2}\\=\frac{\pi}{6}

Question:28 Evaluate the definite integrals in Exercises 25 to 33.

\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}

Answer:

\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}

Here first let convert sin2x as the angle of x ( sinx, and cosx)

\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{2sinxcosx}}

Now let's remove the square root form function by making a perfect square inside the square root

\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{-(-1+1-2sinxcosx)}}

\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{(1-(sin^2x+cos^2x-2sinxcosx)}}

\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{(1-(sinx-cosx)^2}}

Now let

, \\t=sinx-cosx \\dt=(cosx+sinx)dx

since we are changing the variable, limit of integration will change

\\when\: x=\pi/6, t=sin\pi/6-cos\pi/6=(1-\sqrt{3})/2 \\ when x= \pi/3,t=sin\pi/3-cos/pi/3=(\sqrt{3}-1)/2

our function in terms of t :

\\=\int_\frac{1-\sqrt{3}}{2}^\frac{\sqrt{3}-1}{2} \frac{1}{\sqrt{(1-t^2)}}dt

\\=\left [ sin^{-1}t \right ]_\frac{1-\sqrt{3}}{2}^\frac{\sqrt{3}-1}{2} \\ \\ \\=2sin^{-1}\left (\frac{\sqrt{3}-1}{2} \right )

Question:29 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}

Answer:

\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}

First, let's get rid of the square roots from the denominator,

\\=\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}*\frac{\sqrt{1+x} +\sqrt x}{\sqrt{1+x} +\sqrt x}

\\=\int_0^1\frac{\sqrt{1+x}+\sqrt{x}}{{1+x} -x}dx

\\=\int_0^1({\sqrt{1+x}+\sqrt{x}})dx

\\=\int_0^1({\sqrt{1+x})dx+\int_0^1({\sqrt{x}})dx

\\=\int_0^1(1+x)^\frac{1}{2}dx+\int_0^1x^\frac{1}{2}dx

\\=\left [ \frac{2}{3}(1+x)^{\frac{3}{2}} \right ]_0^1+\left [ \frac{2}{3}(x)^{\frac{3}{2}} \right ]_0^1

\\=\left [ \frac{2}{3}(1+1)^{\frac{3}{2}} \right ]-\left [ \frac{2}{3} \right ]+\left [ \frac{2}{3}(1)^{\frac{3}{2}} \right ]-\left [ 0 \right ]

\\=\frac{4\sqrt{2}}{3}

Question:30 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx

Answer:

\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx

First let's assume t = cosx - sin x so that (sinx +cosx)dx=dt

So,

Now since we are changing the variable, the new limit of the integration will be,

when x = 0, t = cos0-sin0=1-0=1

when x=\pi/4 t=\cos\pi/4-\sin\pi/4=0

Now,

(\cos x-\sin x)^2=t^2

\cos ^2x+\sin^2 x-2\cos x \sin x =t^2

1-\sin 2x =t^2

\sin 2x =1-t^2

Hence our function in terms of t becomes,

\int_{-1}^{0}\frac{dt}{9+16(1-t^2)}=\int_{-1}^{0}\frac{dt}{9+16-16t^2}=\int_{-1}^{0}\frac{dt}{25-16t^2}=\int_{-1}^{0}\frac{dt}{5^2-(4t)^2)}

= \frac{1}{4}\left [\frac{1}{2(5)}\log \frac{5+4t}{5-4t} \right ]_{-1}^0

= \frac{1}{40}\left[ \log (1)-\log (\frac{1}{9})\right ]

=\frac{\log 9}{40}

Question:31 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^\frac{\pi}{2}\sin 2x\tan^{-1}(\sin x)dx

Answer:

Let I =

\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx

=\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx

Here, we can see that if we put sinx = t, then the whole function will convert in term of t with dx being changed to dt.so

\\t=sinx \\dt=cosxdx

Now the important step here is to change the limit of the integration as we are changing the variable.so,

\\when\:x=0,t=sin0=0 \\when\:x=\frac{\pi}{2},t=sin\frac{\pi}{2}=1

So our function becomes,

I=2\int_{0}^{1}(tan^{-1}t)tdt

Now, let's integrate this by using integration by parts method,

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\int\frac{1}{1+t^2}\cdot\frac{t^2}{2}dt \right ]_0^1

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\frac{t^2}{1+t^2}\cdot dt \right ]_0^1

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\frac{(1+t^2)-1}{1+t^2}\cdot dt \right ]_0^1

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\left ( 1-\frac{1}{1+t^2} \right )\cdot dt \right ]_0^1

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}(t-tan^{-1}t) \right ]_0^1

I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}(t)+\frac{1}{2}tan^{-1}t) \right ]_0^1

I=2\left [ \frac{1}{2} \left (tan^{-1}t\cdot(t^2+1)-t \right )\right ]_0^1

I=\left [ \left (tan^{-1}t\cdot(t^2+1)-t \right )\right ]_0^1

I=\left [ \left (tan^{-1}(1)\cdot(1^2+1)-1 \right )\right ]-\left [ \left (tan^{-1}(0)\cdot(0^2+1)-0 \right )\right ] I=2tan^{-1}1-1=2\times \frac{\pi}{4}-1

I=\frac{\pi}{2}-1

Question:32 Evaluate the definite integrals in Exercises 25 to 33.

\int_0^\pi\frac{x\tan x}{\sec x + \tan x} dx

Answer:

Let I = \int_0^\pi\frac{x\tan x}{\sec x + \tan x} dx -(i)

Replacing x with ( \pi -x),

\\ I = \int_\pi^0\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x) + \tan (\pi -x)} (-dx) \\ = -\int_\pi^0\frac{(\pi -x)(-)\tan x}{-\sec x - \tan x} dx

\\ \implies I = \int^\pi_0\frac{(\pi -x)\tan x}{\sec x + \tan x} dx - (ii)

Adding (i) and (ii)

I + I = \int^\pi_0\left(\frac{x\tan x}{\sec x + \tan x} + \frac{(\pi -x)\tan x}{\sec x + \tan x} \right) dx

\implies 2I = \int^\pi_0\frac{\pi\tan x}{\sec x + \tan x} dx

\\ \implies 2I = \int^\pi_0\frac{\pi \frac{sin x}{cos x} }{\frac{1}{cos x} + \frac{sin x}{cos x}} dx \\ \implies 2I =\pi \int^\pi_0\frac{ sin x }{1+sin x} dx \\ \implies 2I =\pi \int^\pi_0\frac{ (1 +sin x ) -1}{1+sin x} dx \\ \implies 2I =\pi \int^\pi_0\left [1- \frac{1}{1+sin x} \right ]dx

\\ \implies 2I =\pi \int^\pi_0\left [1- \frac{1}{1+sin x} \right ]dx \\ \implies 2I =\pi \int^\pi_01 dx - \pi \int^\pi_0\frac{1}{1+sin x}.\frac{(1-sin x)}{(1 - sin x)}dx \\ \implies 2I =\pi\int^\pi_01 dx - \pi \int^\pi_0[\sec^2 x - \sec x \tan x]dx \\ \implies 2I =\pi[x]^\pi_0 - \pi[\sec x - \tan x]^\pi_0

\\ \implies 2I =\pi[\pi - 0] - \pi[tan \pi - sec \pi- tan \pi + sec 0] \\ \implies 2I =\pi[\pi -2] \\ \implies I =\frac{\pi}{2}[\pi -2]

Question:33 Evaluate the definite integrals in Exercises 25 to 33.

\int_1^4[|x-1| + |x-2| + |x-3|]dx

Answer:

Given integral \int_1^4[|x-1| + |x-2| + |x-3|]dx

So, we split it in according to intervals they are positive or negative.

= \int_{1}^4 |x-1| dx + \int_{1}^4 |x-2| dx + \int^4_{1} |x-3| dx

= I_{1}+I_{2}+I_{3}

Now,

I_{1} = \int^4_{1}|x-1| dx = \int^4_{1} (x-1)dx

\because as (x-1) is positive in the given x -range [1,4]

=\left [ \frac{x^2}{2}-x\right ]^4_{1} = \left [ \frac{4^2}{2}-4 \right ] - \left [ \frac{1^2}{2}-1 \right ]

= \left [ 8-4 \right ] - [-\frac{1}{2}] = 4+\frac{1}{2} = \frac{9}{2}

Therefore, I_{1} = \frac{9}{2}

I_{2} = \int^4_{1}|x-2| dx = \int^2_{1} (2-x)dx +\int^4_{2} (x-2)dx

\because as (x-2)\geq 0 is in the given x -range [2,4] and \leq 0 in the range [1,2]

=\left [ 2x - \frac{x^2}{2}\right ] ^2_{1} + \left [ \frac{x^2}{2} -2x\right ] ^4_{2}

= \left \{ \left [ 2(2)-\frac{2^2}{2} \right ] - \left [ 2(1)-\frac{1^2}{2} \right ] \right \} + \left \{ \left [ \frac{4^2}{2}-2(4) \right ] - \left [ \frac{2^2}{2}-2(2) \right ] \right \}

= [4-2-2+\frac{1}{2}] +[8-8-2+4]

= \frac{1}{2}+2 =\frac{5}{2}

Therefore, I_{2} = \frac{5}{2}

I_{3} = \int^4_{1}|x-3| dx = \int^3_{1} (3-x)dx +\int^4_{3} (x-3)dx

\because as (x-3)\geq 0 is in the given x -range [3,4] and \leq 0 in the range [1,3]

=\left [ 3x - \frac{x^2}{2}\right ] ^3_{1} + \left [ \frac{x^2}{2} -3x\right ] ^4_{3}

= \left \{ \left [ 3(3)-\frac{3^2}{2} \right ] - \left [ 3(1)-\frac{1^2}{2} \right ] \right \} + \left \{ \left [ \frac{4^2}{2}-3(4) \right ] - \left [ \frac{3^2}{2}-3(3) \right ] \right \}

= [9-\frac{9}{2}-3+\frac{1}{2}]+[8-12-\frac{9}{2}+9]

= [6-4]+\frac{1}{2} =\frac{5}{2}

Therefore, I_{3} = \frac{5}{2}

So, We have the sum = I_{1}+I_{2}+I_{3}

I = \frac{9}{2}+\frac{5}{2}+\frac{5}{2} = \frac{19}{2}

Question:34 Prove the following (Exercises 34 to 39)

. \int_1^3\frac{dx}{x^2(x+1)} = \frac{2}{3}+ \log \frac{2}{3}

Answer:

L.H.S = \int_1^3\frac{dx}{x^2(x+1)}

We can write the numerator as [(x+1) -x]

\therefore \int_1^3\frac{dx}{x^2(x+1)} = \int_1^3\frac{(x+1)-x}{x^2(x+1)}dx

\\ = \int_1^3\left [ \frac{1}{x^2} - \frac{1}{x(x+1)} \right ]dx \\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\frac{(x+1)-x}{x(x+1)}dx

\\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\left [ \frac{1}{x} - \frac{1}{(x+1)} \right ]dx \\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\frac{1}{x}dx + \int_1^3\frac{1}{(x+1)}dx \\ = \left [ -\frac{1}{x} \right ]^3_1 - \left [ \log x \right ]^3_1 +\left [ \log(x+1) \right ]^3_1

\\ = \left [ -\frac{1}{3} +1 \right ] - \left [ \log 3 - \log 1 \right ] +\left [\log 4 - \log 2 \right ] \\ = \frac{2}{3} + \log \left ( \frac{4}{3.2}\right ) \\

= \log \left(\frac{2}{3} \right ) +\frac{2}{3} = RHS

Hence proved.

Question:35 Prove the following (Exercises 34 to 39)

\int_0^1 xe^xdx =1

Answer:

Let\ I=\int xe^{x}dx

Integrating I by parts

\\I=x\int e^{x}dx-\int ( (\frac{\mathrm{d} (x)}{\mathrm{d} x})\int e^{x}dx)dx\\ I=xe^{x}-\int e^{x}dx\\ I=xe^{x}-e^{x}+c

Applying Limits from 0 to 1

\\\int_{0}^{1}xe^{x}dx=[xe^{x}-e^{x}+c]_{0}^{1}\\ I=[e-e+c]-[0-1+c]\\ I=1

Hence proved I = 1

Question:36 Prove the following (Exercises 34 to 39)

\int_{-1}^1x^{17}\cos^4 x dx=0

Answer:

Let \ x^{17}cos^{4}x=g(x)

g(-x)= (-x)^{17}cos^{4}(-x)=-x^{17}cos^{4}x=-g(x)

The Integrand g(x) therefore is an odd function and therefore

\int_{-1}^{1}g(x)dx=0

Question:37 Prove the following (Exercises 34 to 39)

\int_0^\frac{\pi}{2}\sin^3 x dx =\frac{2}{3}

Answer:

\\Let\ I= \int_{0}^{\frac{\pi }{2}}sin^{3}xdx\\ I=\int_{0}^{\frac{\pi }{2}}sinx(1-sin^{2}x)dx\\ I=\int_{0}^{\frac{\pi }{2}}sinxdx-\int_{0}^{\frac{\pi }{2}}cos^{2}xsinxdx\\ I=I_{1}-I_{2}

\\I_{1}=[-cosx]_{0}^{\frac{\pi }{2}}\\ I_{1}=-0-(-1)=1

For I 2 let cosx=t, -sinxdx=dt

The limits change to 0 and 1

\\I_{2}=-\int_{1}^{0}t^{2}dt\\ I_{2}=-[\frac{t^{3}}{3}]{_{1}}^{0}\\ I_{2}=0-(-\frac{1}{3})\\ I_{2}=\frac{1}{3}

I 1 -I 2 =2/3

Hence proved.

Question:€‹38 Prove the following (Exercises 34 to 39)

\int_0^\frac{\pi}{4}2\tan^3 x dx = 1 - \log 2

Answer:

The integral is written as

\\Let\ I=\int 2tan^{3}xdx\\ I=\int 2tan^{2}x\cdot tanxdx\\ I=\int 2(sec^{2}x-1)tanxdx\\ I=2\int tanxsec^{2}xdx-2\int tanxdx\\ I=2\int tdt-2log(cosx)+c\ \ \ \ \ \ \ (t=tanx) \\I=t^{2}-2log(cosx)+c\\ I=tan^{2}x-2log(cosx)+c

[I]_{0}^{\frac{\pi }{4}}=[tan^{2}x-2log(cosx)]_{0}^{\frac{\pi }{4}}\\

[I]_{0}^{\frac{\pi }{4}}=(1-2log\sqrt{2})-(0-2log1)

[I]_{0}^{\frac{\pi }{4}}=1-log2

Hence Proved

Question:39 Prove the following (Exercises 34 to 39)\

\int_0^1\sin^{-1}xdx = \frac{\pi}{2}-1

Answer:

Let \ I=\int sin^{-}xdx

Integrating by parts we get

\\ I= sin^{-}x\int 1\cdot dx-\int (\frac{\mathrm{d} (sin^{-}x)}{\mathrm{d} x}\int 1\cdot dx)\\ I=xsin^{-}x+c-\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx\\ I=I_{1}-I_{2}

For I 2 take 1-x 2 = t 2 , -xdx=tdt

\\I_{2}=\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx\\ I_{2}=-\int\frac{1}{t}tdt \\ I_{2}=-t+c\\ I_{2}=-\sqrt{1-x^{2}}+c

[I]_{0}^{1}=[I_{1}-I_{2}]_{0}^{1}\\

\\=[xsin^{-}x-(-\sqrt{1-x^{2}})]_{0}^{1}\\ =[xsin^{-}x+\sqrt{1-x^{2}}]_{0}^{1}\\ =[1\cdot \frac{\pi }{2}+0]-[0+1]\\ =\frac{\pi }{2}-1

Hence Proved

Question:40 Evaluate \int_0^1e^{2-3x}dx as a limit of a sum.

Answer:

As we know

\int_{a}^{b}f(x)dx=(b-a)\lim_{n\rightarrow \infty }\frac{1}{n}[f(a)+f(a+h)+f(a+2h)........+f(a+(n-1)h)]

where b-a=hn

In the given problem b=1, a=0 and f(x)=e^{2-3x}
\\\int_{0}^{1}e^{2-3x}dx=(1-0)\lim_{n\rightarrow \infty }\frac{1}{n}(e^{2}+e^{2-3h}+e^{2-3(2h)}.....+e^{2-3(n-1)h})\\ =e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(1+e^{-3h}+e^{-6h}....+e^{-3(n-1)h})\\ =e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1-(e^{-3h})^{n}}{1-e^{-3h}})\\ =e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1-e^{-\frac{3}{n}\times n}}{1-e^{-\frac{3}{n}}})\\

\\=e^{2}\lim_{n\rightarrow \infty }\frac{1}{n}(\frac{1-e^{-3}}{1-e^{-\frac{3}{n}}})\\ =\frac{e^{2}(1-e^{-3})}{3}\lim_{n\rightarrow \infty }\frac{-\frac{3}{n}}{e^{-\frac{3}{n}}-1}\\ =\frac{e^{2}(1-e^{-3})}{3}

=\frac{e^{2}-e^{-1}}{3}

Question:41 Choose the correct answers in Exercises 41 to 44.

. \int\frac{dx}{e^x + e^{-x}} is equal to

(A) \tan^{-1}(e^x) + c

(B) \tan^{-1}(e^{-x}) + c

(C) \log (e^x - e^{-x}) + C

(D) \log (e^x + e^{-x}) + C

Answer:

\int\frac{dx}{e^x + e^{-x}}

the above integral can be re arranged as

\\=\int \frac{e^{x}}{e^{2x}+1}dx\\

let e x =t, e x dx=dt

\int\frac{dx}{e^x + e^{-x}}

\\=\int \frac{1}{t^{2}+1}dt\\ =tan^{-1}t+c\\ =tan^{-1}(e^{x})+c

(A) is correct

Question:42​​​​​​​ Choose the correct answers in Exercises 41 to 44.

. \int\frac{\cos 2x}{(\sin x + \cos x)^2}dx is equal to

(A) \frac{-1}{\sin x + \cos x} + C

(B) \log |{\sin x + \cos x} |+ C

(C) \log |{\sin x- \cos x} |+ C

(D) \frac{1}{(\sin x + \cos x)^2} + C

Answer:

\\\frac{\cos 2x}{(\sin x + \cos x)^2}\\ =\frac{cos^{2}x-sin^{2}x}{(\sin x + \cos x)^2}\\ =\frac{(\sin x + \cos x)(\cos x-\sin x)}{(\sin x + \cos x)^2} \\=\frac{(\cos x-\sin x)}{(\sin x + \cos x)} cos2x=cos 2 x-sin 2 x

let sinx+cosx=t,(cosx-sinx)dx=dt

hence the given integral can be written as

\\\int\frac{\cos 2x}{(\sin x + \cos x)^2}dx\\ =\int \frac{dt}{t}\\ =log|t|+c \\=log|cosx+sinx|+c

B is correct

Question:43​​​​​​​ Choose the correct answers in Exercises 41 to 44.

If f(a+b-x) = f(x) , then \int_a^bxf(x)dx is equal to

(A) \frac{a+b}{2}\int^b_af(b-x)dx

(B) \frac{a+b}{2}\int^b_af(b+x)dx

(C) \frac{b-a}{2}\int^b_af(x)dx

(D) \frac{a+b}{2}\int^b_af(x)dx

Answer:

Let\ \int_a^bxf(x)dx=I

As we know \int_a^bf(x)dx=\int_a^bf(a+b-x)dx

Using the above property we can write the integral as

\\I=\int_{a}^{b}(a+b-x)f(a+b-x)dx\\ I=\int_{a}^{b}(a+b-x)f(x)dx\\ I=(a+b)\int_{a}^{b}f(x)dx-\int_{a}^{b}xf(x)dx\\ I=(a+b)\int_{a}^{b}f(x)dx-I\\ 2I=(a+b)\int_{a}^{b}f(x)dx\\ I=\frac{a+b}{2}\int_{a}^{b}f(x)dx

Answer (D) is correct

Question:€‹44 Choose the correct answers in Exercises 41 to 44.

The value of \int_0^1\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )dx is

(A) 1

(B) 0

(C) -1

(D) \frac{\pi}{4}

Answer:

\\Let\ I=\int_0^1\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )dx\\

\\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )\\ =tan^{-1}\left ( \frac{x-(1-x)}{1+x(1-x)} \right )\\ =tan^{-1}x-tan^{-1}(1-x) as tan^{-1}\left ( \frac{a-b}{1+ab} \right )=tan^{-1}a-tan^{-1}b

Now the integral can be written as

\\I=\int_{0}^{1} \left ( tan^{-1}x-tan^{-1}(1-x) \right )dx\\ I=\int_{0}^{1} \left ( tan^{-1}(1-x)-tan^{-1}(1-(1-x)) \right )dx\\ I=\int_{0}^{1} \left ( tan^{-1}(1-x)-tan^{-1}x\right )dx\\ I=-I\\ 2I=0\\ I=0

(B) is correct.

If you are looking for integrals class 12 NCERT solutions of exercises then they are listed below.

About NCERT solutions for class 12 maths chapter 7 integrals

The word integration literally means summation. When you have to find the sum of finite numbers you can do by simply adding these numbers. But when you are finding the sum of a certain number of elements as the number of elements tends to infinity and at the same time each term becomes infinitesimally small, you can use a prosses to find its limit called integration.

Integrals has 13 % weightage in 12 board final examinations. Next chapter "applications of integrals" is also dependent on this chapter. So you should try to solve every problem of this chapter on your own. If you are not able to do, you can take the help of these NCERT solutions for class 12 maths chapter 7 integrals. In this chapter, there are 11 exercises with 227 questions and also 44 questions are there in miscellaneous exercise. Here, the NCERT solutions for class 12 maths chapter 7 integrals are solved and explained in detail to develop a grip on the topic. Here, you will learn two types of integrals: Definite integral and Indefinite integral and also learn their properties and formulas.


Definite Integral

Indefinite Integral

Definition
A definite Integral has upper and lower limits if 'a' and 'b' are the limits or boundaries. The definite integral of f(x) is a number, not function .

An integral without upper limit and lower limit. It is also an antiderivative. The indefinite integral of f(x) is a function not number.

Expression
\int_{a}^{b}f(x)dx

F(x)=\int f(x)dx

NCERT Solutions for Class 12 Maths Chapter 7 Integrals - Topics

7.1 Introduction

7.2 Integration as an Inverse Process of Differentiation

7.2.1 Geometrical interpretation of indefinite integral

7.2.2 Some properties of indefinite integral

7.2.3 Comparison between differentiation and integration

7.3 Methods of Integration

7.3.1 Integration by substitution

7.3.2 Integration using trigonometric identities

7.4 Integrals of Some Particular Functions

7.5 Integration by Partial Fractions

7.6 Integration by Parts

7.7 Definite Integral

7.7.1 Definite integral as the limit of a sum

7.8 Fundamental Theorem of Calculus

7.8.1 Area function

7.8.2 First fundamental theorem of integral calculus

7.8.3 Second fundamental theorem of integral calculus

7.9 Evaluation of Definite Integrals by Substitution

7.10 Some Properties of Definite Integral

NCERT solutions for class 12 maths - Chapter wise

Key Features of NCERT Solutions for Class 12 Maths Chapter 7 Integrals

The NCERT chapter 7 class 12 maths offers several key features to aid students in their understanding and mastery of this topic. Some of these features include

  1. Detailed explanation: The chapter 7 class 12th maths solutions provide a comprehensive and in-depth explanation of the concepts of integrals, making it easier for students to grasp the subject.

  2. Step-by-Step Solutions: The integration class 12 ncert solutions break down complex problems into simpler steps, making it easier for students to follow and understand the process.

  3. Clear Diagrams: The integrals class 12 solutions make use of clear and informative diagrams to help students visualize the concepts, making it easier for them to comprehend the material.

  4. Plenty of Practice Problems: The ch 7 maths class 12 include a large number of practice problems to help students strengthen their knowledge and skills.

  5. Accurate answers: The class 12 ch 7 maths ncert solutions are checked and verified by experts to ensure accuracy, helping students avoid mistakes and improve their grades.

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More about NCERT Solutions for Class 12 Maths Chapter 7 Integrals

The word integration literally means summation. When you have to find the sum of finite numbers you can do by simply adding these numbers. But when you are finding the sum of a certain number of elements as the number of elements tends to infinity and at the same time each term becomes infinitesimally small, you can use a prosses to find its limit called integration.

  • Integrals have 13 % weightage in 12 board final examinations. Next chapter "applications of integrals" is also dependent on this chapter. So you should try to solve every problem of this chapter on your own.

  • If you are not able to, you can take the help of these NCERT solutions for class 12 maths chapter 7 integrals.

  • In this NCERT Class 12 Maths solutions chapter 7, there are 11 exercises with 227 questions and also 44 questions are there in miscellaneous exercises. Here, the Class 12 Maths Chapter 7 NCERT solutions are solved and explained in detail to develop a grip on the topic.

NCERT solutions for class 12 - subject wise

How to use NCERT solutions for class 12 maths chapter 7 Integrals

  • You are expected to remember all formulas of differentiation, and then can start with basic integration

  • Try to relate differentiation formulas with integrations formulas because it will help you to remember all the integration formulas

  • When you are done with basic integration, you should learn different methods of integration for different types of functions. If you find difficulties in learning the method you should learn one method at a time to solve a particular type of problem

  • After learning the different methods of integration, you should be able to determine which method will be used to solve a particular type of problem

  • When you become good with indefinite integrals, move to the definite integrals and learn some properties to solve definite integrals. NCERT Class 12 Maths solutions chapter 7 integrals will help you for the same

  • This chapter requires a lot of practice. First, solve all the NCERT textbook questions, then, you can take the help of NCERT solutions for class 12 maths chapter 7 integrals.

  • If you have solved all NCERT questions, you can solve CBSE previous year papers also to get familiar with the type of questions which are asked in previous years.

  • NCERT solutions class 12 maths chapter 12 pdf download will be made available soon. Till then you can save the webpage and practice these solutions offline.

NCERT Books and NCERT Syllabus

Frequently Asked Question (FAQs)

1. Which are the most difficult chapters of NCERT Class 12 Maths syllabus?

Students consider Integration which is integrals and applications of integration are the most difficult chapter in CBSE class 12 maths but with the regular practice of NCERT problems you will be able to have a strong grip on this chapter also. it's true that there is no substitute for hard work but the right strategy and quality study material are also essential to get command of this chapter, therefore, NCERT exercises are recommended for practice.  

2. How are the NCERT solutions helpful in the board exam?

NCERT solutions are created by the expert team of careers360 who know how best to write answers in the board exam in order to get good marks. Integrating their techniques can benefit in obtaining meritorious marks. Sometimes students do not understand where s/he are making a mistake, NCERT solutions can help them to understand that. the practice of a lot of questions and their solution make you confident and help you in getting an in-depth understanding of concepts. Therefore NCERT solutions are very helpful for students.

3. What is the weightage of the chapter Integrals for CBSE board exam?

Integrals have 13 % weightage in CBSE class 12th board final examination.  having 13% weightage, Integral become very students for CBSE aspirant but it demand lot of practice. Interested students can refer to integrals class 12 solutions.

4. Is the NCERT Solutions for class 12 maths integration challenging to comprehend?

The NCERT Solutions for maths ch 7 maths class 12 is not complicated to understand. It is a fascinating topic in Class 12 that is also relevant at higher education levels. A solid understanding of integral formulas will enable students to solve integration problems effectively. A thorough knowledge of derivatives is crucial for comprehending the concepts of integral calculus with ease. For ease, students can study integrals class 12 ncert solutions pdf both online and offline.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


Believe in Yourself! You can make anything happen


All the very best.

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
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