NCERT Solutions for Class 12 Maths Chapter 7 Integrals

# 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

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

Class 12 Integrals NCERT solutions Exercise: 7.1

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. $\cos 3x$

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. $e ^{2x}$

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. $( ax + b )^2$

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. $\sin 2x - 4 e ^{3x}$

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

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

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 $\int ( ax ^2 + bx + c ) dx$

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 $\int \left ( 2x^2 + e ^x \right ) dx$

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

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

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 $\int \frac{x^3+ 3x +4 }{\sqrt x } dx$

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

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 $\int (1-x) \sqrt x dx$

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 $\int \sqrt x ( 3x^2 + 2x +3 )dx$

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 $\int ( 2x - 3 \cos x + e ^x ) dx$

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 $\int ( 2 x ^2 - 3 \sin x + 5 \sqrt x ) dx$

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 $\int \sec x ( \sec x + \tan x ) dx$

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 $\int \frac{sec ^2 x }{cosec ^2 x } dx$

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 $\int \frac{2- 3 \sin x }{\cos ^ 2 x } dx$

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

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

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 $\frac{2x}{1+ x ^2}$

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 $\frac{( \log x )^2}{x}$

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 $\frac{1}{x+ x \log x }$

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 $\sin x \sin ( \cos x )$

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 $\sin ( ax + b ) \cos ( ax + b )$

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 $\sqrt { ax + b }$

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 $x \sqrt { x +2 }$

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 $x \sqrt { 1+ 2 x^2 }$

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

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 $\frac{1}{x - \sqrt x }$

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

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 $( x ^3 - 1 ) ^{1/3} x ^ 5$

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 $\frac{x ^2 }{(2+3x^3)^3}$

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

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 $\frac{x}{9- 4 x ^2 }$

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 $e ^{ 2 x +3 }$

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 $\frac{x }{e^{x^{2}}}$

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

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

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 $\frac{e ^{2x}- e ^{-2x }}{e ^ {2x }+ e ^{ -2 x }}$

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 $\tan ^2 ( 2x-3 )$

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 $\sec ^2 ( 7- 4x )$

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 $\frac{\sin ^{-1}x}{\sqrt { 1- x^2 }}$

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 $\frac{2 \cos x - 3\sin x }{6 \cos x + 4 \sin x }$

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 $\frac{1 }{ \cos ^2 x (1-\tan x )^2}$

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 $\frac{\cos \sqrt x }{\sqrt x }$

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 $\sqrt { \sin 2x } \cos 2x$

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 $\frac{\cos x }{\sqrt { 1+ \sin x }}$

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 $\cot x \: log \sin x$

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 $\frac{\sin x }{1+ \cos x }$

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 $\frac{\sin x }{( 1+ \cos x )^2}$

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 $\frac{1}{1+ \cot x }$

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 $\frac{1}{1- \tan x }$

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 $\frac{\sqrt { \tan x } }{\sin x \cos x }$

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

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

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

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 $\int \frac{10 x^ 9 + 10 ^x \log _ e 10 dx }{x ^{10}+ 10 ^x }dx\: \: \: equals$

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 $\int \frac{dx }{\sin ^ 2 x \cos ^2 x }\: \: \: equals$

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 $\sin ^ 2 ( 2x+ 5 )$

$\sin ^ 2 ( 2x+ 5 )$

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 $\sin 3x \cos 4x$

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 $\cos 2x \cos 4x \cos 6x$

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 $\sin ^ 3 ( 2x +1 )$

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

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 $\sin ^3 x \cos ^ 3 x$

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

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 $\sin x \sin 2x \sin 3x$

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 $\sin 4x \sin 8x$

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 $\frac{1- \cos x }{1+ \cos x }$

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 $\frac {\cos x }{1 + \cos x }$

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 $\sin ^ 4 x$

$\sin ^ 4 x$ 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 $\cos ^ 4 2x$

$cos^42x=cos^32xcos2x$

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

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 $\frac{\cos 2x - \cos 2 \alpha }{\cos x - \cos \alpha }$

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 $\frac{\cos x - \sin x }{1+ \sin 2x }$

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

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

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 $\tan ^ 3 2x \sec 2x$

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

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 $\tan ^ 4x$

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

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

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

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 $\frac{1}{\sin x \cos ^3 x }$

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 $\frac{\cos 2x }{( \cos x + \sin x )^2}$

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 $\sin ^ { -1} ( \cos x )$

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 $\frac{1}{\cos ( x-a ) \cos ( x-b )}$

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

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 $\int \frac{e ^x ( 1+x)}{\cos ^ 2 ( e ^xx )} dx \: \: equals$

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 $\frac{3x^ 2 }{x^6 + 1 }$

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 $\frac{1}{\sqrt { 1+ 4 x^2 }}$

$\frac{1}{\sqrt { 1+ 4 x^2 }}$
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 $\dpi{100} \int\frac{1}{\sqrt{x^2+a^2}}dt = \log\left | x+\sqrt{x^2+a^2} \right |$

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

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

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 $\frac{1}{\sqrt {9 - 25 x^2 }}$

$\frac{1}{\sqrt {9 - 25 x^2 }}$
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 $\frac{3x }{1+ 2 x ^ 4 }$

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

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

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

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 $\frac{x-1 }{\sqrt { x^2 -1 }}$

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 $\frac{x ^ 2 }{\sqrt { x^6 + a ^ 6 }}$

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 $\frac{\sec ^ 2 x }{\sqrt { \tan ^ 2 x+ 4 }}$

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 $\frac{1 }{ \sqrt { x ^ 2 + 2 x + 2 }}$

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

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 $\frac{1}{\sqrt { ( x-1)( x-2 )}}$

(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 $\frac{1}{\sqrt { 8 + 3 x - x ^ 2 }}$

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 $\frac{1}{\sqrt {(x-a)( x-b )}}$

(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 $\frac{4x+1 }{\sqrt {2x ^ 2 + x -3 }}$

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 $\frac{x+ 2 }{\sqrt { x ^2 -1 }}$

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

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 $\frac{6x + 7 }{\sqrt {( x-5 )( x-4)}}$

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 $\frac{x +2 }{\sqrt { 4x - x ^ 2 }}$

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 $\frac{x +2 }{\sqrt { x^ 2 + 2x +3 }}$

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

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 $\frac{5x + 3 }{\sqrt { x^2 + 4x +10 }}$

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

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 $\int \frac{dx }{\sqrt { 9x - 4x ^2 }} \: \: equals$

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

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 $\frac{1}{x^2 -9 }$

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 $\frac{3x -1}{( x-1)(x-2)(x-3)}$

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 $\frac{x }{( x-1)(x-2)(x-3)}$

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 $\frac{2x}{x^2 + 3x +2 }$

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 $\frac{1- x^2 }{ x ( 1- 2x )}$

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

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 $\frac{x }{( x+1)^2 ( x+2)}$

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

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

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

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

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

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

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 $\frac{1}{x^4 -1 }$

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 $\frac{1}{x ( x^n+1)}$

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 $\frac{\cos x }{(1- \sin x )( 2- \sin x )}$

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

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

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 $\frac{1}{x (x^4 -1)}$

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 $e ^x= t$ ]

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 $\int \frac{x dx }{( x-1)(x-2) } \: \: equals$

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 $\int \frac{dx}{x ( x ^2+1)} \: \: equals$

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

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 $x \sin 3x$

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 $x ^ 2 e ^x$

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 $x \log x$

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 $x \log 2x$

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 $x^ 2 \log x$

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 $x \sin ^{ -1} x$

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 $x \tan ^{-1} x$

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 $x\cos ^{ -1} x$

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 $( \sin ^{-1}x ) ^ 2$

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 $\frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}$

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,

### Question:12 Integrate the functions $x \sec ^2 x$

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 $\tan ^{-1} x$

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 $x ( \log x )^ 2$

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 $( x^2 + 1 ) \log x$

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 $e ^ x ( \sin x + \cos x )$

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 $\frac{x e ^x }{( 1+ x )^2}$

$\frac{x e ^x }{( 1+ x )^2}$
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 $e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )$

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 $e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )$

$e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )$
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 $\frac{( x-3)e ^x }{( x-1)^3}$

$\frac{( x-3)e ^x }{( x-1)^3}$
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 $e ^{ 2x } \sin x$

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

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

$\int \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$
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

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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

(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) $\frac{x^2}{2}\sqrt{1+x^2} + \frac{1}{2}x^2\log\left |x + \sqrt{1+x^2} \right| +C$

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.

(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) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} -\frac{9}{2}\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C$

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.

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.

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.

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.

$\int_{1}^4(x^2-x)dx$

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

$\int_2^3 \frac{dx}{x^2 -1 }$

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{12}$

(C) $\frac{\pi}{24}$

(D) $\frac{\pi}{4}$

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.

$\int_0^1\frac{x}{x^2 +1}dx$
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.

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$
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.

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$
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

$\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta-\int \tan\theta\ d\theta]^{\pi/4}_0\\ =2[\theta\tan\theta+\log\left | \cos\theta \right |]^{\pi/4}_0\\ =2[\pi/4+\log(1/\sqrt{2})]\\ =\pi/4-\log 2$

### $\int_0^2x\sqrt{x+2}$ . (Put ${x+2} = t^2$ )

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.

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx =I$
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.

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.

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$
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.

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$
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.

(A) 6

(B) 0

(C) 3

(D) 4

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.

(A) $\cos x + x\sin x$

(B) $x\sin x$

(C) $x\cos x$

(D) $\sin x + x\cos x$

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 $g(x) + g(a-x) = 4$

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.

(A) 0

(B) 2

(C) $\pi$

(D) 1

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.

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.

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.

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.

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.

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 $x = t^6$ ]

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.

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.

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.

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.

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.

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.

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.

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.

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

$\implies I = \frac{1}{3}tan^{-1}x - \frac{1}{6}tan^{-1}\frac{x}{2} + C$

### Question:15 Integrate the functions in Exercises 1 to 24.

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

$\therefore \int cos^3xe^{\log sinx}dx = \int cos^3x.sinx dx$

$I = -\frac{cos^4x}{4} + C$

### Question:16Integrate the functions in Exercises 1 to 24.

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 $x^4 = t \implies 4x^3 dx = dt$

$I = \int e^{3\log x} (x^4 + 1)^{-1} = \int \frac{x^3}{x^4 + 1}$

$\implies I = \frac{1}{4}\log(x^4 +1) + C$

### Question:17 Integrate the functions in Exercises 1 to 24.

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.

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

$\therefore \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}} = \frac{1}{\sqrt{\sin^3 x (\sin x \cos \alpha + \cos x \sin \alpha)}}$

$= \frac{1}{\sqrt{\sin^3 x . \sin x(\cos \alpha + \cot x \sin \alpha)}} = \frac{1}{\sqrt{\sin^4 x (\cos \alpha + \cot x \sin \alpha)}}$

$\frac{cosec^2 x}{\sqrt{(\cos \alpha + \cot x \sin \alpha)}}$

### Question:19 Integrate the functions in Exercises 1 to 24.

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.

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

$\\ I = \int \sqrt{\frac{1-\sqrt {cos^2\theta}}{1 +\sqrt {cos^2\theta}}}(-2\sin\theta\cos\theta)d\theta \\ = -\int \sqrt{\frac{1-cos\theta}{1 +cos\theta}}(2\sin\theta\cos\theta)d\theta$

$\\ = -\int \sqrt{tan^2\frac{\theta}{2}}(2\sin\theta\cos\theta)d\theta \\ = -\int \sqrt{tan^2\frac{\theta}{2}}(2. 2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}\cos\theta)d\theta \\ = -4\int \sin^2\frac{\theta}{2}\cos\theta d\theta$

$\\ = -4\int \sin^2\frac{\theta}{2}(2cos^2\frac{\theta}{2} -1) d\theta$

### Question:21 Integrate the functions in Exercises 1 to 24.

Given to evaluate

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

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

now the integral becomes

Let tan x = f(x)

$\fn_cm \implies f'(x) = sec^2x dx$

### Question:22 Integrate the functions in Exercises 1 to 24.

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

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

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