NCERT Solutions for Class 12 Maths Chapter 7 - Integrals

NCERT Solution for Class 12 Maths Chapter 7 Solutions: Download PDF

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NCERT Solutions for Class 12 Maths Chapter 7: Exercise Questions

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

Answer:

Given $\sin 2x$;

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

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

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

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

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

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

Answer:

Given $\cos 3x$;

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

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

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

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

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

Answer:

Given $e ^{2x}$;

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

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

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

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

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

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

Answer:

Given $( ax + b )^2$;

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

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

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

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

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

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

Answer:

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

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

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

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

Question 6: Find the following integrals $\int ( 4e ^{3x}+1) dx$

Answer:

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

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

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

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

Answer:

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

$=\int x^2 dx - \int1dx$

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

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

Answer:

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

Answer:

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

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

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

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

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

Answer:

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

Answer:

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

It can be written as

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

Taking $(x-1)$ common out

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

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

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

Splitting the terms inside the brackets

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

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

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

Answer:

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

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

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

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

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

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

Answer:

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

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

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

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

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

Answer:

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

Splitting the integral as the sum of three integrals

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

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

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

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

Answer:

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

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

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

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

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

Answer:

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

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

Using the integral of trigonometric functions

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

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

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

Answer:

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

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

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

$= \int \tan^2x \ dx$

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

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

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

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

Answer:

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

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

Using the antiderivative of trigonometric functions

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

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

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

Answer:

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

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

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

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

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

Hence, the correct option is (C).

Question 22: Choose the correct answer:

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

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

Answer:

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

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

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

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

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

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

Now, to find the constant C;

Putting the condition, f (2) = 0

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

$⇒ 16+\frac{1}{8} +C = 0$

$⇒ C = \frac{-129}{8}$

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

Therefore, the correct answer is A.

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

Answer:

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

Let us assume $1+x^2 =t$

We get, $2xdx = dt$

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

$= \log|t| +C$

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

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

$= \log(1+x^2) +C$, where C is any constant value.

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

Answer:

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

Let us assume $\log |x| = t$

We get, $\frac{1}{x}dx= dt$

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

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

$= \frac{(\log|x|)^3}{3}+C$, where C is any constant value.

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

Answer:

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

Let us assume $1+\log x = t$

We get, $\frac{1}{x}dx= dt$

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

$= \log|t| +C$

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

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

Answer:

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

Let us assume $\cos x =t$

We get, $-\sin x dx =dt$

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

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

$= \cos t +C$

Back substituting the value of $t$ we get,

$= \cos (\cos x ) +C$, where C is any constant value.

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

Answer:

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

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

Let us assume $2(ax+b) = t$

We get, $2adx =dt$

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

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

Now, by back substituting the value of t,

$= \frac{-1}{4a}[cos 2(ax+b)] +C$, where C is any constant value.

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

Answer:

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

Let us assume $(ax+b) = t$

We get, $adx =dt$

$dx = \frac{1}{a}dt$

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

Now, by back substituting the value of t,

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

$= \frac{2(ax+b)^\frac{3}{2}}{3a} +C$, where C is any constant value.

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

Answer:

Given function $x \sqrt { x +2 }$ ,

$\int x\sqrt{x+2}$

Assume the $(x+2) = t$ 19634

$\therefore dx =dt$

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

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

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

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

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

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

Substituting the value of $t$ in the above equation,

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

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

Answer:

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

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

Assume the $1+2x^2= t$

$\therefore 4xdx =dt$

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

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

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

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

Answer:

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

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

Assume the $1+x+x^2 = t$

$\therefore (2x+1)dx =dt$

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

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

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

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

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

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

Answer:

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

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

Can be written in the form:

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

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

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

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

$= 2\log|t| +C$

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

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

Answer:

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

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

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

$\therefore dx=dt$

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

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

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

$= \frac{2}{3}(x+4)^{\frac{3}{2}} -16(x+4)^{\frac{1}{2}}+C$, where C is any constant value.

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

Answer:

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

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

Assume the $x^3-1 = t$

$\therefore 3x^2dx=dt$

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

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

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

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

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

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

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

Answer:

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

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

Assume the $2+3x^3 =t$

$\therefore 9x^2dx=dt$

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

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

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

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

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

Answer:

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

Assume the $\log x =t$

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

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

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

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

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

Answer:

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

Assume the $9-4x^2 =t$

$\therefore -8xdx =dt$

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

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

Now, substituting the value of t;

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

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

Answer:

Given function $e ^{ 2 x +3 }$ ,

Assume the $2x+3 =t$

$\therefore 2dx =dt$

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

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

Now, substituting the value of t;

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

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

Answer:

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

Assume the $x^2=t$

$\therefore 2xdx =dt$

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

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

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

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

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

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

Answer:

Given,

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

Let's do the following substitution

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

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

$= e^{tan^{-1}x} + C$, where C is any constant value.

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

Answer:

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

Simplifying it by dividing both the numerator and the 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 $e^{x}+e^{-x} =t$

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

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

$= \int \frac{dt}{t}$

$= \log |t| +C$

Now, back substituting the value of t,

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

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

Answer:

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

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

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

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

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

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

Now, back substituting the value of t,

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

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

Answer:

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

Assume the $2x-3 =t$

$\therefore 2dx =dt$

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

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

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

Now, back substituting the value of t,

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

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

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

Answer:

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

Assume the $7-4x=t$

$\therefore -4dx =dt$

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

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

Now, substituting the value of t,

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

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

Answer:

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

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

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

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

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

Now, substituting the value of t,

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

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

Answer:

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

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

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

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

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

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

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

Now, substituting the value of t,

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

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

Answer:

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

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

Assume the $(1-\tan x)=t$

$\therefore -\sec^2xdx =dt$

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

$= -\int t^{-2} dt$

$= \frac{1}{t} +C$

Now, substituting the value of t,

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

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

Answer:

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

Assume the $\sqrt x =t$

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

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

$= 2\sin t +C$

Now, substituting the value of t,

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

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

Answer:

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

Assume the $\sin 2x = t$

$\therefore 2\cos 2x dx =dt$

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

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

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

Now, substituting the value of t,

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

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

Answer:

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

Assume the $1+\sin x =t$

$\therefore \cos x dx = dt$

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

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

$= 2\sqrt t +C$

Now, substituting the value of t,

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

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

Answer:

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

Assume the $\log \sin x =t$

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

$\cot x dx =dt$

$⇒ \int \cot x \log \sin x dx =\int t dt$

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

Now, substituting the value of t,

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

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

Answer:

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

Assume the $1+\cos x =t$

$\therefore -\sin x dx =dt$

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

$= -\log|t| +C$

Now, substituting the value of t,

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

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

Answer:

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

Assume the $1+\cos x =t$

$\therefore -\sin x dx =dt$

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

$= -\int t^{-2}dt$

$= \frac{1}{t}+C$

Now, substituting the value of t,

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

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

Answer:

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

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

Now solving the assumed integral,

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

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

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

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

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

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

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

$⇒ (\cos x -\sin x)dx =dt$

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

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

Now, substituting the value of t,

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

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

Answer:

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

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

Now solving the assumed integral,

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

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

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

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

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

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

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

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

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

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

Now, back substituting the value of t,

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

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

Answer:

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

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

Now solving the assumed integral,

Multiplying the 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,

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

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

Answer:

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

Assume that $1+\log x =t$

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

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

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

Now, back substituting the value of t,

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

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

Answer:

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

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

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

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,

$= \frac{(x+\log x )^3}{3}+C$, where C is any constant value.

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

Answer:

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

Assume that $x^4 =t$

$\therefore 4x^3 dx =dt$

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

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

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

So, from 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,

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

Substituting the value of t,

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

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

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

Answer:

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

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

Now, differentiating both sides, we get

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

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

$= \log t +C$

Substituting the value of t,

$= \log (x^{10}+10^x) +C$, where C is any constant value.

Therefore the correct answer is D.

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

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

Answer:

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

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

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

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

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

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

Therefore, the correct answer is B

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

Answer:

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

Using the trigonometric identity $\sin^2x=\frac{1-\cos2x}{2}$

We can write the given question as

$=\frac{1-\cos 2(2x+5)}{2}$ $= \frac{1-\cos (4x+10)}{2}$

Now, $\int \frac{1-\cos (4x+10)}{2}dx\\ =\frac{1}{2}\int dx - \frac{1}{2}\int \cos(4x+10)dx$

$=\frac{x}{2}-\frac{1}{2}[\frac{\sin(4x+10)}{4}]\\ =\frac{x}{2}-\frac{\sin(4x+10)}{8}+C$

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

Answer:

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

The given integral can be written as

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

$=\frac{1}{2}\int \sin(7x)-\sin(x)\ dx\\ =1/2[\int \sin (7x) dx-\int \sin x\ dx]$

$=\frac{1}{2}[(-1/7)\cos 7x+\cos x+ C]$

$= \frac{\cos x}{2}-\frac{\cos 7x}{14}+C$

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

Answer:

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

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

Again use the same identity mentioned in the first line

$= \frac{1}{2}\int (\cos 2x.\cos 10x+\cos 2x. \cos 2x)dx$

$=\frac{1}{2}\int\frac{1}{2}({\cos12x +\cos 8x})dx+\frac{1}{2}\int (\frac{1+\cos 4x}{2})dx$

$=\frac{\sin 12x}{48}+\frac{\sin 8x}{32}+\frac{\sin 4x}{16}+ \frac{x}{4}+C$

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

Answer:

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

The integral can be written as

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

Let $\\\cos (2x+1) =t\\ ⇒\sin (2x+1)dx = -dt/2$

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

Now, replacing the value of t, we get;

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

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

Answer:

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

Rewrite the integral as follows

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

Let $\cos x = t \Rightarrow \sin x dx =-dt$

$\\=-\int t^3(1-t^2)dt\\ =-\int(t^3-t^5)dt\\ =-[\frac{t^4}{4}]+[\frac{t^6}{6}] +C$

$=\frac{\cos^6x}{6}-\frac{cos^4x}{4}+C$ ......(replacing the value of t as $\cos\ x$)

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

Answer:

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

We can write the integral as follows

$\int \sin x.\sin 2x\sin 3x\ dx = \int \sin x\frac{1}{2}[\cos x-\cos 5x]dx$

$=\frac{1}{2} \int [\sin x.\cos x-\sin x.\cos 5x]dx$

$=\frac{1}{2}\int \frac{\sin 2x}{2}dx-\frac{1}{2}\int \sin x. \cos 5x\ dx$

$=-\frac{\cos 2x}{8}-\frac{1}{4}\int[\sin 6x -\sin 4x]$

$=-\frac{\cos 2x}{8}-\frac{1}{4}[\frac{-\cos 6x}{6}+\frac{\cos 4x}{4}]$

$=-\frac{\cos 2x}{8}+\frac{\cos 6x}{24}-\frac{\cos 4x}{16}+C$

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

Answer:

Using identity

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

We can write the following integral as

$\sin 4x \sin 8x=\frac{1}{2}\int(\cos 4x - \cos 12x) dx$

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

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

Answer:

We know the identities

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

Using the above relations we can write

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

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

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

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

Answer:

The integral is rewritten using trigonometric identities

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

Now, $\int \frac{1}{2}[1-\tan^2x/2] dx =\frac{1}{2}\int 1-[\sec^2\frac{x}{2}-1]=\frac{1}{2}\int 2-\sec^2\frac{x}{2}\\=x-\tan\frac{x}{2}+c$

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

Answer:
$\sin ^ 4 x$ can be written as follows using trigonometric identities

$=\sin^2x.\sin^2x =\frac{1}{4}(1-\cos 2x)^{2}\\ =\frac{1}{4}(1+\cos^22x-2\cos 2x)$

$=\frac{1}{4}(1+\frac{1}{2}(1+\cos 4x)-2\cos 2x)\\ =\frac{3}{8}+\frac{\cos 4x}{8}-\frac{\cos 2x}{2}$

$\Rightarrow \int \sin^4x\ dx = \int \frac{3}{8}dx+\frac{1}{8}\int\cos 4x\ dx -\frac{1}{2}\int\cos 2x\ dx$

$= \frac{3x}{8}+\frac{\sin 4x}{32} -\frac{\sin 2x}{4}+C$

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

Answer:

$\cos^42x=\cos^32x\cos2x$

Now using the identity

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

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

Now using the two identities below,

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

the value

$\cos^42x=\cos^32x\cos2x\\=\frac{\cos6x\cos2x+3\cos^22x}{4}=\frac{\cos 4x+\cos8x}{8}+\frac{3}{4}\frac{1+\cos4x}{2}$ .

The integral of the given function can be written as

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

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

Answer:

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

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

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

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

$\\= \int 1dx-\int\cos x\ dx\\ =x-\sin x+C$

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

Answer:

We know that,

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

Using this identity we can rewrite the given integral as

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

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

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

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

Answer:

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

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

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

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

Answer:

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

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

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

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

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

Answer:

The given question can be rearranged using trigonometric identities

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

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

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

now the final solution is,

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

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

Answer:

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

now splitting the terms we can write

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

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

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

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

Answer:

The integral of the above equation is

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

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

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

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

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

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

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

Answer:

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

$\therefore \int \frac{\cos 2x}{(\cos x+\sin x)^2}=\frac{1}{2}\int \frac{1}{t}dt$

$\\\Rightarrow \frac{1}{2}\log\left | t \right |+C$
$ \Rightarrow \frac{1}{2}\log\left | 1+\sin 2x \right |+C$

$=log|sin^2x+cos^2x+2sinxcosx|+C$
$=\frac{1}{2}log|(sinx+cosx)^2|+C$
$=log|sinx+cosx|+C$

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

Answer:

Using the trigonometric identities, we can evaluate the following integral as follows

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

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

Answer:

Using the trigonometric identities following integrals can be simplified as follows

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

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

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

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

$=\frac{1}{\sin(a-b)}\int tan(x-b)-\tan (x-a)dx$

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

Question 23: Choose the correct answer

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

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

Answer:

The correct option is (A)

On reducing, the above integral becomes $\sec^2x-csc^2x$

$\int\sec^2x-csc^2x\ dx = \tan x+ \cot x+C$

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

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

Answer:

The correct option is (B)

Let $e^xx = t$ .

So, $(e^x.x+ 1.e^x)dx = dt$

(1+ $x$ ) $e^x\ dx =dt$

Therefore, $\int \frac{e^x(1+x)}{\cos^2(e^x.x)}dx =\int\frac{dt}{\cos^2t}$

$\\=\int \sec^2t dt\\ =\tan t +C\\ =\tan(e^x.x)+C$

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

Answer:

The given integral can be calculated as follows

Let $x^3 = t$
Therefore, $3x^2 dx =dt$

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

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

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

Answer:

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

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

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

Answer:

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

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

using the identity

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

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

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

Answer:

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

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

The above result is obtained using the identity

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

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

Answer:

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

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

The integration can be done as follows

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

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

Answer:

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

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

using the special identities we can simplify the integral as follows

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

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

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

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

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

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

Answer:

The integration can be down as follows

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

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

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

Answer:

The integral can be evaluated as follows

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

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

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

Answer:

$\frac{1 }{ \sqrt { x ^ 2 + 2 x + 2 }}$
the above equation can be also written as,
$=\int\frac{1}{\sqrt{(1+x)^2+1^2}}dx$
let 1+x = t
then dx = dt
Therefore,
$=\int\frac{1}{\sqrt{t^2+1^2}}dx\\ =\log\left | t+\sqrt{t^2+1} \right |+C$
$ =\log\left | (1+x)+\sqrt{(1+x)^2+1} \right |+C\\ =\log\left | (1+x)+\sqrt{(x^2+2x+2} \right |+C$

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

Answer:

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

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

Answer:

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

Therefore

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

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

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

Answer:

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

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

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

Answer:

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

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

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

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

Answer:

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

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

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

Answer:

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

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

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

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

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

Answer:

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

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

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

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

Answer:

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

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

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

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

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

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

Answer:

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

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

Considering $I_1$

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

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

Now consider $I_2$

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

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

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

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

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

Answer:

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

A = -1/2 and B = 4

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

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

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

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

put the value of $I_1$ and $I_2$

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

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

Answer:

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

take $I_1$

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

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

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

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

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

Answer:

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

By comparing the coefficients and constant term, we get;

A = 1/2 and B =4

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

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

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

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

$I=I_1+I_2$

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

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

Answer:

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

A =5/2 and B = -7

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

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

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

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

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

Question 24: Choose the correct answer

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

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

Answer:

The correct option is (B)

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

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

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

Answer:

The following integration can be done as

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

The correct option is (B).

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

Answer:

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

Partial function of this function:

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

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

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

$A+B =1$

$2A+B =0$

On solving, we get

$A=-1\ and\ B =2$

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

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

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

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

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

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

Answer:

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

The partial function of this function:

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

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

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

$A+B =1$

$-3A+3B =1$

On solving, we get

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

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

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

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

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

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

Answer:

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

Partial function of this function:

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

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

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

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

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

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

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

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

Answer:

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

Partial function of this function:

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

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

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

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

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

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

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

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

Answer:

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

Partial function of this function:

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

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

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

$A ={-2},\ B=4$

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

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

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

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

Answer:

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

Integral is not a proper fraction so,

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

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

Partial function of this function:

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

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

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

$A =2,\ B=3$

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

No, substituting in equation (1) we get

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

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

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

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

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

Answer:

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

Partial function of this function:

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

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

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

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

$A+C = 0$

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

On solving these equations, we get

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

From equation (1), we get

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

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

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

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

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

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

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

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

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

Answer:

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

Partial function of this function:

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

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

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

$B =\frac{1}{3}$

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

$A+C=0$

$-2A+2B+C = 0$

then after solving, we get

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

Therefore,

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

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

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

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

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

Answer:

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

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

Partial function of this function:

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

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

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

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

$B =4$

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

$A+C=0$

$B-2C =3$

then after solving, we get

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

Therefore,

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

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

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

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

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

Answer:

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

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

The partial function of this function:

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

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

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

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

Therefore,

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

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

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

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

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

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

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

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

Answer:

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

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

The partial function of this function:

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

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

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

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

Therefore,

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

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

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

Answer:

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

As the given integral is not a proper fraction.

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

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

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

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

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

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

Therefore,

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

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

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

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

Answer:

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

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

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

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

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

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

Solving these equations, we get

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

Therefore,

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

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

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

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

Answer:

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

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

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

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

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

Solving these equations, we get

$B=-7$

Therefore,

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

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

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

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

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

Answer:

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

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

The partial fraction of above equation,

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

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

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

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

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

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

Solving these equations, we get

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

Therefore,

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

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

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

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

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

Answer:

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

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

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

Putting $x^n =t$

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

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

Partial fraction of above equation,

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

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

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

$A=1\ and\ B=-1$

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

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

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

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

$= \frac{1}{n} \log|\frac{x^n}{x^n+1}| +C$

Question 17: Integrate the rational functions $\frac{\cos x }{(1- \sin x )( 2- \sin x )}$

[Hint : Put $\sin x = t$ ]

Answer:

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

Applying the given hint: putting $\sin x =t$

We get, $\cos x dx =dt$

$\therefore \int \frac{\cos x }{(1- \sin x )( 2- \sin x )}dx = \int \frac{dt}{(1-t)(2-t)}$

Partial fraction of above equation,

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

$1 = A(2-t)+B(1-t)$ ................(1)

Now, substituting $t = 2\ and\ 1$ in equation (1), we get

$A=1\ and\ B=-1$

$\therefore \frac{1}{(1-t)(2-t)} = \frac{1}{(1-t)} - \frac{1}{(2-t)}$

$\implies \int \frac{\cos x }{(1-\sin x)(2-\sin x )}dx = \int \left \{ \frac{1}{1-t}-\frac{1}{(2-t)} \right \}dt$

$= -\log|1-t| +\log|2-t| +C$

$= \log\left | \frac{2-t}{1-t} \right |+C$

Back substituting the value of t in the above equation, we get

$= \log\left | \frac{2-\sin x}{1- \sin x} \right |+C$

Question 18: Integrate the rational functions $\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )}$

Answer:

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

We can rewrite it as: $\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \frac{(4x^2+10)}{(x^2+3)(x^2+4)}$

Partial fraction of above equation,

$\frac{(4x^2+10)}{(x^2+3)(x^2+4)} =\frac{Ax+B}{(x^2+3)}+\frac{Cx+D}{(x^2+4)}$

$4x^2+10 = (Ax+B)(x^2+4)+(Cx+D)(x^2+3)$

$4x^2+10 = Ax^3+4Ax+Bx^2+4B+Cx^3+3Cx+Dx^2+3D$

$4x^2+10 = (A+C)x^3+(B+D)x^2+(4A+3C)x+(3D+4B)$

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

$A+C=0$ , $B+D = 4$ , $4A+3C = 0$ , $4B+3D =10$

After solving these equations, we get

$A= 0, B =-2, C=0,$ and $D=6$

$\therefore \frac{4x^2+10}{(x^2+3)(x^2+4)} = \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)}$

$\frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} = 1- \left ( \frac{-2}{(x^2+3)} + \frac{6}{(x^2+4)} \right )$

$\implies \int \frac{( x^2 +1 )( x^2 +2 )}{( x^2 +3 )( x^2 +4 )} dx= \int \left \{ 1+ \frac{2}{(x^2+3)} - \frac{6}{(x^2+4)} \right \}dx$

$= \int \left \{ 1+ \frac{2}{(x^2+(\sqrt3)^2)} - \frac{6}{(x^2+2^2)} \right \}dx$

$= x+2\left ( \frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt 3} \right ) - 6\left ( \frac{1}{2}\tan^{-1}\frac{x}{2} \right )+C$

$= x+\frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3} -3\tan^{-1}\frac{x}{2}+C$

Question 19: Integrate the rational functions $\frac{2x }{( x^2 +1)( x^2 +3)}$

Answer:

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

Taking $x^2 = t \Rightarrow 2xdx=dt$

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

The partial fraction of above equation,

$\frac{1}{(t+3)(t+3)} = \frac{A}{(t+1)}+\frac{B}{(t+3)}$

$1= A(t+3)+B(t+1)$ ..............(1)

Now, substituting $t = -3\ and\ t = -1$ in equation (1), we get

$A =\frac{1}{2}\ and\ B = -\frac{1}{2}$

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

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

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

$= \frac{1}{2}\log\left | \frac{t+1}{t+3} \right | +C$

$= \frac{1}{2}\log\left | \frac{x^2+1}{x^2+3} \right | +C$

Question 20: Integrate the rational functions $\frac{1}{x (x^4 -1)}$

Answer:

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

So, we multiply numerator and denominator by $x^3$ , to obtain

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

$\therefore \int \frac{1}{x(x^4-1)}dx =\int\frac{x^3}{x^4(x^4-1)}dx$

Now, putting $x^4 = t$

we get, $4x^3dx =dt$

Taking $x^2 = t \Rightarrow 2xdx=dt$

$\therefore \int \frac{1}{x(x^4-1)}dx =\frac{1}{4}\int \frac{dt}{t(t-1)}$

Partial fraction of above equation,

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

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

Now, substituting $t = 0\ and\ t = 1$ in equation (1), we get

$A = -1\ and\ B=1$

$\Rightarrow \frac{1}{t(t+1)} = -\frac{1}{t}+\frac{1}{t-1}$

$\Rightarrow \int \frac{1}{x(x^4+1)}dx =\frac{1}{4}\int \left \{ \frac{-1}{t}+\frac{1}{t-1} \right \}dt$

$= \frac{1}{4} \left [ -\log|t|+\log|t-1| \right ]+C$

$= \frac{1}{4}\log\left | \frac{t-1}{t} \right |+C$

Back substituting the value of t,

$=\frac{1}{4}\log \left | \frac{x^4-1}{x^4} \right | +C$

Question 21: Integrate the rational functions $\frac{1}{( e ^x-1)}$ [Hint : Put $e ^x= t$ ]

Answer:

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

So, applying the hint: Putting $e^x = t$

Then $e^x dx= dt$

$\int \frac{1}{( e ^x-1)}dx = \int\frac{1}{t-1}\times\frac{dt}{t} = \int \frac{1}{t(t-1)}dt$

Partial fraction of above equation,

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

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

Now, substituting $t = 0\ and\ t = 1$ in equation (1), we get

$A = -1\ and\ B=1$

$\Rightarrow \frac{1}{t(t+1)} = -\frac{1}{t}+\frac{1}{t-1}$

$\implies \int \frac{1}{t(t-1)}dt = \log \left | \frac{t-1}{t} \right |+C$

Now, back substituting the value of t,

$= \log \left | \frac{e^x-1}{e^x} \right |+C$

Question 22: Choose the correct answer $\int \frac{x dx }{( x-1)(x-2) } \: \: equals$

$A ) \log |\frac{(x-1)^2}{x-2}| + C \\\\ B) \log |\frac{(x-2)^2}{x-1}| + C \\\\ C ) \log |(\frac{x-1}{x-2})^2| + C \\\\ D ) \log |{(x-1)^2}({x-2})| + C$

Answer:

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

Partial fraction of above equation,

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

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

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

$A = -1\ and\ B=2$

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

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

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

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

Therefore, the correct answer is B.

Question 23: Choose the correct answer $\int \frac{dx}{x ( x ^2+1)} \: \: equals$

$A ) \log |x| - \frac{1}{2} \log ( x^2 +1 ) + C$
$B ) \log |x|+ \frac{1}{2} \log ( x^2 +1 ) + C$
$C )- \log |x| + \frac{1}{2} \log ( x^2 +1 ) + C$
$D ) \frac{1}{2}\log |x| +\log ( x^2 +1 ) + C$

Answer:

Given integral $\int \frac{dx}{x ( x ^2+1)}$

Partial fraction of above equation,

$\frac{1}{x ( x ^2+1)} = \frac{A}{x}+\frac{Bx+c}{x^2+1}$

$1= A(x^2+1)+(Bx+C)x$

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

$A+B = 0$ , $C=0$ , $A=1$

We have the values, $A = 1\ and\ B=-1,\ and\ C=0$

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

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

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

Therefore, the correct answer is A.

Question 1: Integrate the functions $x \sin x$

Answer:

The given function is
$f(x)=x \sin x$
We will use the integration by parts method.
$\int x\sin x = x.\int \sin xdx - \int(\frac{d(x)}{dx}.\int sin x dx)dx$
$\int x\sin x = x.(-\cos x)- \int (1.(-\cos x))dx\\ \\ \int x\sin x= -x\cos x+\sin x + C$
Therefore, the answer is $-x\cos x+\sin x + C$

Question 2: Integrate the functions $x \sin 3x$

Answer:

The given function is
$f(x)=x \sin 3x$
We will use the integration by parts method.
$\int x\sin 3x = x.\int \sin 3xdx - \int(\frac{d(x)}{dx}.\int sin 3x dx)dx$
$\int x\sin 3x = x.(\frac{-\cos 3x}{3})- \int (1.(\frac{-\cos 3x}{3}))dx$
$\int x\sin 3x= -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C$
Therefore, the answer is $-\frac{x\cos 3x}{3}+\frac{\sin 3x}{9} + C$

Question 3: Integrate the functions $x ^ 2 e ^x$

Answer:

The given function is
$f(x)=x^2e^x$
We will use the 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.n
We will get,
$\int x^2.e^x dx =x^2e^x -2xe^x+ 2e^x + C\\ \int x^2.e^x dx = e^x(x^2-2x+2)+ C$
Therefore, answer is $e^x(x^2-2x+2)+ C$

Question 4: Integrate the functions $x \log x$

Answer:

The given function is
$f(x)=x.\log x$
We will use the integration by parts method.
$\int x.\log xdx= \log x.\int xdx - \int(\frac{d(\log x)}{dx}.\int x dx)dx$
$\int x\log xdx = \log x.\frac{x^2}{2}- \int (\frac{1}{x}.\frac{x^2}{2})dx$
$ \int x\log xdx = \log x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log xdx = \log x.\frac{x^2}{2}- \frac{x^2}{4}+ C$
Therefore, the answer is $\frac{x^2}{2}\log x- \frac{x^2}{4}+ C$

Question 5: Integrate the functions $x \log 2x$

Answer:

The given function is
$f(x)=x.\log 2 x$
We will use the integration by parts method.
$\int x.\log 2xdx= \log 2x.\int xdx - \int(\frac{d(\log 2x)}{dx}.\int x dx)dx$
$ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \int (\frac{2}{2x}.\frac{x^2}{2})dx$
$\int x\log 2xdx = \log 2x.\frac{x^2}{2}- \int \frac{x}{2}dx\\ \int x\log 2xdx = \log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C$
Therefore, the answer is $\log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C$

Question 6: Integrate the functions $x^ 2 \log x$

Answer:

The given function is
$f(x)=x^2.\log x$
We will use the integration by parts method.
$\int x^2.\log xdx= \log x.\int x^2dx - \int(\frac{d(\log x)}{dx}.\int x^2 dx)dx$
$ \int x^2\log xdx = \log x.\frac{x^3}{3}- \int (\frac{1}{x}.\frac{x^3}{3})dx$
$ \int x^2\log xdx = \log x.\frac{x^3}{3}- \int \frac{x^2}{3}dx$
$ \int x^2\log xdx = \log x.\frac{x^3}{3}- \frac{x^3}{9}+ C$
Therefore, the answer is $\log x.\frac{x^3}{3}- \frac{x^3}{9}+ C$

Question 7: Integrate the functions $x \sin ^{ -1} x$

Answer: The given function is
$f(x)=x.\sin^{-1} x$
We will use the integration by parts method.
$\int x.\sin^{-1} xdx= \sin^{-1} x.\int xdx - \int(\frac{d(\sin^{-1} x)}{dx}.\int x dx)dx$
$\int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
Now, we need to integrate $\int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \right )dx$
$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} \right )dx$
$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\left ( \int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx \right)$
$ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{1}{2}\left ( \frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x-\sin^{-1}x \right )$
$ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{x\sqrt{1-x^2}}{4} -\frac{\sin^{-1}x}{4}+C$
Put this value in our equation.
Therefore, the answer is $\int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x\sqrt{1-x^2}}{4}$

Question 8: Integrate the functions $x \tan ^{-1} x$

Answer:

The given function is
$f(x)=x.\tan^{-1} x$
We will use the integration by parts method.
$\int x.\tan^{-1} xdx= \tan^{-1} x.\int xdx - \int(\frac{d(\tan^{-1} x)}{dx}.\int x dx)dx$
$ \int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}- \int (\frac{1}{1+x^2}.\frac{x^2}{2})dx$
$\int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\int \left ( \frac{x^2+1}{1+x^2}-\frac{1}{1+x^2} \right )dx$
$\int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\int \left ( 1-\frac{1}{1+x^2} \right )dx$
$\int x\tan^{-1} xdx = \tan^{-1} x.\frac{x^2}{2}-\frac{1}{2}\left ( x- \tan^{-1}x \right )+C$
$\int x\tan^{-1}xdx = \frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C$
Put this value in our equation.
$\int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \frac{x}{4\sqrt{1-x^2}}-\frac{\sin^{-1}x}{4}+C$
$\int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x}{4\sqrt{1-x^2}}$
Therefore, the answer is $\frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C$

Question 9: Integrate the functions $x\cos ^{ -1} x$

Answer:

The given function is
$f(x)=x.\cos^{-1} x$
We will use the integration by parts method.
$\int x.\cos^{-1} xdx= \cos^{-1} x.\int xdx - \int(\frac{d(\cos^{-1} x)}{dx}.\int x dx)dx$
$\int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}- \int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
Now, we need to integrate $\int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$
$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} \right )dx$
$ \int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\int \left ( \sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} \right )dx$
$\int \frac{-x^2}{2\sqrt{1-x^2}}dx=\frac{1}{2}\left ( \int \sqrt{1-x^2}dx-\int \frac{1}{\sqrt{1-x^2}}dx \right )$
$\int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{1}{2}\left ( \frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}\cos^{-1}x+\cos^{-1}x \right )$
$ \int \frac{-x^2}{2\sqrt{1-x^2}}dx = \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C$
Put this value in our equation.
$\int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}-\left ( \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C \right )$
$ \int x\cos^{-1} xdx =\frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}$
Therefore, the answer is $\frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}$

Question 10: Integrate the functions $( \sin ^{-1}x ) ^ 2$

Answer:

The given function is
$f(x)=( \sin ^{-1}x ) ^ 2$
We will use the integration by parts method.
$\int (\sin^{-1}x)^2= (\sin^{-1}x)^2.\int 1dx-\int \left ( \frac{d( (\sin^{-1}x)^2)}{dx} .\int 1dx\right )dx$
$\int (\sin^{-1}x)^2 = (\sin^{-1}x)^2.x-\int \left ( \sin^{-1}.\frac{2x}{\sqrt{1-x^2}} \right )dx$
$\int (\sin^{-1}x)^2 = (\sin^{-1}x)^2.x + \left [ \sin^{-1}x.\int \frac{-2x}{\sqrt{1-x^2}}dx-\int \left ( \frac{d(\sin^{-1}x)}{dx}. \int \frac{-2x}{\sqrt{1-x^2}}dx\right ) \right ]$
$= (\sin^{-1}x)^2.x + \left [ \sin^{-1}x.2\sqrt{1-x^2}- \int \frac{1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]$
$= (\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C$
Therefore, answer is $(\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C$

Question 11: Integrate the functions $\frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}$

Answer:

Consider $\int \frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}dx =I$

So, we have then: $I = \frac{-1}{2}\int \frac{-2x}{\sqrt{1-x^2}}. \cos^{-1}x dx$

After taking $\cos ^{-1}x$ as a first function and $\left ( \frac{-2x}{\sqrt{1-x^2}} \right )$ as second function and integrating by parts, we get

$I =-\frac{1}{2}\left [ \cos^{-1}x\int\frac{-2x}{\sqrt{1-x^2}}dx - \int\left \{ \left ( \frac{d}{dx}\cos^{-1}x \right )\int \frac{-2x}{\sqrt{1-x^2}}dx \right \}dx \right ]$
$=-\frac{1}{2}\left [ \cos^{-1}x.2{\sqrt{1-x^2}} + \int \frac{-1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]$

$=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-\int2dx \right ]$

$=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-2x \right ]+C$

Or, $-\sqrt{1-x^2} \cos ^{-1} x+x+C$

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

Answer:

Consider $x \sec ^2 x$

So, we have then: $I =\int x\sec^2 x dx$

After taking $x$ as a first function and $\sec^2x$ as a second function and integrating by parts, we get

$I =x\int \sec^2 x dx -\int \left \{ \left ( \frac{d}{dx}x \right )\int \sec^2 x dx \right \}dx$

$= x\tan x -\int1.\tan x dx$

$= x\tan x +\log|\cos x | +C$

Question 13: Integrate the functions $\tan ^{-1} x$

Answer:

Consider $\tan ^{-1} x$

So, we have then: $I =\int 1.\tan^{-1}x dx$

After taking $\tan^{-1}x$ as a first function and $1$ as second function and integrating by parts, we get

$I = \tan^{-1}x \int 1dx -\int \left \{ \left ( \frac{d}{dx}\tan^{-1}x \right )\int1.dx \right \}dx$

$= \tan^{-1}x.x -\int \frac{1}{1+x^2}.xdx$

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

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

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

Question 14: Integrate the functions $x ( \log x )^ 2$

Answer:

Consider $x ( \log x )^ 2$

So, we have then: $I = \int x(\log x)^2 dx$

After taking $(\log x )^2$ as a first function and $x$ as second function and integrating by parts, we get

$I = (\log x )^2 \int xdx -\int \left \{ \left ( \frac{d}{dx} (\log x)^2 \right )\int x.dx \right \}dx$

$= (\log x)^2 .\frac{x^2}{2} - \int \frac{2\log x }{x}.\frac{x^2}{2} dx$

$= (\log x)^2 .\frac{x^2}{2} - \int x\log x dx$

$= (\log x)^2 .\frac{x^2}{2} - \left ( \frac{x^2 \log x }{2} -\frac{x^2}{4} \right )+C$

Question 15: Integrate the functions $( x^2 + 1 ) \log x$

Answer:

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

So, we have then: $I = \int (x^2+1) \log x dx = \int x^2 \log x dx +\int \log x dx$

Let us take $I = I_{1} +I_{2}$ ....................(1)

Where, $I_{1} = \int x^2\log x dx$ and $I_{2} = \int \log x dx$

So, $I_{1} = \int x^2\log x dx$

After taking $\ logx$ as a first function and $x^2$ as a 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 a second function and integrating by parts, we get

$I_{2} = \log x \int 1.dx - \int \left \{ \left ( \frac{d}{dx}\log x \right ) \int 1.dx \right \}dx$

$= \log x .x -\int \frac{1}{x}. xdx$

$= x\log x -\int 1 dx$

$= x\log x -x +C_{2}$ ................(3)

Now, using the two equations (2) and (3) in (1), we get,

$I = \frac{x^3}{3}\log x -\frac{x^3}{9} +C_{1} +x\log x - x +C_{2}$

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

$=\left ( \frac{x^3}{3}+x \right ) \log x -\frac{x^3}{9} -x+C$

Question 16: Integrate the functions $e ^ x ( \sin x + \cos x )$

Answer:

Let suppose
$I =$ $e ^ x ( \sin x + \cos x )$
$f(x) = \sin x \Rightarrow f'(x) = \cos x$
We know that,
$I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C$
Thus, the solution of the given integral is given by

$\therefore I = e^x\sin x +C$

Question 17: Integrate the functions $\frac{x e ^x }{( 1+ x )^2}$

Answer:

$\frac{x e ^x }{( 1+ x )^2}$
Let suppose
$I = \int \frac{e^x(x)}{(1+x)^2}dx$
By rearranging the equation, we get
$\Rightarrow \int e^x[\frac{1}{1+x}-\frac{1}{(1+x)^2}]dx$
let
$f(x)=\frac{1}{1+x} \Rightarrow f'(x)= -\frac{1}{(1+x)^2}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$
Therefore, the solution of the given integral is

$I = \frac{e^x}{1+x}+C$

Question 18: Integrate the functions $e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )$

Answer:

Let
$I =e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )$
substitute $1 =\sin ^2\frac{x}{2}+\cos^2\frac{x}{2}$ and $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$

$\\\Rightarrow e^x(\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}})\\ =e^x(\frac{1}{2}\sec^2\frac{x}{2}+\tan\frac{x}{2})\\$
let
$f(x) =\tan\frac{x}{2} \Rightarrow f'(x)=\frac{1}{2}\sec^2\frac{x}{2}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$
Therefore, the solution of the given integral is

$I = e^x\tan\frac{x}{2} +C$

Question 19: Integrate the functions $e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )$

Answer:

$e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )$
It is known that
$\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$

let
$f(x)=\frac{1}{x}\Rightarrow f'(x)=-\frac{1}{x^2}$
Therefore, the required solution of the integral given above is
$I = e^x.\frac{1}{x}+C$

Question 20: Integrate the functions $\frac{( x-3)e ^x }{( x-1)^3}$

Answer:

$\frac{( x-3)e ^x }{( x-1)^3}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$

So, by adjusting the given equation, we get
$\int\frac{( x-3)e ^x }{( x-1)^3} =\int e^x(\frac{x-1-2}{(x-1)^3}) =\int e^x({\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3})}dx$

to let
$f(x)=\frac{1}{(x-1)^2}\Rightarrow f'(x)=-\frac{2}{(x-1)^3}$
Therefore the required solution of the given $I=\frac{e^x}{(x-1)^2}+C$ integral is

Question 21: Integrate the functions $e ^{ 2x } \sin x$

Answer:

Let
$I =e ^{ 2x } \sin x$
By using integration by parts, we get

$\\=\sin x\int e ^{ 2x }dx-\int(\frac{d}{dx}\sin x.\int e^{2x}dx)\ dx$
$=\frac{\sin x.e^{2x}}{2}-\frac{1}{2}\int e^{2x}.\cos x\ dx$
$=\frac{\sin x.e^{2x}}{2}-\frac{1}{2}[\cos x\int e^{2x}dx-\int (\frac{d}{dx}\cos x.\int e^{2x}dx)\ dx]$
$=\frac{\sin x.e^{2x}}{2}-\frac{1}{2}[\cos x.\frac{e^x}{2}+\frac{1}{2}\int e^{2x}\sin x dx]$
$=\frac{\sin x.e^{2x}}{2}-\frac{1}{4}\cos x.e^{2x}-\frac{1}{4}I$
$\Rightarrow \frac{5}{4}I =\frac{\sin x.e^{2x}}{2}-\frac{1}{4}\cos x.e^{2x}$
$I = \frac{e^{2x}}{5}[2\sin x-\cos x]+C$

Question 22: Integrate the functions $\sin ^ { -1} \left ( \frac{2x}{1+x^2 } \right )$

Answer:

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

$\int \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$
let $x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta$

$\\=\int\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta$
$ =\int\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta\\ =\int2\theta \sec^2\theta d\theta\\$
Taking $\theta$ as a first function and $\sec^2\theta$ as a second function, by using by parts method

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

Question 23: Choose the correct answer

$\int x ^ 2 e ^{x ^3 } dx \: \: equals$

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

Answer:

The integration can be done ass follows

Let $x^3 =t\Rightarrow 3x^2dx=dt$
$\Rightarrow I =\frac{1}{3}\int e^tdt =\frac{1}{3}e^t+C=\frac{1}{3}e^{x^3}+C$

Question 24: Choose the correct answer

$\int e ^ x \sec ( 1+ \tan x ) dx \: \: \: equals$

$A ) e ^ x \cos x + C \\\\ B) e ^ x \sec x + C \\\\ C ) e ^ x \sin x + C\\\\D ) e ^ x \tan x + C$

Answer:

We know that,
$I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C$
from the 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

Question 1: Integrate the functions in Exercises 1 to 9.

$\sqrt{4 - x^2}$

Answer:

Given function $\sqrt{4 - x^2}$ ,

So, let us consider the function to be;

$I = \int \sqrt{4-x^2}dx$

$= \int \sqrt{(2)^2-x^2}dx$

Then it is known that, $= \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

Therefore, $I = \frac{x}{2}\sqrt{4-x^2} +\frac{4}{2}\sin^{-1}{\frac{x}{2}}+C$

$= \frac{x}{2}\sqrt{4-x^2} +2\sin^{-1}{\frac{x}{2}}+C$

Question 2: Integrate the functions in Exercises 1 to 9.

$\sqrt{1 - 4x^2}$

Answer:

Given function to integrate $\sqrt{1 - 4x^2}$

Now we can rewrite as

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

As we know the integration of this form is $\left [ \because \int \sqrt{a^2-x^2}dx =\frac{x}{2}\sqrt{a^2-x^2} +\frac{a^2}{2}\sin^{-1}\frac{x}{a} \right ]$

$= \frac{(\frac{2x}{2})\sqrt{1^2-(2x)^2}+\frac{1^2}{2}\sin^{-1}\frac{2x}{1}}{2\rightarrow Coefficient\ of\ x\ in\ 2x} +C$

$= \frac{1}{2}\left [ x\sqrt{1-4x^2}+\frac{1}{2}\sin^{-1}2x \right ]+C$

$= \frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\sin^{-1}2x+C$

Question 3: Integrate the functions in Exercises 1 to 9.

$\sqrt{x^2 + 4x + 6}$

Answer:

Given function $\sqrt{x^2 + 4x + 6}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2 + 4x + 6}dx$

$= \int\sqrt{(x^2 + 4x + 4)+2}dx = \int\sqrt{(x + 2)^2 +(\sqrt2)^2}dx$

And we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

$\Rightarrow I = \frac{x+2}{2}\sqrt{x^2+4x+6}+\frac{2}{2}\log\left | (x+2)+\sqrt{x^2+4x+6} \right |+C$

$= \frac{x+2}{2}\sqrt{x^2+4x+6}+\log\left | (x+2)+\sqrt{x^2+4x+6} \right |+C$

Question 4: Integrate the functions in Exercises 1 to 9.

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

Answer:

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

So, let us consider the function to be;

$I = \int\sqrt{x^2 + 4x + 1}dx$

$= \int\sqrt{(x^2 + 4x + 4)-3}dx = \int\sqrt{(x + 2)^2 -(\sqrt3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$\therefore I = \frac{x+2}{2}\sqrt{x^2+4x+1}-\frac{3}{2}\log\left | (x+2)+\sqrt{x^2+4x+1} \right |+C$

Question 5: Integrate the functions in Exercises 1 to 9.

$\sqrt{1-4x-x^2}$

Answer:

Given function $\sqrt{1-4x-x^2}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1-4x-x^2}dx$

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

$= \int\sqrt{(\sqrt5)^2 -(x+2)^2}dx$

And we know that, $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

$\therefore I = \frac{x+2}{2}\sqrt{1-4x-x^2}+\frac{5}{2}\sin^{-1}\left ( \frac{x+2}{\sqrt5} \right )+C$

Question 6: Integrate the functions in Exercises 1 to 9.

$\sqrt{x^2 + 4x - 5}$

Answer:

Given function $\sqrt{x^2 + 4x - 5}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2+4x-5}dx$

a $= \int\sqrt{(x^2+4x+4)-9}dx = \int\sqrt{(x+2)^2 -(3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}|+C$

$\therefore I = \frac{x+2}{2}\sqrt{x^2+4x-5}-\frac{9}{2}\log\left | (x+2)+ \sqrt{x^2+4x-5} \right |+C$

Question 7: Integrate the functions in Exercises 1 to 9.

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

Answer:

Given function $\sqrt{1 + 3x - x^2}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1+3x-x^2}dx$

$= \int\sqrt{(1-\left ( x^2-3x+\frac{9}{4}-\frac{9}{4} \right )}dx = \int \sqrt{\left ( 1+\frac{9}{4} \right )-\left ( x-\frac{3}{2} \right )^2}dx$ $= \int \sqrt{\left ( \frac{\sqrt{13}}{2} \right )^2-\left ( x-\frac{3}{2} \right )^2}dx$

And we know that, $\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$

$\therefore I = \frac{x-\frac{3}{2}}{2}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\left ( \frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}} \right )+C$

$= \frac{2x-3}{4}\sqrt{1+3x-x^2}+\frac{13}{8}\sin^{-1}\left ( \frac{2x-3}{\sqrt{13}} \right )+C$

Question 8: Integrate the functions in Exercises 1 to 9.

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

Answer:

Given function $\sqrt{x^2 + 3x}$ ,

So, let us consider the function to be;

$I = \int\sqrt{x^2+3x}dx$

$= \int\sqrt{x^2+3x+\frac{9}{4}-\frac{9}{4}}dx$

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

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

$\therefore I = \frac{x+\frac{3}{2}}{2}\sqrt{x^2+3x}-\frac{\frac{9}{4}}{2}\log \left | \left ( x+\frac{3}{2} \right )+\sqrt{x^2+3x} \right |+C$

$= \frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}\log\left | \left ( x+\frac{3}{2} \right )+\sqrt{x^2+3x} \right |+C$

Question 9: Integrate the functions in Exercises 1 to 9.

$\sqrt{1 + \frac{x^2}{9}}$

Answer:

Given function $\sqrt{1 + \frac{x^2}{9}}$ ,

So, let us consider the function to be;

$I = \int\sqrt{1+\frac{x^2}{9}}dx = \frac{1}{3}\int \sqrt{9+x^2}dx$

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

And we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

$\therefore I = \frac{1}{3}\left [ \frac{x}{2}\sqrt{x^2+9} +\frac{9}{2}\log|x+\sqrt{x^2+9}| \right ]+C$

$= \frac{x}{6}\sqrt{x^2+9} +\frac{3}{2}\log\left | x+\sqrt{x^2+9} \right |+C$

Question 10: Choose the correct answer in Exercises 10 to 11.

$\int \sqrt{1+x^2}dx$ is equal to

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

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

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

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

Answer:

As we know that, $\int \sqrt{x^2+a^2}dx = \frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log|x+\sqrt{x^2+a^2}| +C$

So, $\int \sqrt{1+x^2}dx = \frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\log|x+\sqrt{x^2+1}| +C$

Therefore, the correct answer is A.

Question 11: Choose the correct answer in Exercises 10 to 11.

$\int \sqrt{x^2 - 8x+7}dx$ is equal to

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

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

(C) $\frac{1}{2}(x-4)\sqrt{x^2-8x+7} -3\sqrt2\log\left|x-4+\sqrt{x^2 -8x+7}\right| +C$

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

Answer:

Given integral $\int \sqrt{x^2 - 8x+7}dx$

So, let us consider the function to be;

$I = \int\sqrt{x^2-8x+7}dx =\int\sqrt{(x^2-8x+16)-(9)}dx$

$=\int\sqrt{(x-4)^2-(3)^2}dx$

And we know that, $\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log|x+\sqrt{x^2-a^2}| +C$

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

Therefore, the correct answer is D.

Question 1: Evaluate the definite integrals in Exercises 1 to 20.

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

Answer:

Given integral: $I = \int_{-1}^{1} (x+1)dx$

Consider the integral $\int (x+1)dx$

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

So, we have the function of $x$ , $f(x) = \frac{x^2}{2}+x$

Now, by the Second fundamental theorem of calculus, we have

$I = f(1)-f(-1)$

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

Question 2: Evaluate the definite integrals in Exercises 1 to 20.

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

Answer:

Given integral: $I = \int_2^3\frac{1}{x}dx$

Consider the integral $\int_2^3\frac{1}{x}dx$

$\int \frac{1}{x}dx = \log|x|$

So, we have the function of $x$ , $f(x) = \log|x|$

Now, by the Second fundamental theorem of calculus, we have

$I = f(3)-f(2)$

$=\log|3|-\log|2| = \log \frac{3}{2}$

Question 3: Evaluate the definite integrals in Exercises 1 to 20.

$\int_1^2(4x^3-5x^2 + 6x +9)dx$

Answer:

Given integral: $I = \int_1^2(4x^3-5x^2 + 6x +9)dx$

Consider the integral $I = \int (4x^3-5x^2 + 6x +9)dx$

$\int (4x^3-5x^2 + 6x +9)dx = 4\frac{x^4}{4} -5\frac{x^3}{3}+6\frac{x^2}{2}+9x$

$= x^4 -\frac{5x^3}{3}+3x^2+9x$

So, we have the function of $x$ , $f(x) = x^4 -\frac{5x^3}{3}+3x^2+9x$

Now, by the Second fundamental theorem of calculus, we have

$I = f(2)-f(1)$

$=\left \{ 2^4-\frac{5(2)^3}{3}+3(2)^2+9(2)\right \} - \left \{ 1^4-\frac{5(1)^3}{3}+3(1)^2+9(1) \right \}$

$=\left \{ 16-\frac{40}{3}+12+18\right \} - \left \{ 1-\frac{5}{3}+3+9 \right \}$

$=\left \{ 46-\frac{40}{3}\right \} - \left \{ 13-\frac{5}{3}\right \}$

$=\left \{ 33-\frac{35}{3} \right \} = \left \{ \frac{99-35}{3} \right \}$

$= \frac{64}{3}$

Question 4: Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{4}\sin 2x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{4}\sin 2x dx$

Consider the integral $\int \sin 2x dx$

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

So, we have the function of $x$ , $f(x) = \frac{-\cos 2x }{2}$

Now, by the Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4})-f(0)$

$= \frac{-\cos 2(\frac{\pi}{4})}{2} + \frac{\cos 0}{2}$

$=\frac{1}{2} - 0$

$= \frac{1}{2}$

Question 5: Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{2}\cos 2x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{2}\cos 2x dx$

Consider the integral $\int \cos 2x dx$

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

So, we have the function of $x$ , $f(x) = \frac{\sin 2x }{2}$

Now, by the Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{2})-f(0)$

$= \frac{1}{2}\left \{ \sin 2(\frac{\pi}{2}) - \sin 0 \right \}$

$= \frac{1}{2}\left \{ 0 - 0 \right \} = 0$

Question 6: Evaluate the definite integrals in Exercises 1 to 20.

$\int_4^5 e^x dx$

Answer:

Given integral: $\int_4^5 e^x dx$

Consider the integral $\int e^x dx$

$\int e^x dx = e^x$

So, we have the function of $x$ , $f(x) = e^x$

Now, by the Second fundamental theorem of calculus, we have

$I = f(5) -f(4)$

$= e^5 -e^4$

$= e^4(e-1)$

Question 7: Evaluate the definite integrals in Exercises 1 to 20.

$\int^\frac{\pi}{4}_0 \tan x dx$

Answer:

Given integral: $\int^\frac{\pi}{4}_0 \tan x dx$

Consider the integral $\int \tan x dx$

$\int \tan x dx = -\log|\cos x |$

So, we have the function of $x$ , $f(x) = -\log|\cos x |$

Now, by the Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(0)$

$= -\log\left | \cos \frac{\pi}{4} \right | +\log|\cos 0|$

$= -\log\left | \cos \frac{1}{\sqrt2} \right | +\log|1|$

$= -\log\left | \frac{1}{\sqrt2} \right | + 0 = -\log (2)^{-\frac{1}{2}}$

$= \frac{1}{2}\log (2)$

Question 8: Evaluate the definite integrals in Exercises 1 to 20.

$\int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec}xdx$

Answer:

Given integral: $\int_\frac{\pi}{6}^\frac{\pi}{4}\textup{cosec}xdx$

Consider the integral $\int\textup{cosec}xdx$

$\int\textup{cosec}xdx = \log|cosec x -\cot x |$

So, we have the function of $x$ , $f(x) =\log|cosec x -\cot x |$

Now, by the Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(\frac{\pi}{6})$

$= \log|cosec \frac{\pi}{4} -\cot \frac{\pi}{4} | - \log|cosec \frac{\pi}{6} -\cot \frac{\pi}{6} |$

$= \log|\sqrt2 -1 | - \log|2 -\sqrt3 |$

$= \log \left ( \frac{\sqrt2 -1}{2-\sqrt3} \right )$

Question 9: Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{dx}{\sqrt{1-x^2}}$

Answer:

Given integral: $\int_0^1\frac{dx}{\sqrt{1-x^2}}$

Consider the integral $\int \frac{dx}{\sqrt{1-x^2}}$

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

So, we have the function of $x$ , $f(x) = \sin^{-1}x$

Now, by the Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \sin^{-1}(1) -\sin^{-1}(0)$

$= \frac{\pi}{2} - 0$

$= \frac{\pi}{2}$

Question 10: Evaluate the definite integrals in Exercises 1 to 20.

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

Answer:

Given integral: $\int_0^1\frac{dx}{1 + x^2}$

Consider the integral $\int\frac{dx}{1 + x^2}$

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

So, we have the function of $x$ , $f(x) =\tan^{-1}x$

Now, by the Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \tan^{-1}(1) -\tan^{-1}(0)$

$= \frac{\pi}{4} - 0$

$= \frac{\pi}{4}$

Question 11: Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3 \frac{dx}{x^2 -1 }$

Answer:

Given integral: $\int_2^3 \frac{dx}{x^2 -1 }$

Consider the integral $\int \frac{dx}{x^2 -1 }$

$\int \frac{dx}{x^2 -1 } = \frac{1}{2}\log\left | \frac{x-1}{x+1} \right |$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\log\left | \frac{x-1}{x+1} \right |$

Now, by the Second fundamental theorem of calculus, we have

$I = f(3) -f(2)$

$= \frac{1}{2}\left \{ \log\left | \frac{3-1}{3+1} \right | - \log\left | \frac{2-1}{2+1} \right | \right \}$

$= \frac{1}{2}\left \{ \log\left | \frac{2}{4} \right | -\log\left | \frac{1}{3} \right | \right \}$

$= \frac{1}{2}\left \{ \log \frac{1}{2} -\log \frac{1}{3} \right \} = \frac{1}{2}\log\frac{3}{2}$

Question 12: Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{2}\cos^2 x dx$

Answer:

Given integral: $\int_0^\frac{\pi}{2}\cos^2 x dx$

Consider the integral $\int \cos^2 x dx$

$\int \cos^2 x dx = \int \frac{1+\cos 2x}{2} dx = \frac{x}{2}+\frac{\sin 2x }{4}$

$= \frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\left ( x+\frac{\sin 2x}{2} \right )$

Now, by the Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{2}) -f(0)$

$= \frac{1}{2}\left \{ \left ( \frac{\pi}{2}-\frac{\sin \pi}{2} \right ) -\left ( 0+\frac{\sin 0}{2} \right ) \right \}$

$= \frac{1}{2}\left \{ \frac{\pi}{2}+0-0-0 \right \}$

$= \frac{\pi}{4}$

Question 13: Evaluate the definite integrals in Exercises 1 to 20.

$\int_2^3\frac{xdx}{x^2+1}$

Answer:

Given integral: $\int_2^3\frac{xdx}{x^2+1}$

Consider the integral $\int \frac{xdx}{x^2+1}$

$\int \frac{xdx}{x^2+1} = \frac{1}{2}\int \frac{2x}{x^2+1}dx =\frac{1}{2}\log(1+x^2)$

So, we have the function of $x$ , $f(x) =\frac{1}{2}\log(1+x^2)$

Now, by the Second fundamental theorem of calculus, we have

$I = f(3) -f(2)$

$= \frac{1}{2}\left \{ \log(1+(3)^2)-\log(1+(2)^2) \right \}$

$= \frac{1}{2}\left \{ \log(10)-\log(5) \right \} = \frac{1}{2}\log\left ( \frac{10}{5} \right ) = \frac{1}{2}\log2$

Question 14: Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1\frac{2x+3}{5x^2+1}dx$

Answer:

Given integral: $\int_0^1\frac{2x+3}{5x^2+1}dx$

Consider the integral $\int \frac{2x+3}{5x^2+1}dx$

Multiplying by 5 both in the numerator and the 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 the Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \left \{ \frac{1}{5}\log(1+5)+\frac{3}{\sqrt5}\tan^{-1}(\sqrt5) \right \} - \left \{ \frac{1}{5}\log(1)+\frac{3}{\sqrt5}\tan^{-1}(0) \right \}$

$= \frac{1}{5}\log 6 +\frac{3}{\sqrt 5}\tan^{-1}{\sqrt5}$

Question 15: Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1xe^{x^2}dx$

Answer:

Given integral: $\int_0^1xe^{x^2}dx$

Consider the integral $\int xe^{x^2}dx$

Putting $x^2 = t$ which gives, $2xdx =dt$

As, $x\rightarrow0 ,t \rightarrow0$ and as $x\rightarrow1 ,t \rightarrow1$ .

So, we have now:

$\therefore I = \frac{1}{2}\int_0^1 e^t dt$

$= \frac{1}{2}\int e^t dt = \frac{1}{2} e^t$

So, we have the function of $x$ , $f(x) = \frac{1}{2} e^t$

Now, by the Second fundamental theorem of calculus, we have

$I = f(1) -f(0)$

$= \frac{1}{2}e^1 -\frac{1}{2}e^0 = \frac{1}{2}(e-1)$

Question 16: Evaluate the definite integrals in Exercises 1 to 20.

$\int_1^2\frac{5x^2}{x^2 + 4x +3}$

Answer:

Given integral: $I = \int_1^2\frac{5x^2}{x^2 + 4x +3}$

So, we can rewrite the integral as;

$I = \int_1^2 \frac{5x^2}{x^2 + 4x +3}= \int_1^2 \left ( 5 - \frac{20x+15}{x^2 + 4x +3} \right ) dx$

$= \int_1^2 5 dx - \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

$= [5x]_1^2 - \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

$I = 5-I_1$ where $I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx$ . ................(1)

Now, consider $I = \int_1^2 \frac{20x+15}{x^2+4x+3}dx$

Take numerator $20x+15 = A \frac{d}{dx}\left ( x^2+4x+3 \right )+B$

$= 2A x+(4A+B)$

We now equate the coefficients of x and the constant term, and we get

$A= 10$ and $B =-25$

$\Rightarrow I_1 = 10\int_1^2 \frac{2x+4}{x^2+4x+3}dx -25\int_1^2 \frac{dx}{x^2+4x+3}$

Now take denominator $x^2+4x+3 = t$

Then we have $(2x+4)dx =dt$

$\Rightarrow I_{1} =10\int \frac{dt}{t} -25\int \frac{dx}{(x+2)^2-1^2}$

$= 10\log t -25\left [ \frac{1}{2}\log\left ( \frac{x+2-1}{x+2+1} \right ) \right ]$

$=[10\log(x^2+4x+3)]_1^2 -25 \left [ \frac{1}{2}\log\left ( \frac{x+1}{x+3} \right ) \right ]_1^2$

$= \left [ 10\log15 -10\log 8 \right ] -25 \left [ \frac{1}{2}\log\frac{3}{5} -\frac{1}{2}\log\frac{2}{4} \right ]$

$= \left [ 10\log5 +10\log3 -10\log4-10\log2 \right ] -\frac{25}{2}\left [ \log3 -\log5-\log2+\log4 \right ]$ $= \left ( 10+\frac{25}{2} \right )\log5 + \left ( -10-\frac{25}{2} \right )\log 4 + \left ( 10-\frac{25}{2} \right )\log 3 + \left ( -10+\frac{25}{2} \right )\log 2$ $= \frac{45}{2}\log5 -\frac{45}{2}\log4 - \frac{5}{2}\log3 +\frac{5}{2}\log2$

$= \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log \frac{3}{2}$

Then substituting the value of $I_{1}$ in equation (1), we get

$I= 5 -\left ( \frac{45}{2}\log\frac{5}{4}-\frac{5}{2}\log\frac{3}{2} \right )$

$= 5 -\frac{5}{2}\left ( 9\log\frac{5}{4}-\log\frac{3}{2} \right )$

Question 17: Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx$

Answer:

Given integral: $\int_0^\frac{\pi}{4}(2\sec^2x + x^3 + 2)dx$

Consider the integral $\int (2\sec^2x + x^3 + 2)dx$

$\int (2\sec^2x + x^3 + 2)dx = 2\tan x +\frac{x^4}{4}+2x$

So, we have the function of $x$ , $f(x) = 2\tan x +\frac{x^4}{4}+2x$

Now, by the Second fundamental theorem of calculus, we have

$I = f(\frac{\pi}{4}) -f(0)$

$= \left \{ \left ( 2\tan\frac{\pi}{4}+\frac{1}{4}\left ( \frac{\pi}{4} \right )^4+2\frac{\pi}{4} \right ) - \left ( 2\tan 0 +0 +0 \right ) \right \}$

$=2\tan\frac{\pi}{4} +\frac{\pi^4}{4^5} +\frac{\pi}{2}$

$2+\frac{\pi}{2}+\frac{\pi^4}{1024}$

Question 18: Evaluate the definite integrals in Exercises 1 to 20.

$\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

Answer:

Given integral: $\int^\pi_0(\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

Consider the integral $\int (\sin^2\frac{x}{2} - \cos^2\frac{x}{2})dx$

can be rewritten as: $-\int (\cos^2\frac{x}{2} - \sin^2\frac{x}{2})dx = -\int_0^{\pi} \cos x dx$

$= \sin x$

So, we have the function of $x$ , $f(x) =\sin x$

Now, by the Second fundamental theorem of calculus, we have

$I = f(\pi) - f(0)$

$\Rightarrow \sin \pi - \sin 0 = 0-0 =0$

Question 19: Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^2\frac{6x+3}{x^2+ 4}$

Answer:

Given integral: $\int_0^2\frac{6x+3}{x^2+ 4}$

Consider the integral $\int \frac{6x+3}{x^2+ 4}$

can be rewritten as: $\int \frac{6x+3}{x^2+ 4} = 3\int \frac{2x+1}{x^2+4}dx$

$= 3\int \frac{2x}{x^2+4}dx +3\int \frac{1}{x^2+4}dx$

$= 3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}$

So, we have the function of $x$ , $f(x) =3\log (x^2+4) +\frac{3}{2}\tan^{-1}\frac{x}{2}$

Now, by the Second fundamental theorem of calculus, we have

$I = f(2) - f(0)$

$= \left \{ 3\log(2^2+4)+\frac{3}{2}\tan^{-1}\left ( \frac{2}{2} \right ) \right \}- \left \{ 3\log(0+4)+\frac{3}{2}\tan^{-1}\left ( \frac{0}{2} \right ) \right \}$ $=3\log 8 +\frac{3}{2}\tan^{-1}1 -3\log 4 -\frac{3}{2}\tan^{-1} 0$

$=3\log 8 +\frac{3}{2}\times\frac{\pi}{4} -3\log 4 -0$

$=3\log \frac{8}{4} +\frac{3\pi}{8}$

or we have $=3\log 2 +\frac{3\pi}{8}$

Question 20: Evaluate the definite integrals in Exercises 1 to 20.

$\int_0^1(xe^x + sin\frac{\pi x}{4})dx$

Answer:

Given integral: $\int_0^1(xe^x + sin\frac{\pi x}{4})dx$

Consider the integral $\int (xe^x + sin\frac{\pi x}{4})dx$

can be rewritten as: $x\int e^x dx - \int \left \{ \left ( \frac{d}{dx}x \right )\int e^x dx \right \}dx +\left \{ \frac{-\cos \frac{\pi x}{4}}{\frac{\pi}{4}} \right \}$

$= xe^x -\int e^x dx -\frac{4\pi}{\pi} \cos \frac{x}{4}$

$= xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}$

So, we have the function of $x$ , $f(x) = xe^x -e^x -\frac{4\pi}{\pi} \cos \frac{x}{4}$

Now, by the Second fundamental theorem of calculus, we have

$I = f(1) - f(0)$

$= \left (1.e^t-e^t - \frac{4}{\pi}\cos \frac{\pi}{4} \right ) - \left ( 0.e^0 -e^0 -\frac{4}{\pi}\cos 0 \right )$

$= e-e -\frac{4}{\pi}\left ( \frac{1}{\sqrt2} \right )+1+\frac{4}{\pi}$

Question 21: Choose the correct answer in Exercises 20 and 21.

$\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}$

(A) $\frac{\pi}{3}$

(B) $\frac{2\pi}{3}$

(C) $\frac{\pi}{6}$

(D) $\frac{\pi}{12}$

Answer:

Given definite integral $\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2}$

Consider $\int \frac{dx}{1 +x^2} = \tan^{-1}x$

we have then the function of x, as $f(x) = \tan^{-1}x$

By applying the second fundamental theorem of calculus, we will get

$\int^{\sqrt{3}}_{1} \frac{dx}{1 +x^2} = f(\sqrt3) - f(1)$

$= \tan^{-1}\sqrt{3} - \tan^{-1}1$

$=\frac{\pi}{3} - \frac{\pi}{4}$

$= \frac{\pi}{12}$

Therefore, the correct answer is D.

Question 22: Choose the correct answer in Exercises 21 and 22.

$\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2}$ equals

(A) $\frac{\pi}{6}$

(B) $\frac{\pi}{12}$

(C) $\frac{\pi}{24}$

(D) $\frac{\pi}{4}$

Answer:

Given definite integral $\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2}$

Consider $\int \frac{dx}{4+ 9x^2} = \int \frac{dx}{2^2+(3x)^2}$

Now, putting $3x = t$

We get $3dx=dt$

Therefore we have, $\int \frac{dx}{2^2+(3x)^2} = \frac{1}{3}\int \frac{dt}{2^2+t^2}$

$= \frac{1}{3}\left ( \frac{1}{2}\tan^{-1}\frac{t}{2} \right ) = \frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )$

we have the function of x , as $f(x) =\frac{1}{6}\tan^{-1}\left ( \frac{3x}{2} \right )$

So, by applying the second fundamental theorem of calculus, we get

$\int_0^\frac{2}{3}\frac{dx}{4+ 9x^2} = f(\frac{2}{3}) - f(0)$

$= \frac{1}{6}\tan^{-1}\left ( \frac{3}{2}.\frac{2}{3} \right ) -\frac{1}{6}\tan^{-1}0$

$= \frac{1}{6}\tan^{-1}1 - 0$

$= \frac{1}{6}\times \frac{\pi}{4} = \frac{\pi}{24}$

Therefore, the correct answer is C.

Question 1: Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^1\frac{x}{x^2 +1}dx$

Answer:

$\int_0^1\frac{x}{x^2 +1}dx$
Let $x^2+1 = t \Rightarrow xdx =dt/2$
when x = 0 then t = 1 and when x =1 then t = 2
$\therefore \int_{o}^{1}\frac{x}{x^2+1}dx=\frac{1}{2}\int_{1}^{2}\frac{dt}{t}$
$\\=\frac{1}{2}[\log\left | t \right |]_{1}^{2}\\ =\frac{1}{2}\log 2$

Question 2: Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$

Answer:

$\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi$
let $\sin \phi = t \Rightarrow \cos \phi d\phi = dt$
when $\phi =0,t\rightarrow 0$ and $\phi =\pi/2,t\rightarrow 1$

Using the above substitution, we can evaluate the integral as

$\\\therefore \int_{0}^{1}\sqrt{t}(1-t^2)dt\\ =\int_{0}^{1} t^\frac{1}{2}(1+t^4-2t^2)dt$
$=\int_{0}^{1}t^\frac{1}{2}dt+\int_{0}^{1}t^{9/2}dt-2\int_{0}^{1}t^{5/2}dt$
$=[2t^{3/2}/3+2t^{11/2}/11+4t^{7/2}/7]^1_0\\ =\frac{64}{231}$

Question 3: Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$

Answer:

$\int_0^1 \sin^{-1}\left(\frac{2x}{1+x^2} \right )dx$
let $x = \tan\theta\Rightarrow dx =\sec^2\theta d\theta$
when x = 0 then $\theta= 0$ and when x = 1 then $\theta= \pi/4$

$\\=\int_{0}^{\pi/4}\sin^{-1}(\frac{2\tan\theta}{1+\tan\theta})\sec^2\theta d\theta$
$=\int_{0}^{\pi/4}\sin^{-1}(\sin 2\theta)\sec^2\theta d\theta$
$=\int_{0}^{\pi/4}2\theta \sec^2\theta d\theta\\$
Taking $\theta$ as a first function and $\sec^2\theta$ as a second function, by using by parts method

$\\=2[\theta\int \sec^2\theta d\theta-\int(\frac{d}{d\theta}\theta.\int \sec^2\theta\ d\theta)d\theta]^{\pi/4}_0$
$=2[\theta\tan\theta-\int \tan\theta\ d\theta]^{\pi/4}_0$
$=2[\theta\tan\theta+\log\left | \cos\theta \right |]^{\pi/4}_0$
$=2[\pi/4+\log(1/\sqrt{2})]\\ =\pi/4-\log 2$

Question 4: Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^2x\sqrt{x+2}$ . (Put ${x+2} = t^2$ )

Answer:

Let $x+2 = t^2\Rightarrow dx =2tdt$
when x = 0 then t = $\sqrt{2}$ and when x=2 then t = 2

$I=\int_{0}^{2}x\sqrt{x+2}dx$

$\\=2\int_{\sqrt{2}}^{2}(t^2-2)t^2dt\\ =2\int_{\sqrt{2}}^{2}(t^4-2t^2)dt$
$=2[t^5/5-\frac{2}{3}t^3]^2_{\sqrt{2}}$
$=2[\frac{32}{5}-\frac{16}{3}-\frac{4\sqrt{2}}{5}+\frac{4\sqrt{2}}{3}]\\ =\frac{16\sqrt{2}(\sqrt{2}+1)}{15}$

Question 5: Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx$

Answer:

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}dx =I$
Let $\cos x =t\Rightarrow -\sin x dx = dt$
when x=0 then t = 1 and when x= $\pi/2$ then t = 0

$\\I=\int_{1}^{0}\frac{dt}{1+t^2}\\ =[\tan ^{-1}t]^0_1\\ =\pi/4$

Question 6: Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_0^2\frac{dx}{x + 4 - x^2}$

Answer:

By adjusting, the denominator can also be written as $(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2 =x+4-x^2$
Now,
$\Rightarrow \int_{0}^{2}\frac{dx}{(\frac{\sqrt{17}}{2})^2-(x-\frac{1}{2})^2}$
let $x-1/2 = t\Rightarrow dx=dt$
when x= 0 then t =-1/2 and when x =2 then t = 3/2

$\\\Rightarrow\int_{-1/2}^{3/2}\frac{dt}{(\frac{\sqrt{17}}{2})^2-t^2}$
$=\frac{1}{2.\frac{\sqrt{17}}{2}}\log\frac{\frac{\sqrt{17}}{2}+t}{\frac{\sqrt{17}}{2}-t}$
$=\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}/2+3/2}{\sqrt{17}/2-3/2}-\log\frac{\sqrt{17}/2-1/2}{\sqrt{17}/2+1/2}]$
$=\frac{1}{\sqrt{17}}[\log\frac{\sqrt{17}+3}{\sqrt{17}-3/}.\frac{\sqrt{17}+1}{\sqrt{17}+1}]$
$=\frac{1}{\sqrt{17}}[\log (\frac{17+3+4\sqrt{17}}{17+3-4\sqrt{17}})]$
$=\frac{1}{\sqrt{17}}[\log (\frac{5+\sqrt{17}}{5-\sqrt{17}})]$
On rationalisation, we get

$=\frac{1}{\sqrt{17}}\log \frac{21+5\sqrt{17}}{4}$

Question 7: Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$

Answer:

$\int_{-1}^1\frac{dx}{x^2 +2x + 5}$
the Dr can be written as $x^2+2x+5 = (x+1)^2+2^2$
and put x+1 = t then dx =dt

when x= -1 then t = 0 and when x = 1 then t = 2

$\\\Rightarrow \int_{0}^{2}\frac{dt}{t^2+2^2}\\ =\frac{1}{2}[\tan^{-1}\frac{t}{2}]^2_0\\ =\frac{1}{2}( \pi/4)\\ =\frac{\pi}{8}$

Question 8: Evaluate the integrals in Exercises 1 to 8 using substitution.

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$

Answer:

$\int_1^2\left(\frac{1}{x} - \frac{1}{2x^2} \right )e^{2x}dx$
let $2x =t \Rightarrow 2dx =dt$
when x = 1 then t = 2 and when x = 2 then t= 4

$\\=\frac{1}{2}\int_{2}^{4}(\frac{2}{t}-\frac{2}{t^2})e^tdt\\$
let
$\frac{1}{t} = f(t)\Rightarrow f'(t)=-\frac{1}{t^2}$
$\Rightarrow \int_{2}^{4}(\frac{1}{t}-\frac{1}{t^2})e^tdt =\int_{2}^{}4e^t[f(t)+f'(t)]dt$
$\\=[e^tf(t)]^4_2\\ =[e^t.\frac{1}{t}]^4_2\\ =\frac{e^4}{4}-\frac{e^2}{2}\\ =\frac{e^2(e^2-2)}{4}$

Question 9: Choose the correct answer in Exercises 9 and 10.

The value of the integral $\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$ is

(A) 6

(B) 0

(C) 3

(D) 4

Answer:

The value of the integral is (A) = 6

$\int_{\frac{1}{3}}^1\frac{(x-x^3)^\frac{1}{3}}{x^4}dx$
$\int_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1)^\frac{1}{3}}{x^3}dx\\$
let
$\frac{1}{x^2}-1 = t\Rightarrow \frac{dx}{x^3}=-dt/2$
now, when x = 1/3, t = 8 and when x = 1 , t = 0

therefore

$\\=-\frac{1}{2}\int_{8}^{0}t^{1/3}dt\\ =-\frac{1}{2}.\frac{3}{4}[t^4/3]^0_8\\ =-\frac{3}{8}[-2^4]\\ =6$

Question 10: Choose the correct answer in Exercises 9 and 10.

If $f(x) = \int_0^x t \sin t dt$ , then $f'(x)$ is

(A) $\cos x + x\sin x$

(B) $x\sin x$

(C) $x\cos x$

(D) $\sin x + x\cos x$

Answer:

The correct answer is (B) = $x\sin x$

$f(x) = \int_0^x t \sin t dt$
by using the 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, $f'(x)= -\cos x+x\sin x+\cos x\\ =x\sin x$

Question 1: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\frac{\pi}{2}\cos^2 x dx$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{2}\cos^2 x dx$ ............................................................. (i)

By using

$\int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get :-

$I\ =\ \int_0^\frac{\pi}{2}\cos^2 x dx\ =\ \int_0^\frac{\pi}{2}\cos^2\ (\frac{\pi}{2}- x) dx$

or

$I\ =\ \int_0^\frac{\pi}{2}\sin^2 x dx$ ................................................................ (ii)

Adding both (i) and (ii), we get :-

$\int_0^\frac{\pi}{2}\cos^2 x dx$ $+\ \int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2I$

or $\int_0^\frac{\pi}{2}\ (cos^2 x\ +\ sin^2 x) dx\ =\ 2I$

or $\int_0^\frac{\pi}{2}1. dx\ =\ 2I$

or $2I\ =\ \left [ x \right ] ^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}$

or $I\ =\ \frac{\Pi }{4}$

Question 2: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx$ .............. (i)

By using,

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin (\frac{\pi}{2}-x)}}{\sqrt{\sin (\frac{\pi}{2}-x)}+ \sqrt{\cos (\frac{\pi}{2}-x)}}dx$

or $I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx$ .......................................................(ii)

Adding (i) and (ii), we get,

$2I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}\ +\ \sqrt{\cos x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx$

or $2I\ =\ \int_0^\frac{\pi}{2}1.dx$

or $2I\ =\ \left [ x \right ]^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}$

Thus $I\ =\ \frac{\Pi }{4}$

Question 3: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$

Answer:

We have $I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$......(i)

By using :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}(\frac{\pi}{2}-x)dx}{\sin^\frac{3}{2}(\frac{\pi}{2}-x) + \cos^{\frac{3}{2}}(\frac{\pi}{2}-x)}$

or $I\ =\ \int^{\frac{\pi}{2}}_0\frac{\cos^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$...........(ii)

Adding (i) and (ii), we get :

$2I\ =\ \int^{\frac{\pi}{2}}_0\frac{\ (sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x)dx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}$

or $2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx$

or $2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}$

Thus $I\ =\ {\frac{\pi}{4}}$

Question 4: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}$ ................(i)

By using :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 (\frac{\pi}{2}-x)dx}{\sin^5(\frac{\pi}{2}-x) + \cos^5(\frac{\pi}{2}-x)}$

or $I\ =\ \int_0^\frac{\pi}{2} \frac{\sin^5 xdx}{\sin^5x + \cos^5x}$ . ............................................................(ii)

Adding (i) and (ii), we get :

$2I\ =\ \int_0^\frac{\pi}{2} \frac{\ ( sin^5x \ +\ cos^5 x)dx}{\sin^5x + \cos^5x}$

or $2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx$

or $2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}$

Thus $I\ =\ {\frac{\pi}{4}}$

Question 5: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_{-5}^5|x+2|dx$

Answer:

We have, $I\ =\ \int_{-5}^5|x+2|dx$

For opening the modules we need to define the bracket :

If (x + 2) < 0 then x belongs to (-5, -2). And if (x + 2) > 0 then x belongs to (-2, 5).

So the integral becomes :-

$I\ =\ \int_{-5}^{-2} -(x+2)dx\ +\ \int_{-2}^{5} (x+2)dx$

or $I\ =\ -\left [ \frac{x^2}{2}\ +\ 2x \right ]^{-2} _{-5}\ +\ \left [ \frac{x^2}{2}\ +\ 2x \right ]^{5} _{-2}$

This gives $I\ =\ 29$

Question 6: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_2^8|x-5|dx$

Answer:

We have, $I\ =\ \int_{2}^8|x-5|dx$

For opening the modules we need to define the bracket :

If (x - 5) < 0 then x belongs to (2, 5). And if (x - 5) > 0 then x belongs to (5, 8).

So the integral becomes:-

$I\ =\ \int_{2}^{5} -(x-5)dx\ +\ \int_{5}^{8} (x-5)dx$

or $I\ =\ -\left [ \frac{x^2}{2}\ -\ 5x \right ]^{5} _{2}\ +\ \left [ \frac{x^2}{2}\ -\ 5x \right ]^{8} _{5}$

This gives $I\ =\ 9$

Question 7: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int^1_0x(1-x)^ndx$

Answer:

We have $I\ =\ \int^1_0x(1-x)^ndx$

Using the property: -

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get: -

$I\ =\ \int^1_0x(1-x)^ndx\ =\ \int^1_0(1-x)(1-(1-x))^ndx$

or $I\ =\ \int^1_0(1-x)x^n\ dx$

or $I\ =\ \int^1_0(x^n\ -\ x^{n+1}) \ dx$

or $=\ \left [ \frac{x^{n+1}}{n+1}\ -\ \frac{x^{n+2}}{n+2} \right ]^1_0$

or $=\ \left [ \frac{1}{n+1}\ -\ \frac{1}{n+2} \right ]$

or $I\ =\ \frac{1}{(n+1)(n+2)}$

Question 8: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\frac{\pi}{4}\log(1+\tan x)dx$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx$

By using the identity

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx\ =\ \int_0^\frac{\pi}{4}\log(1+\tan (\frac{\pi}{4}-x))dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log(1+\frac{1-\tan x}{1+\tan x})dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log(\frac{2}{1+\tan x})dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ \int_0^\frac{\pi}{4}\log(1+ \tan x)dx$

or $I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ I$

or $2I\ =\ \left [ x\log2 \right ]^{\frac{\Pi }{4}}_0$

or $I\ =\ \frac{\Pi }{8}\log2$

Question 9: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^2x\sqrt{2-x}dx$

Answer:

We have $I\ =\ \int_0^2x\sqrt{2-x}dx$

By using the identity

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get:

$I\ =\ \int_0^2x\sqrt{2-x}dx\ =\ \int_0^2(2-x)\sqrt{2-(2-x)}dx$

or $I\ =\ \int_0^2(2-x)\sqrt{x}dx$

or $I\ =\ \int_0^2(2\sqrt{x}\ -\ x^\frac{3}{2} dx$

or $=\ \left [ \frac{4}{3}x^\frac{3}{2}\ -\ \frac{2}{5}x^\frac{5}{2} \right ]^2_0$

or $=\ \frac{4}{3}(2)^\frac{3}{2}\ -\ \frac{2}{5}(2)^\frac{5}{2}$

or $I\ =\ \frac{16\sqrt{2}}{15}$

Question 10: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx$

Answer:

We have $I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx$

or $I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log(2\sin x\cos x))dx$

or $I\ =\ \int_0^\frac{\pi}{2} (\log\sin x- \log\cos x\ -\ \log2)dx$ ..............................................................(i)

By using the identity :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get :

$I\ =\ \int_0^\frac{\pi}{2} (\log\sin (\frac{\pi}{2}-x)- \log\cos (\frac{\pi}{2}-x)\ -\ \log2)dx$

or $I\ =\ \int_0^\frac{\pi}{2} (\log\cos x- \log\sin x\ -\ \log2)dx$ ....................................................................(ii)

Adding (i) and (ii) we get :-

$2I\ =\ \int_0^\frac{\pi}{2} (- \log 2 -\ \log 2)dx$

or $I\ =\ -\log 2\left [ \frac{\Pi }{2} \right ]$

or $I\ =\ \frac{\Pi }{2}\log\frac{1}{2}$

Question 11: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx$

Answer:

We have $I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx$

We know that sin 2 x is an even function. i.e., sin 2 (-x) = (-sinx) 2 = sin 2 x.

Also,

$I\ =\ \int_{-a}^af(x) dx\ =\ 2\int_{0}^af(x) dx$

So,

$I\ =\ 2\int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2\int_0^\frac{\pi}{2}\frac{(1-\cos2x)}{2} dx$

or $=\ \left [ x\ -\ \frac{\sin2x}{2} \right ]^{\frac{\Pi }{2}}_0$

or $I\ =\ \frac{\Pi }{2}$

Question 12: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\pi\frac{xdx}{1+\sin x}$

Answer:

We have $I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}$ ..........................................................................(i)

By using the identity :-

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin (\Pi -x)}$

or $I\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin x}$ ............................................................................(ii)

Adding both (i) and (ii) we get,

$2I\ =\ \int_0^\pi\frac{\Pi}{1+\sin x} dx$

or $2I\ =\ \Pi \int_0^\pi\frac{1-\sin x}{(1+\sin x)(1-\sin x)} dx\ =\ \Pi \int_0^\pi\frac{1-\sin x}{\cos^2 x} dx$

or $2I\ =\ \Pi \int_0^\pi (\sec^2\ -\ \tan x \sec x) x dx$

or $I\ =\ \Pi$

Question 13: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx$

Answer:

We have $I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx$

We know that $\sin^7x$ is an odd function.

So the following property holds here:-

$\int_{-a}^{a}f(x)dx\ =\ 0$

Hence

$I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx\ =\ 0$

Question 14: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^{2\pi}\cos^5xdx$

Answer:

We have $I\ =\ \int_0^{2\pi}\cos^5xdx$

It is known that :-

$\int_0^{2a}f(x)dx\ =\ 2\int_0^{a}f(x)dx$ If f(2a - x) = f(x)

$=\ 0$ If f (2a - x) = - f(x)

Now, using the above property

$\cos^5(\Pi - x)\ =\ - \cos^5x$

Therefore, $I\ =\ 0$

Question 15: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$

Answer:

We have $I\ =\ \int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx$ ................(i)

By using the property :-

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int^\frac{\pi}{2} _0\frac{\sin (\frac{\pi}{2}-x) - \cos (\frac{\pi}{2}-x) }{1+\sin (\frac{\pi}{2}-x)\cos (\frac{\pi}{2}-x)}dx$

or $I\ =\ \int^\frac{\pi}{2} _0\frac{\cos x - \sin x }{1+\sin x\cos x}dx$ ..........................(ii)

Adding both (i) and (ii), we get

$2I\ =\ \int^\frac{\pi}{2} _0\frac{0 }{1+\sin x\cos x}dx$

Thus I = 0

Question 16: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^\pi\log(1 +\cos x)dx$

Answer:

We have $I\ =\ \int_0^\pi\log(1 +\tan x)dx$ .....................................................................................(i)

By using the property:-

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\pi\log(1 +\cos (\Pi -x))dx$

$I\ =\ \int_0^\pi\log(1 -\cos x)dx$ ....................................................................(ii)

Adding both (i) and (ii) we get,

$2I\ =\ \int_0^\pi\log(1 +\cos x)dx\ +\ \int_0^\pi\log(1 -\cos x)dx$

or $2I\ =\ \int_0^\pi\log(1 -\cos^2 x)dx\ =\ \int_0^\pi\log \sin^2 xdx$

or $2I\ =\ 2\int_0^\pi\log \sin xdx$

or $I\ =\ \int_0^\pi\log \sin xdx$ ........................................................................(iii)

or $I\ =\ 2\int_0^ \frac{\pi}{2} \log \sin xdx$ ........................................................................(iv)

or $I\ =\ 2\int_0^ \frac{\pi}{2} \log \cos xdx$ .....................................................................(v)

Adding (iv) and (v) we get,

$I\ =\ -\pi \log2$

Question 17: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx$

Answer:

We have $I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx$ ...........................(i)

By using, we get

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx\ =\ \int_0^a \frac{\sqrt {(a-x)}}{\sqrt {(a-x)} + \sqrt{x}}dx$ .........(ii)

Adding (i) and (ii) we get :

$2I\ =\ \int_0^a \frac{\sqrt x\ +\ \sqrt{a-x}}{\sqrt x + \sqrt{a-x}}dx$

or $2I\ =\ \left [ x \right ]^a_0 = a$

or $I\ =\ \frac{a}{2}$

Question 18: By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

$\int_0^4 |x-1|dx$

Answer:

We have, $I\ =\ \int_{0}^4|x-1|dx$

For opening the modules we need to define the bracket :

If (x - 1) < 0 then x belongs to (0, 1). And if (x - 1) > 0 then x belongs to (1, 4).

So the integral becomes:-

$I\ =\ \int_{0}^{1} -(x-1)dx\ +\ \int_{1}^{4} (x-1)dx$

or $I\ =\ \left [ x\ -\ \frac{x^2}{2}\ \right ]^{1} _{0}\ +\ \left [ \frac{x^2}{2}\ -\ x \right ]^{4} _{1}$

This gives $I\ =\ 5$

Question 19: Show that $\int_0^a f(x)g(x)dx = 2\int_0^af(x)dx$ if $f$ and $g$ are defined as $f(x) = f(a-x)$ and $g(x) + g(a-x) = 4$

Answer:

Let $I\ =\ \int_0^a f(x)g(x)dx$ ........................................................(i)

This can also be written as :

$I\ =\ \int_0^a f(a-x)g(a-x)dx$

or $I\ =\ \int_0^a f(x)g(a-x)dx$ ................................................................(ii)

Adding (i) and (ii), we get,

$2I\ =\ \int_0^a f(x)g(a-x)dx +\ \int_0^a f(x)g(x)dx$

$2I\ =\ \int_0^a f(x)4dx$

or $I\ =\ 2\int_0^a f(x)dx$

Question 20: Choose the correct answer in Exercises 20 and 21.

The value of is $\int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx$ is

(A) 0

(B) 2

(C) $\pi$

(D) 1

Answer:

We have

$I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx$

This can be written as :

$I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}x^3dx +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} x\cos x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} \tan^5 x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} 1dx$

Also if a function is even function then $\int_{-a}^{a}f(x)\ dx\ =\ 2\int_{0}^{a}f(x)\ dx$

And if the function is an odd function then : $\int_{-a}^{a}f(x)\ dx\ =\ 0$

Using the above property I become:-

$I\ =\ 0+0+0+ 2\int_{0}^{\frac{\Pi }{2}}1.dx$

or $I\ =\ 2\left [ x \right ]^{\frac{\Pi }{2}}_0$

or $I\ =\ \Pi$

Question 21: Choose the correct answer in Exercises 20 and 21.

The value of $\int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx$ is

Answer:

We have

$I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx$ ..............................(i)

By using :

$\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx$

We get,

$I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin (\frac{\pi}{2}-x)}{4+3\cos (\frac{\pi}{2}-x)} \right )dx$

or $I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx$ ..............(ii)

Adding (i) and (ii), we get:

$2I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ +\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx$

or $2I\ =\ \int_0^\frac{\pi}{2}\log1.dx$

Thus $I\ =\ 0$

Question 1: Integrate the functions in Exercises 1 to 23.

$\frac{1}{x - x^3}$

Answer:

Firstly we will simplify the given equation :-

$\frac{1}{x - x^3}\ =\ \frac{1}{(x)(1-x)(1+x)}$

Let

$\frac{1}{(x)(1-x)(1+x)} =\ \frac{A}{x}\ +\ \frac{B}{1-x}\ +\ \frac{C}{1+x}$

By solving the equation and equating the coefficients of x2, 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 23.

$\frac{1}{\sqrt{x+a} + \sqrt{x+b}}$

Answer:

At first we will simplify the given expression,

$\frac{1}{\sqrt{x+a} + \sqrt{x+b}}\ =\ \frac{1}{\sqrt{x+a} + \sqrt{x+b}}\times\frac{\sqrt{x+a} - \sqrt{x+b}}{\sqrt{x+a} - \sqrt{x+b}}$

or $=\ \frac{\sqrt{x+a} - \sqrt{x+b}}{a-b}$

Now taking its integral we get,

$\int \frac{1}{\sqrt{x+a} + \sqrt{x+b}}\ =\ \frac{1}{a-b}\int (\sqrt{x+a} -\sqrt{x+b})dx$

or $=\ \frac{1}{a-b}\left [ \frac{(x+a)^{\frac{3}{2}}} {\frac{3}{2}}\ -\ \frac{(x+b)^{\frac{3}{2}}} {\frac{3}{2}} \right ]$

or $=\ \frac{2}{3(a-b)}\left [ (x+a)^{\frac{3}{2}}\ -\ (x+b)^{\frac{3}{2}} \right ]\ +\ C$

Question 3: Integrate the functions in Exercises 1 to 23.

$\frac{1}{x\sqrt{ax-x^2}}$ [Hint: Put $x = \frac{a}{t}$ ]

Answer:

Let

$x = \frac{a}{t}\ dx\ \Rightarrow \ dx\ =\ \frac{-a}{t^2}dh$

Using the above substitution we can write the integral is

$\int \frac{1}{x\sqrt{ax-x^2}}\ =\ \int \frac{1}{\frac{a}{t}\sqrt{a.\frac{a}{t}\ -\ (\frac{a}{t})^2}} \frac{-a}{t^2}dt$

or

$=\ \frac{-1}{a}\int \frac{1}{\sqrt{(t-1)}}dt$

or

$=\ \frac{-1}{a}\ (2\sqrt{t-1})\ +\ C$

or $=\ \frac{-1}{a}\ (2\sqrt{\frac{a}{x}\ -\ 1})\ +\ C$

or $=\ \frac{-2}{a}\ \sqrt{\frac{a-x}{x}}\ +\ C$

Question 4: Integrate the functions in Exercises 1 to 23.

$\frac{1}{x^2(x^4 + 1)^\frac{3}{4}}$

Answer:

For the simplifying the expression, we will multiply and dividing it by x-3 .

We then have,

$\frac{x^{-3}}{x^2 x^{-3}(x^4 + 1)^\frac{3}{4}}\ =\ \frac{1}{x^5}\left [ \frac{x^4\ +\ 1}{x^4} \right ]^{\frac{-3}{4}}$

Now, let

$\frac{1}{x^4}\ =\ t\ \Rightarrow \ \frac{1}{x^5}dx\ =\ \frac{-dt}{4}$

Thus,

$\int \frac{1}{x^2(x^4 + 1)^\frac{3}{4}}\ =\ \int \frac{1}{x^5}\left ( 1+\ \frac{1}{x^4}^{\frac{-3}{4}}\ \right )dx$

or $=\ \frac{-1}{4} \int (1+t)^{\frac{-3}{4}}dt$

$=\ \frac{-1}{4} \frac{(1+\frac{1}{x^4})^{\frac{1}{4}}}{\frac{1}{4}}\ +\ C$

$=\ - \left [ 1+\frac{1}{x^4} \right ]^{\frac{1}{4}}\ +\ C$

Question 5: Integrate the functions in Exercises 1 to 23.

$\frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}}$ [Hint: $\frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}} = \frac{1}{x^\frac{1}{3}(1 + \ x^\frac{1}{6})}$ , put $x = t^6$ ]

Answer:

Put $x = t^6\ \Rightarrow \ dx = 6t^5dt$

We get,

$\int \frac{1}{x^{\frac{1}{2}}+ x^\frac{1}{3}}dx\ =\ \int \frac{6t^5}{t^3+t^2}dt$

or $=\ 6\int \frac{t^3}{1+t}dt$

or $=\ 6\int \left \{ (t^2-t+1)-\frac{1}{1+t} \right \}dt$

or $=\ 6 \left [ \left ( \frac{t^3}{3} \right ) -\left ( \frac{t^2}{2} \right )+t - \log(1+t) \right ]$

Now put $x = t^6$ in the above result :

$=\ 2\sqrt{x} -3x^{\frac{1}{3}}+ 6x^{\frac{1}{6}} - 6 \log \left ( 1-x^\frac{1}{6} \right )\ +\ C$

Question 6: Integrate the functions in Exercises 1 to 23.

$\frac{5x}{(x+1)(x^2 + 9)}$

Answer:

Let us assume that :

$\frac{5x}{(x+1)(x^2 + 9)}\ =\ \frac{A}{(x+1)}\ +\ \frac{Bx + c}{x^2 + 9}$

Solving the equation and comparing coefficients of x2, x and the constant term.

We get,

$A\ =\ \frac{-1}{2}\ ;\ B\ =\ \frac{1}{2}\ ;\ C\ =\ \frac{9}{2}$

Thus, the equation becomes :

$\frac{5x}{(x+1)(x^2 + 9)}\ =\ \frac{-1}{2(x+1)}\ +\ \frac{\frac{x}{2}+\frac{9}{2}}{x^2 + 9}$

or

$\int \frac{5x}{(x+1)(x^2 + 9)}\ =\ \int \left [ \frac{-1}{2(x+1)}\ +\ \frac{x+9}{2(x^2 + 9}) \right ]dx$

or $=\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{2} \int \frac{x}{x^2 +9}dx +\frac{9}{2} \int \frac{1}{x^2+9}dx$

or $=\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{4} \int \frac{2x}{x^2 +9}dx +\frac{9}{2} \int \frac{1}{x^2+9}dx$

or $=\ \frac{-1}{2} \log \left | x+1 \right | + \frac{1}{4} \log {(x^2 +9)} +\frac{3}{2} \tan^{-1}\frac{x}{3}\ +\ C$

Question 7: Integrate the functions in Exercises 1 to 24.

$\frac{\sin x}{\sin (x-a)}$

Answer:

We have,

$I\ =\ \frac{\sin x}{\sin (x-a)}$

Assume :- $(x-a)\ =\ t \Rightarrow \ dx=dt$

Putting this in above integral :

$\int \frac{\sin x}{\sin (x-a)}dx\ =\ \int \frac{\sin (t+a)}{\sin t}dt$

or $=\ \int \frac{\sin t \cos a\ +\ \cos t \sin a }{\sin t}dt$

or $=\ \int (\cos a\ +\ \cot t \sin a)dt$

or $=\ t\cos a\ +\ \sin a \log |\sin t|\ +\ C$

or $=\ \sin a \log \left | \sin(x-a) \right | + x\cos a\ +\ C$

Question 9: Integrate the functions in Exercises 1 to 23.

$\frac{\cos x}{\sqrt{4 - \sin^2 x}}$

Answer:

We have the given integral

$I\ =\ \frac{\cos x}{\sqrt{4 - \sin^2 x}}$

Assume $\sin x = t\ \Rightarrow \cos x dx = dt$

So, this substitution gives,

$\int \frac{\cos x}{\sqrt{4 - \sin^2 x}}\ =\ \int \frac{dt}{\sqrt{(2)^2 - (t)^2}}$

$=\ \sin^{-1}\frac{t}{2}\ +\ C$

or $=\ \sin^{-1}\left ( \frac{\sin x}{2} \right )\ +\ C$

Question 10: Integrate the functions in Exercises 1 to 23.

$\frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}$

Answer:

We have

$I\ =\ \int \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}$

Simplifying the given expression, we get :

$\frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}\ =\ \frac{(\sin^4x + \cos^4x)(\sin^4x - \cos^4x) }{1- 2\sin^ x\cos^2 x}$

or $=\ \frac{(\sin^4x + \cos^4x)(\sin^2x - \cos^2x)(\sin^2x + \cos^2x) }{1- 2\sin^ x\cos^2 x}$

or $=\ -\frac{(\sin^4x + \cos^4x)(\cos^2x - \sin^2x) }{1- 2\sin^ x\cos^2 x}$

or $=\ -\cos^2x - \sin^2x\ =\ -\cos 2x$

Thus,

$I\ =\ \int \frac{\sin^8 x - \cos^8 x}{1- 2\sin^ x\cos^2 x}\ =\ -\int \cos 2x\ dx$

and $=\ - \frac{\sin 2x}{2}\ +\ C$

Question 11: Integrate the functions in Exercises 1 to 23.

$\frac{1}{\cos(x+a)\cos(x+b)}$

Answer:

For simplifying the given equation, we need to multiply and divide the expression by $\sin (a-b)$ .

Thus, we obtain :

$\frac{1}{\cos(x+a)\cos(x+b)}\ =\ \frac{1}{\sin(a-b)}\times\frac{\sin (a-b)}{\cos(x+a)\cos(x+b)}$

or $= \frac{1}{ \sin (a-b)}\times \frac{\sin{\left [ (x+a) - (x+b) \right ]}}{\cos (x+a) \cos (x+b)}$

or $= \frac{1}{ \sin (a-b)}\times \left ( \frac{\sin (x+a) }{\cos (x+a) } - \frac{\sin(x+b)}{\cos (x+b)} \right )$

or $= \frac{1}{ \sin (a-b)}\times \left ( \tan(x+a)\ -\ \tan(x+b) \right )$

Thus integral becomes :

$\int \frac{1}{\cos(x+a)\cos(x+b)}\ =\ \frac{1}{\sin (a-b)} \times \int \left ( \tan(x+a)\ -\ \tan(x+b) \right )dx$

or $=\ \frac{1}{\sin (a-b)} \times \left [ -\log \left | \cos (x+a) \right | + \log \left | \cos(x+b) \right | \right ]\ +\ C$

or $=\ \frac{1}{\sin (a-b)} \times \log \left [ \frac{\cos(x+b) }{cos(x+a)} \right ]\ +\ C$

Question 12: Integrate the functions in Exercises 1 to 23.

$\frac{x^3}{\sqrt{1-x^8}}$

Answer:

Given that to integrate

$\frac{x^3}{\sqrt{1-x^8}}$

Let $x^4 = t \implies 4x^3dx = dt$

$\therefore \int \frac{x^3}{\sqrt{1-x^8}}dx = \frac{1}{4}\int\frac{1}{\sqrt {1-t^2}}dt$

$= \frac{1}{4}sin^{-1}t + C= \frac{1}{4}sin^{-1}{x^4} + C$

the required solution is $\frac{1}{4}sin^{-1}{(x^4)} + C$

Question 13: Integrate the functions in Exercises 1 to 23.

$\frac{e^x}{(1 + e^x)(2 + e^x)}$

Answer:

We have to integrate the following function

$\frac{e^x}{(1 + e^x)(2 + e^x)}$

Let $1+e^x = t \implies e^xdx = dt$

using this we can write the integral as

$\therefore \int\frac{e^x}{(1 + e^x)(2 + e^x)}dx = \int\frac{1}{t(1+t)}dt = \int\frac{(1+t)-t}{t(1+t)}dt$

$\\ = \int\left ( \frac{1}{t}-\frac{1}{t+1} \right )dt$

$\\ = \int\frac{1}{t}dt - \int\frac{1}{t+1}dt$

$\\ = \log t - \log (1+t) + C \\ = \log (1+e^x) - \log (2+e^x) + C \\ = \log\left ( \frac{e^x + 1}{e^x + 2} \right ) + C$

Question 14: Integrate the functions in Exercises 1 to 23.

$\frac{1}{(x^2 + 1)(x^2 +4)}$

Answer:

Given,

$\frac{1}{(x^2 + 1)(x^2 +4)}$

Let $I = \int\frac{1}{(x^2 + 1)(x^2 +4)}$

Now, Using partial differentiation,

$\frac{1}{(x^2 + 1)(x^2 +4)} = \frac{Ax + B}{(x^2 + 1)} + \frac{Cx +D}{(x^2 +4)}$

$\implies \frac{1}{(x^2 + 1)(x^2 +4)} = \frac{(Ax + B)(x^2 +4) + (Cx +D)(x^2 + 1)}{(x^2 + 1)(x^2 +4)}$
$\\ \implies1 = (Ax + B)(x^2 + 4)+(Cx + D)(x^2 + 1)$
$ \implies 1 = Ax^3 +4Ax+ Bx^2 + 4B+ Cx^3 + Cx + Dx^2 + D$
$ \implies (A+C)x^3 +(B+D)x^2 +(4A+C)x + (4B+D) = 1$

Equating the coefficients of $x, x^2, x^3$ and constant value,

A + C = 0 $\implies$ C = -A

B + D = 0 $\implies$ B = -D

4A + C =0 $\implies$ 4A = -C $\implies$ 4A = A $\implies$ A = 0 = C

4B + D = 1 $\implies$ 4B – B = 1 $\implies$ B = 1/3 = -D

Putting these values in equation, we have

$\implies I = \frac{1}{3}tan^{-1}x - \frac{1}{6}tan^{-1}\frac{x}{2} + C$

Question 15: Integrate the functions in Exercises 1 to 23.

$\cos^3 x \;e^{\log\sin x}$

Answer:

Given,

$\cos^3 x \;e^{\log\sin x}$

$I = \int \cos^3 x \;e^{\log\sin x}$ (let)

Let $cos x = t \implies -sin x dx = dt \implies sin x dx = -dt$

Using the above substitution the integral is written as

$\therefore \int cos^3xe^{\log sinx}dx = \int cos^3x.sinx dx$

$\begin{aligned} & =\int \mathrm{t}^3 \cdot(-\mathrm{dt}) \\ & =-\int \mathrm{t}^3 \cdot \mathrm{dt}\end{aligned}$

$I = -\frac{cos^4x}{4} + C$

Question 16: Integrate the functions in Exercises 1 to 23.

$e^{3\log x} (x^4 + 1)^{-1}$

Answer:

Given the function to be integrated as $e^{3 \log x}\left(x^4+1\right)^{-1}$

$=e^{\log x^3}\left(x^4+1\right)^{-1}=\frac{x^3}{x^4+1}$

$I=\int e^{3 \log x}\left(x^4+1\right)^{-1}$

Let $x^4=t \Longrightarrow 4 x^3 d x=d t$

$\begin{aligned} & I=\int e^{3 \log x}\left(x^4+1\right)^{-1}=\int \frac{x^3}{x^4+1} \\ & =\frac{1}{4} \cdot \int \frac{1}{\mathrm{t}+1} \cdot \mathrm{dt} \\ & =\frac{1}{4} \log (\mathrm{t}+1)+\mathrm{C} \\ & \Longrightarrow I=\frac{1}{4} \log \left(x^4+1\right)+C\end{aligned}$

Question 17: Integrate the functions in Exercises 1 to 23.

$f'(ax +b)[f(ax +b)]^n$

Answer:

Given,

$f'(ax +b)[f(ax +b)]^n$

Let $I = \int f'(ax +b)[f(ax +b)]^n$

Let f(ax +b) = t ⇒ a .f ' (ax + b)dx = dt

Now we can write the integral 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 23.

$\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}$

Answer:

Given, $\frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}$
Let $I=\int \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}$
We know the identity that

$\begin{aligned} & \sin (A+B)=\sin A \cos B+\cos A \sin B \\ & \therefore \frac{1}{\sqrt{\sin ^3 x \sin (x+\alpha)}}=\frac{1}{\sqrt{\sin ^3 x(\sin x \cos \alpha+\cos x \sin \alpha)}} \\ & =\frac{1}{\sqrt{\sin ^3 x \cdot \sin x(\cos \alpha+\cot x \sin \alpha)}}=\frac{1}{\sqrt{\sin ^4 x(\cos \alpha+\cot x \sin \alpha)}} \\ & \frac{\operatorname{cosec}^2 x}{\sqrt{(\cos \alpha+\cot x \sin \alpha)}}\end{aligned}$

$\begin{aligned} & \Rightarrow \int \frac{1}{\sqrt{\sin ^3 \mathrm{x} \sin (\mathrm{x}+\alpha)}} \mathrm{dx}=\int \frac{\operatorname{cosec}^2 \mathrm{x}}{\sqrt{(\cos \alpha+\cot \mathrm{x} \sin \alpha)}} \mathrm{dx} \\ & =\int \frac{1}{\sqrt{\mathrm{t}}} \cdot-\frac{\mathrm{dt}}{\sin \alpha} \\ & =-\frac{1}{\sin \alpha} \int \frac{1}{\sqrt{\mathrm{t}}} \cdot \mathrm{dt} \\ & =-\frac{1}{\sin \alpha} \int \mathrm{t}^{-\frac{1}{2}} \cdot \mathrm{dt} \\ & =-\frac{1}{\sin \alpha}\left[\frac{\mathrm{t}^{\frac{1}{2}}}{\frac{1}{2}}\right]+\mathrm{C} \\ & =-\frac{2}{\sin \alpha}[\sqrt{\mathrm{t}}]+\mathrm{C} \\ & =-\frac{2}{\sin \alpha}[\sqrt{(\cos \alpha+\cot \mathrm{x} \sin \alpha)}]+\mathrm{C}\end{aligned}$

$=-\frac{2}{\sin \alpha}\left[\sqrt{\frac{(\cos \alpha \sin x+\cos x \sin \alpha)}{\sin x}}\right]+C$

Question 19: Integrate the functions in Exercises 1 to 23.

$\sqrt{\frac{1-\sqrt x}{1 +\sqrt x}}$

Answer:

Given,

$\sqrt{\frac{1-\sqrt x}{1 +\sqrt x}}$ = I (let)

Let $x= cos^2\theta \implies dx = -2sin\theta cos\theta d\theta$

$And \sqrt x= cos\theta \implies \theta = \cos^{-1}\sqrt x$

using the above substitution we can write the integral as

$\\ I = \int \sqrt{\frac{1-\sqrt {cos^2\theta}}{1 +\sqrt {cos^2\theta}}}(-2\sin\theta\cos\theta)d\theta $
$= -\int \sqrt{\frac{1-cos\theta}{1 +cos\theta}}(2\sin\theta\cos\theta)d\theta$

$\\ = -\int \sqrt{tan^2\frac{\theta}{2}}(2\sin\theta\cos\theta)d\theta $
$ = -\int \sqrt{tan^2\frac{\theta}{2}}(2. 2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}\cos\theta)d\theta$
$ = -4\int \sin^2\frac{\theta}{2}\cos\theta d\theta$

$\\ = -4\int \sin^2\frac{\theta}{2}(2cos^2\frac{\theta}{2} -1) d\theta$

$=-2 \cdot \int \sin ^2 \theta d \theta+4 \int \sin ^2\left(\frac{\theta}{2}\right) d \theta$

$\begin{aligned} & =\theta+\frac{2 \cdot \sqrt{1-\cos ^2 \theta} \cdot \cos \theta}{2}-2 \sqrt{1-\cos ^2 \theta}+C \\ & =\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}-2 \sqrt{1-x}+C \\ & \Rightarrow I=\cos ^{-1} \sqrt{x}+\sqrt{x-x^2}-2 \sqrt{1-x}+C\end{aligned}$

Question 20: Integrate the functions in Exercises 1 to 23.

$\frac{2 + \sin 2x}{1 + \cos 2x}e^x$

Answer:

Given to evaluate

$\frac{2 + \sin 2x}{1 + \cos 2x}e^x$

$\frac{2 + \sin 2x}{1 + \cos 2x}e^x$

$\begin{aligned} & =\left(\frac{2+2 \sin x \cos x}{2 \cos ^2 x}\right) e^x \\ & =\left(\sec ^2 x+\tan x\right) e^x\end{aligned}$

Now the integral becomes

$\Rightarrow \int \frac{2+\sin 2 x}{1+\cos 2 x} e^x d x=\int\left(\sec ^2 x+\tan x\right) e^x d x$

Let tan x = f(x)

$\implies f'(x) = sec^2x dx$

$\Rightarrow \mathrm{I}=\mathrm{e}^{\mathrm{x}} \tan \mathrm{x}+\mathrm{C}$

Question 21: Integrate the functions in Exercises 1 to 23.

$\frac{x^2 + x + 1}{(x+1)^2 (x+2)}$

Answer:

Given,

$\frac{x^2 + x + 1}{(x+1)^2 (x+2)}$

Using partial fractions, 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, x2, and constant value, we get:

A + C = 1

3A + B + 2C = 1

2A+2B+C =1

Solving these:

A= -2, B=1 and C=3

$\implies \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \frac{-2}{x+1}+\frac{1}{(x+1)^2}+\frac{3}{x+2}$

$\\ \implies \int \frac{x^2 + x + 1}{(x+1)^2 (x+2)} = \int\frac{-2}{x+1}dx+\int\frac{1}{(x+1)^2}dx+\int\frac{3}{x+2}dx$
$= -2\log(x+1) - \frac{1}{(x+1)} + 3\log (x+2) + C$

Question 22: Integrate the functions in Exercises 1 to 23.

$\tan^{-1}\sqrt{\frac{1-x}{1+x}}$

Answer:

We have

$I\ =\ \int \tan^{-1}\sqrt{\frac{1-x}{1+x}}$

Let us assume that : $x\ =\ \cos 2\Theta$

Differentiating wrt x,

$dx\ =\ -2 \sin 2\Theta\ d\Theta$

Substituting this in the original equation, we get

$\int \tan^{-1}\sqrt{\frac{1-x}{1+x}}\ =\ \int \tan^{-1}\sqrt{\frac{1-cos2\Theta }{1+cos2\Theta }}\times -2\sin 2\Theta \ d\Theta$

or $=\ -2\int \tan^{-1} (\frac{sin\Theta }{cos\Theta })\times \sin 2\Theta \ d\Theta$

or $=\ -2\int \Theta \sin 2\Theta \ d\Theta$

Using integration by parts, we get

$=\ -2\left ( \Theta \int \sin 2\Theta \ d\Theta\ - \int \frac{d\Theta }{d\Theta } \int \sin 2\Theta \ d\Theta\ \right )$

or $=\ -2\left ( \Theta \left ( \frac{-\cos 2\Theta }{2} \right ) - \int 1.\frac{-\cos 2\Theta }{2} \ d\Theta\ \right )$

or $=\ -2\left ( \frac{-\Theta \cos 2\Theta }{2}+ \frac{\sin 2\Theta }{4} \right )$

Putting all the assumed values back in the expression,

$=\ -2\left ( -\frac{1}{2}\left ( \frac{1}{2} \cos^{-1} x \right )+ \frac{\sqrt{1-x^2} }{4} \right )$

or $=\ \frac{1}{2}\left ( x \cos^{-1} x\ -\ \sqrt{1-x^2} \right )\ +\ C$

Question 23: Integrate the functions in Exercises 1 to 23.

$\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}$

Answer:

$\frac{\sqrt{x^2 + 1}[\log(x^2+1)-2\log x]}{x^4}$

Here let's first reduce the log function.

$=\frac{\sqrt{x^2+1}}{x^4}\left [ \log (x^2+1)-\log x^2 \right ]dx$

$=\frac{\sqrt{x^2\left ( 1+\frac{1}{x^2} \right )}}{x^4}\left [ \log\frac{ (x^2+1)}{x^2} \right ]dx$

$=\int\frac{\sqrt{\left ( 1+\frac{1}{x^2} \right )}}{x^3}\left [ \log\left ( 1+\frac{1}{x^2} \right ) \right ]dx$

Now, let

$t=1+\frac{1}{x^2}$

$dt=\frac{-2}{x^3}dx$

So our function in terms if new variable t is :

$I=\frac{-1}{2}\int \left [\log t \right ]\cdot t^{\frac{1}{2}}dt$

now let's solve this By using integration by parts

$I=\frac{-1}{2}\int \left [(\log t)\frac{t^\frac{3}{2}}{\frac{3}{2}} -\int \frac{1}{t}\cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}dt\right ]$

$I=\frac{-1}{3}t^\frac{3}{2}\log t+\frac{1}{3}\int t^{\frac{1}{2}}dt$

$I=\frac{-1}{3}t^\frac{3}{2}\log t+\frac{1}{3}\frac{t^{\frac{3}{2}}}{\frac{3}{2}}$

$I=\frac{2}{9}t^{\frac{3}{2}}-\frac{1}{3}t^{\frac{3}{2}}logt+c$

$I=\frac{1}{3}t^{\frac{3}{2}}\left [ \frac{2}{3}-\log t \right ]+c$

$I=\frac{1}{3}\left ( 1+\frac{1}{x^2} \right )^{\frac{3}{2}}\left [ \frac{2}{3}-\log \left ( 1+\frac{1}{x^2} \right ) \right ]+c$

Question 24: Evaluate the definite integrals in Exercises 25 to 33.

$\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx$

Answer:

$\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx$

Since, we have $e^x$ multiplied by some function, let's try to make that function in any function and its derivative.Basically we want to use the property,

$\int e^x(f(x)+f'(x))dx=e^xf(x)$

So,

$\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-\sin x}{1-\cos x} \right )dx$

$=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1-2\sin \frac{x}{2}cos\frac{x}{2}}{2sin^2\frac{x}{2}} \right )dx$

$=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1}{2sin^2\frac{x}{2}} -\frac{2\sin \frac{x}{2}cos\frac{x}{2}}{2sin^2\frac{x}{2}}\right )dx$

$=\int_\frac{\pi}{2}^\pi e^x \left(\frac{1}{2}cosec^2\frac{x}{2}-cot\frac{x}{2}\right )dx$

$=\int_\frac{\pi}{2}^\pi e^x \left(-cot\frac{x}{2}+\frac{1}{2}cosec^2\frac{x}{2}\right )dx$

Here let's use the property

$\int e^x(f(x)+f'(x))dx=e^xf(x)$

so,

$=\int_\frac{\pi}{2}^\pi e^x \left(-cot\frac{x}{2}+\frac{1}{2}cosec^2\frac{x}{2}\right )dx$

$=\left [ -e^xcot\frac{x}{2} \right ]_\frac{\pi}{2}^\pi$

$=\left [ -e^\pi cot\frac{\pi}{2} \right ]-\left [ -e^{\frac{\pi}{2}} cot\frac{\pi}{4} \right ]$

$=e^{\frac{\pi}{2}}$

Question 25: Evaluate the definite integrals in Exercises 24 to 31.

$\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}$

Answer:

$\int_0^\frac{\pi}{4}\frac{\sin x\cos x }{\cos^4 x+\sin^4 x}$

First, let's convert sin and cos into tan and sec. (because we have a good relation in tan and square of sec,)

Let' divide both numerator and denominator by $cos^4x$

$=\int_0^\frac{\pi}{4}\frac{\frac{\sin x\cos x}{cosxcosxsos^2x} }{1+\frac{\sin^4 x}{\cos^4x}}$

$=\int_0^\frac{\pi}{4}\frac{tanxsec^2x}{1+tan^4x}$

Now lets change the variable

$\\t=tan^2x \\dt=2tanxsec^2xdx$

the limits will also change since the variable is changing

$when\:x=0,t=tan^20=0$

$when\:x=\frac{\pi}{4},t=tan^2\frac{\pi}{4}=1$

So, the integration becomes:

$I=\frac{1}{2}\int_{0}^{1}\frac{dt}{1+t^2}$

$I=\frac{1}{2}\left [ tan^{-1}t \right ]_0^1$

$I=\frac{1}{2}\left [ tan^{-1}1 \right ]-\frac{1}{2}\left [ tan^{-1}0\right ]$

$I=\frac{1}{2}\left [ \frac{\pi}{4} \right ]-0$

$I=\frac{\pi}{8}$

Question 26: Evaluate the definite integrals in Exercises 24 to 31.

$\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}$

Answer:

Lets first simplify the function.

$\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4\sin^2 x}=\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{\cos^2 x + 4(1-\cos^2 x)}=\int_0^\frac{\pi}{2}\frac{\cos^2 x dx}{4-3\cos^2 x}$

$\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{4-3\cos^2 x-4\:\: }{4-3\cos^2 x }dx=\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{4-3\cos^2 x\:\: }{4-3\cos^2 x }dx-\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{-4\:\: }{4-3\cos^2 x }dx$

$\\=\frac{-1}{3}\int_0^\frac{\pi}{2}1dx-\frac{-1}{3}\int_0^\frac{\pi}{2}\frac{-4\:\: }{4-3\cos^2 x }dx$
$=\frac{-1}{3} \left [ x \right ]_0^{\frac{\pi}{2}}-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4\:\: }{4-3\cos^2 x }dx$

As we have a good relation in between squares of the tan and square of sec lets try to take our equation there,

$=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{4sec^2x-3 }dx$

AS we can write square of sec in term of tan,

$=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{4(1+tan^2x)-3 }dx$

$=\frac{-1}{3} \left [ \frac{\pi}{2-0} \right ]-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{1+4tan^2x }dx$

Now let's calculate the integral of the second function, (we already have calculated the first function)

$=-\frac{1}{3}\int_0^\frac{\pi}{2}\frac{4sec^2x\:\: }{1+4tan^2x }dx$

let

$\\t=2tanx, \\dt=2sec^2xdx$

here we are changing the variable so we have to calculate the limits of the new variable

when x = 0, t = 2tanx = 2tan(0)=0

when $x=\pi/2,t=2tan{\pi/2}=\infty$

Our function in terms of t is

$=-\frac{2}{3}\int_0^\infty\frac{1 }{1+t^2 }dt$

$=\left [ tan^{-1} t\right ]_0^\infty=\left [ tan^{-1} \infty-tan^{-1} 0\right ]$

$=\frac{\pi}{2}$

Hence, our total solution of the function is

$\\=-\frac{\pi}{6}+\frac{2}{3}×\frac{\pi}{2}\\=\frac{\pi}{6}$

Question 27: Evaluate the definite integrals in Exercises 24 to 31.

$\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}$

Answer:

$\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{\sin 2x}}$

Here first let's convert sin2x as the angle of x ( sinx, and cosx)

$\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{2sinxcosx}}$

Now let's remove the square root form function by making a perfect square inside the square root

$\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{-(-1+1-2sinxcosx)}}$

$\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{(1-(sin^2x+cos^2x-2sinxcosx)}}$

$\\=\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sin x + \cos x }{\sqrt{(1-(sinx-cosx)^2}}$

Now let

, $\\t=sinx-cosx \\dt=(cosx+sinx)dx$

Since we are changing the variable, the limit of integration will change

$\\when\: x=\pi/6, t=sin\pi/6-cos\pi/6=(1-\sqrt{3})/2$
when $x= \pi/3,t=sin\pi/3-cos\pi/3=(\sqrt{3}-1)/2$

Our function in terms of t :

$\\=\int_\frac{1-\sqrt{3}}{2}^\frac{\sqrt{3}-1}{2} \frac{1}{\sqrt{(1-t^2)}}dt$

$\\=\left [ sin^{-1}t \right ]_\frac{1-\sqrt{3}}{2}^\frac{\sqrt{3}-1}{2}$
$=2sin^{-1}\left (\frac{\sqrt{3}-1}{2} \right )$

Question 28: Evaluate the definite integrals in Exercises 24 to 31.

$\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}$

Answer:

$\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}$

First, let's get rid of the square roots from the denominator,

$\\=\int_0^1\frac{dx}{\sqrt{1+x} -\sqrt x}×\frac{\sqrt{1+x} +\sqrt x}{\sqrt{1+x} +\sqrt x}$

$\\=\int_0^1\frac{\sqrt{1+x}+\sqrt{x}}{{1+x} -x}dx$

$\\=\int_0^1({\sqrt{1+x}+\sqrt{x}})dx$

$\\=\int_0^1({\sqrt{1+x})dx+\int_0^1({\sqrt{x}})dx$

$\\=\int_0^1(1+x)^\frac{1}{2}dx+\int_0^1x^\frac{1}{2}dx$

$\\=\left [ \frac{2}{3}(1+x)^{\frac{3}{2}} \right ]_0^1+\left [ \frac{2}{3}(x)^{\frac{3}{2}} \right ]_0^1$

$\\=\left [ \frac{2}{3}(1+1)^{\frac{3}{2}} \right ]-\left [ \frac{2}{3} \right ]+\left [ \frac{2}{3}(1)^{\frac{3}{2}} \right ]-\left [ 0 \right ]$

$\\=\frac{4\sqrt{2}}{3}$

Question 29: Evaluate the definite integrals in Exercises 24 to 31.

$\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx$

Answer:

$\int_0^\frac{\pi}{4}\frac{\sin x +\cos x }{9 + 16 \sin 2x}dx$

First let's assume t = cosx - sin x so that (sinx +cosx)dx=dt

So,

Now since we are changing the variable, the new limit of the integration will be,

when x = 0, t = cos0-sin0=1-0=1

when $x=\pi/4$ $t=\cos\pi/4-\sin\pi/4=0$

Now,

$(\cos x-\sin x)^2=t^2$

$\cos ^2x+\sin^2 x-2\cos x \sin x =t^2$

$1-\sin 2x =t^2$

$\sin 2x =1-t^2$

Hence our function in terms of t becomes,

$\int_{-1}^{0}\frac{dt}{9+16(1-t^2)}=\int_{-1}^{0}\frac{dt}{9+16-16t^2}=\int_{-1}^{0}\frac{dt}{25-16t^2}=\int_{-1}^{0}\frac{dt}{5^2-(4t)^2)}$

$= \frac{1}{4}\left [\frac{1}{2(5)}\log \frac{5+4t}{5-4t} \right ]_{-1}^0$

$= \frac{1}{40}\left[ \log (1)-\log (\frac{1}{9})\right ]$

$=\frac{\log 9}{40}$

Question 30: Evaluate the definite integrals in Exercises 24 to 31.

$\int_0^\frac{\pi}{2}\sin 2x\tan^{-1}(\sin x)dx$

Answer:

Let I =

$\int_{0}^{\frac{\pi}{2}}sin2xtan^{-1}(sinx)dx$

$=\int_{0}^{\frac{\pi}{2}}2sinxcosxtan^{-1}(sinx)dx$

Here, we can see that if we put sinx = t, then the whole function will convert in term of t with dx being changed to dt.so

$\\t=sinx \\dt=cosxdx$

Now the important step here is to change the limit of the integration as we are changing the variable.so,

$\\when\:x=0,t=sin0=0 \\when\:x=\frac{\pi}{2},t=sin\frac{\pi}{2}=1$

So our function becomes,

$I=2\int_{0}^{1}(tan^{-1}t)tdt$

Now, let's integrate this by using integration by parts method,

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\int\frac{1}{1+t^2}\cdot\frac{t^2}{2}dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\frac{t^2}{1+t^2}\cdot dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\frac{(1+t^2)-1}{1+t^2}\cdot dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}\int\left ( 1-\frac{1}{1+t^2} \right )\cdot dt \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}(t-tan^{-1}t) \right ]_0^1$

$I=2\left [ tan^{-1}t\cdot\frac{t^2}{2}-\frac{1}{2}(t)+\frac{1}{2}tan^{-1}t) \right ]_0^1$

$I=2\left [ \frac{1}{2} \left (tan^{-1}t\cdot(t^2+1)-t \right )\right ]_0^1$

$I=\left [ \left (tan^{-1}t\cdot(t^2+1)-t \right )\right ]_0^1$

$I=\left [ \left (tan^{-1}(1)\cdot(1^2+1)-1 \right )\right ]-\left [ \left (tan^{-1}(0)\cdot(0^2+1)-0 \right )\right ]$ $I=2tan^{-1}1-1=2\times \frac{\pi}{4}-1$

$I=\frac{\pi}{2}-1$

Question 31: Evaluate the definite integrals in Exercises 24 to 31

$\int_1^4[|x-1| + |x-2| + |x-3|]dx$

Answer:

Given integral $\int_1^4[|x-1| + |x-2| + |x-3|]dx$

So, we split it in according to intervals they are positive or negative.

$= \int_{1}^4 |x-1| dx + \int_{1}^4 |x-2| dx + \int^4_{1} |x-3| dx$

$= I_{1}+I_{2}+I_{3}$

Now,

$I_{1} = \int^4_{1}|x-1| dx = \int^4_{1} (x-1)dx$

$\because$ as $(x-1)$ is positive in the given x -range $[1,4]$

$=\left [ \frac{x^2}{2}-x\right ]^4_{1} = \left [ \frac{4^2}{2}-4 \right ] - \left [ \frac{1^2}{2}-1 \right ]$

$= \left [ 8-4 \right ] - [-\frac{1}{2}] = 4+\frac{1}{2} = \frac{9}{2}$

Therefore, $I_{1} = \frac{9}{2}$

$I_{2} = \int^4_{1}|x-2| dx = \int^2_{1} (2-x)dx +\int^4_{2} (x-2)dx$

$\because$ as $(x-2)\geq 0$ is in the given x -range $[2,4]$ and $\leq 0$ in the range $[1,2]$

$=\left [ 2x - \frac{x^2}{2}\right ] ^2_{1} + \left [ \frac{x^2}{2} -2x\right ] ^4_{2}$

$= \left \{ \left [ 2(2)-\frac{2^2}{2} \right ] - \left [ 2(1)-\frac{1^2}{2} \right ] \right \} + \left \{ \left [ \frac{4^2}{2}-2(4) \right ] - \left [ \frac{2^2}{2}-2(2) \right ] \right \}$

$= [4-2-2+\frac{1}{2}] +[8-8-2+4]$

$= \frac{1}{2}+2 =\frac{5}{2}$

Therefore, $I_{2} = \frac{5}{2}$

$I_{3} = \int^4_{1}|x-3| dx = \int^3_{1} (3-x)dx +\int^4_{3} (x-3)dx$

$\because$ as $(x-3)\geq 0$ is in the given x -range $[3,4]$ and $\leq 0$ in the range $[1,3]$

$=\left [ 3x - \frac{x^2}{2}\right ] ^3_{1} + \left [ \frac{x^2}{2} -3x\right ] ^4_{3}$

$= \left \{ \left [ 3(3)-\frac{3^2}{2} \right ] - \left [ 3(1)-\frac{1^2}{2} \right ] \right \} + \left \{ \left [ \frac{4^2}{2}-3(4) \right ] - \left [ \frac{3^2}{2}-3(3) \right ] \right \}$

$= [9-\frac{9}{2}-3+\frac{1}{2}]+[8-12-\frac{9}{2}+9]$

$= [6-4]+\frac{1}{2} =\frac{5}{2}$

Therefore, $I_{3} = \frac{5}{2}$

So, We have the sum $= I_{1}+I_{2}+I_{3}$

$I = \frac{9}{2}+\frac{5}{2}+\frac{5}{2} = \frac{19}{2}$

Question 32: Prove the following (Exercises 32 to 37)

$\int_1^3\frac{dx}{x^2(x+1)} = \frac{2}{3}+ \log \frac{2}{3}$

Answer:

L.H.S = $\int_1^3\frac{dx}{x^2(x+1)}$

We can write the numerator as [(x+1) -x]

$\therefore \int_1^3\frac{dx}{x^2(x+1)} = \int_1^3\frac{(x+1)-x}{x^2(x+1)}dx$

$\\ = \int_1^3\left [ \frac{1}{x^2} - \frac{1}{x(x+1)} \right ]dx $
$= \int_1^3\frac{1}{x^2}dx - \int_1^3\frac{(x+1)-x}{x(x+1)}dx$

$\\ = \int_1^3\frac{1}{x^2}dx - \int_1^3\left [ \frac{1}{x} - \frac{1}{(x+1)} \right ]dx $
$= \int_1^3\frac{1}{x^2}dx - \int_1^3\frac{1}{x}dx + \int_1^3\frac{1}{(x+1)}dx $
$= \left [ -\frac{1}{x} \right ]^3_1 - \left [ \log x \right ]^3_1 +\left [ \log(x+1) \right ]^3_1$

$\\ = \left [ -\frac{1}{3} +1 \right ] - \left [ \log 3 - \log 1 \right ] +\left [\log 4 - \log 2 \right ] $
$ = \frac{2}{3} + \log \left ( \frac{4}{3.2}\right ) \\$

$= \log \left(\frac{2}{3} \right ) +\frac{2}{3}$ = RHS

Hence proved.

Question 33: Prove the following (Exercises 32 to 37)

$\int_0^1 xe^xdx =1$

Answer:

$Let\ I=\int xe^{x}dx$

Integrating by parts

$\\I=x\int e^{x}dx-\int ( (\frac{\mathrm{d} (x)}{\mathrm{d} x})\int e^{x}dx)dx$
$I=xe^{x}-\int e^{x}dx\\ I=xe^{x}-e^{x}+c$

Applying Limits from 0 to 1

$\\\int_{0}^{1}xe^{x}dx=[xe^{x}-e^{x}+c]_{0}^{1}$
$I=[e-e+c]-[0-1+c]\\ I=1$

Hence proved I = 1

Question 34: Prove the following (Exercises 32 to 37)

$\int_{-1}^1x^{17}\cos^4 x dx=0$

Answer:

$Let \ x^{17}cos^{4}x=g(x)$

$g(-x)= (-x)^{17}cos^{4}(-x)=-x^{17}cos^{4}x=-g(x)$

The Integrand g(x) therefore is an odd function and therefore

$\int_{-1}^{1}g(x)dx=0$

Question 35: Prove the following (Exercises 32 to 37)

$\int_0^\frac{\pi}{2}\sin^3 x dx =\frac{2}{3}$

Answer:

$\\Let\ I= \int_{0}^{\frac{\pi }{2}}sin^{3}xdx$
⇒ $I=\int_{0}^{\frac{\pi }{2}}sinx(1-sin^{2}x)dx$
⇒ $I=\int_{0}^{\frac{\pi }{2}}sinxdx-\int_{0}^{\frac{\pi }{2}}cos^{2}xsinxdx$
⇒ $I=I_{1}-I_{2}$

$I_{1}=[-cosx]_{0}^{\frac{\pi }{2}}$
$I_{1}=-0-(-1)=1$

For I2 let cosx=t, -sinxdx=dt

The limits change to 0 and 1

$\\I_{2}=-\int_{1}^{0}t^{2}dt\\ I_{2}=-[\frac{t^{3}}{3}]{_{1}}^{0}\\ I_{2}=0-(-\frac{1}{3})\\ I_{2}=\frac{1}{3}$

I1 -I 2 = 2/3

Hence proved.

Question 36: Prove the following (Exercises 32 to 37)

$\int_0^\frac{\pi}{4}2\tan^3 x dx = 1 - \log 2$

Answer:

The integral is written as

$\\Let\ I=\int 2tan^{3}xdx\\ I=\int 2tan^{2}x\cdot tanxdx$
$I=\int 2(sec^{2}x-1)tanxdx\\ I=2\int tanxsec^{2}xdx-2\int tanxdx$
⇒ $I=2\int tdt-2log(cosx)+c\ \ \ \ \ \ \ (t=tanx)$
⇒ $I=t^{2}-2log(cosx)+c$
⇒ $I=tan^{2}x-2log(cosx)+c$

$[I]_{0}^{\frac{\pi }{4}}=[tan^{2}x-2log(cosx)]_{0}^{\frac{\pi }{4}}\\$

⇒ $[I]_{0}^{\frac{\pi }{4}}=(1-2log\sqrt{2})-(0-2log1)$

⇒ $[I]_{0}^{\frac{\pi }{4}}=1-log2$

Hence Proved.

Question 37: Prove the following (Exercises 32 to 37)

$\int_0^1\sin^{-1}xdx = \frac{\pi}{2}-1$

Answer:

$Let \ I=\int sin^{-1}xdx$

Integrating by parts we get

$I= sin^{-1}x\int 1\cdot dx-\int (\frac{\mathrm{d} (sin^{-1}x)}{\mathrm{d} x}\int 1\cdot dx)$
$I=xsin^{-1}x+c-\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx$
$I=I_{1}-I_{2}$

For I2 take 1-x2 = t2 , -xdx=tdt

$\\I_{2}=\int \frac{1}{\sqrt{1-x^{2}}}\cdot xdx$
$I_{2}=-\int\frac{1}{t}tdt \\ I_{2}=-t+c\\ I_{2}=-\sqrt{1-x^{2}}+c$

$[I]_{0}^{1}=[I_{1}-I_{2}]_{0}^{1}\\$

$\\=[xsin^{-1}x-(-\sqrt{1-x^{2}})]_{0}^{1}$
$=[xsin^{-1}x+\sqrt{1-x^{2}}]_{0}^{1}$
$=[1\cdot \frac{\pi }{2}+0]-[0+1]$
$=\frac{\pi }{2}-1$

Hence Proved.

Question 38: $\int\frac{dx}{e^x + e^{-x}}$ is equal to

(A) $\tan^{-1}(e^x) + c$

(B) $\tan^{-1}(e^{-x}) + c$

(C) $\log (e^x - e^{-x}) + C$

(D) $\log (e^x + e^{-x}) + C$

Answer:

$\int\frac{dx}{e^x + e^{-x}}$

The above integral can be rearranged as

$\\=\int \frac{e^{x}}{e^{2x}+1}dx\\$

let e x =t, e x dx=dt

$\int\frac{dx}{e^x + e^{-x}}$

$\\=\int \frac{1}{t^{2}+1}dt\\ =tan^{-1}t+c\\ =tan^{-1}(e^{x})+c$

(A) is correct

Question 39: $\int\frac{\cos 2x}{(\sin x + \cos x)^2}dx$ is equal to

(A) $\frac{-1}{\sin x + \cos x} + C$

(B) $\log |{\sin x + \cos x} |+ C$

(C) $\log |{\sin x- \cos x} |+ C$

(D) $\frac{1}{(\sin x + \cos x)^2} + C$

Answer:

$\\\frac{\cos 2x}{(\sin x + \cos x)^2}\\ =\frac{cos^{2}x-sin^{2}x}{(\sin x + \cos x)^2}\\ =\frac{(\sin x + \cos x)(\cos x-\sin x)}{(\sin x + \cos x)^2} \\=\frac{(\cos x-\sin x)}{(\sin x + \cos x)}$ cos2x=cos 2 x-sin 2 x

let sinx+cosx=t,(cosx-sinx)dx=dt

Hence the given integral can be written as

$\\\int\frac{\cos 2x}{(\sin x + \cos x)^2}dx\\ =\int \frac{dt}{t}\\ =log|t|+c \\=log|cosx+sinx|+c$

B is correct

Question 40: If $f(a+b-x) = f(x)$ , then $\int_a^bxf(x)dx$ is equal to

(A) $\frac{a+b}{2}\int^b_af(b-x)dx$

(B) $\frac{a+b}{2}\int^b_af(b+x)dx$

(C) $\frac{b-a}{2}\int^b_af(x)dx$

(D) $\frac{a+b}{2}\int^b_af(x)dx$

Answer:

$Let\ \int_a^bxf(x)dx=I$

As we know $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$

Using the above property we can write the integral as

$I=\int_{a}^{b}(a+b-x)f(a+b-x)dx$
⇒ $I=\int_{a}^{b}(a+b-x)f(x)dx$
⇒ $I=(a+b)\int_{a}^{b}f(x)dx-\int_{a}^{b}xf(x)dx$
⇒ $I=(a+b)\int_{a}^{b}f(x)dx-I$
⇒ $2I=(a+b)\int_{a}^{b}f(x)dx$
⇒ $I=\frac{a+b}{2}\int_{a}^{b}f(x)dx$

Answer (D) is correct.

Also, read,

• https://school.careers360.com/ncert/ncert-solutions-class-12-maths-chapter-7-exercise-7-1-integrals
• https://school.careers360.com/ncert/ncert-solutions-class-12-maths-chapter-7-exercise-7-2-integrals
• https://school.careers360.com/ncert/ncert-solutions-class-12-maths-chapter-7-exercise-7-3-integrals
• https://school.careers360.com/ncert/ncert-solutions-class-12-maths-chapter-7-exercise-7-4-integrals
• https://school.careers360.com/ncert/ncert-solutions-class-12-maths-chapter-7-exercise-7-5-integrals
• https://school.careers360.com/ncert/ncert-solutions-class-12-maths-chapter-7-exercise-7-6-integrals
• https://school.careers360.com/ncert/ncert-solutions-class-12-maths-chapter-7-exercise-7-7-integrals
• https://school.careers360.com/ncert/ncert-solutions-class-12-maths-chapter-7-exercise-7-8-integrals
• https://school.careers360.com/ncert/ncert-solutions-class-12-maths-chapter-7-exercise-7-9-integrals
• https://school.careers360.com/ncert/ncert-solutions-class-12-maths-chapter-7-exercise-7-10-integrals
• https://school.careers360.com/ncert/ncert-solutions-for-miscellaneous-exercise-chapter-7-class-12-integrals

Class 12 Maths NCERT Chapter 7: Extra Question

Question: $\int \frac{d x}{\cos x+\sqrt{3} \sin x}$ equals

Answer:

We start with the integral:

$I=\int \frac{d x}{\cos x+\sqrt{3} \sin x}$

$\begin{gathered}I=\int \frac{d x}{\frac{1}{2} \cdot 2 \cos x+\frac{\sqrt{3}}{2} \cdot 2 \sin x}= \\ \int \frac{d x}{\frac{1}{2}(\cos x+\sqrt{3} \sin x)}\end{gathered}$

This simplifies to:

$I=2 \int \frac{d x}{\cos x+\sqrt{3} \sin x}$

We can express $\cos x+\sqrt{3} \sin x$ in terms of a single cosine function. Recall the identity:

$\cos a \cos x+\sin a \sin x=\cos (x-a)$

where $a=\frac{\pi}{3}\left(\operatorname{since} \cos \frac{\pi}{3}=\frac{1}{2}\right.$ and $\left.\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right)$. Thus, we have:

$\cos x+\sqrt{3} \sin x=2\left(\frac{1}{2} \cos x+\frac{\sqrt{3}}{2} \sin x\right)$

$I=2 \int \frac{d x}{2 \cos \left(x-\frac{\pi}{3}\right)}=\int \frac{d x}{\cos \left(x-\frac{\pi}{3}\right)}$

The integral of sec is:

$\int \sec u d u=\ln |\sec u+\tan u|+C$

$I=\ln \left|\sec \left(x-\frac{\pi}{3}\right)+\tan \left(x-\frac{\pi}{3}\right)\right|+C$

Now we substitute back:

$I=\ln \left|\frac{1}{\cos \left(x-\frac{\pi}{3}\right)}+\frac{\sin \left(x-\frac{\pi}{3}\right)}{\cos \left(x-\frac{\pi}{3}\right)}\right|+C$

$I=\ln \left|\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)\right|+C$

Hence, answer is $I=\frac{1}{2} \ln \left|\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)\right|+C$.

Integrals Class 12 Chapter 7: Topics

The topics discussed in the NCERT Solutions for class 12, chapter 7, Integrals are::

• Introduction
• Integration as an Inverse Process of Differentiation
• Some properties of the indefinite integral
• Methods of Integration
• Integration using trigonometric identities
• Integrals of Some Particular Functions
• Integration by Partial Fractions
• Integration by Parts
• Definite Integral
• Fundamental Theorem of Calculus
• Evaluation of Definite Integrals by Substitution
• Some Properties of Definite Integrals

Integrals Class 12 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 Integrals:

∫[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

Approach to Solve Questions of Integrals – Class 12

1. Firstly, identify the type of integral: Indefinite Integral, Definite Integral, or Area under the curve.
2. After identifying the type, decide the method to integrate, such as substitution, integration by parts, or partial fractions.
3. For substitution, choose a substitution that simplifies the integrand, and don't forget to change the limits if it's a definite integral.
4. In integration by parts, apply the ILATE rule (Inverse, Log, Algebraic, Trig, Exponential) to choose which function to differentiate and which to integrate.
5. In partial fractions, break the rational function into simpler fractions and then integrate each one individually.
6. For definite integrals, after finding the integral, apply the limits of integration correctly using the Fundamental Theorem of Calculus.
7. Always simplify the final expression, and in definite integrals, express the result with appropriate units if required.

What Extra Should Students Study Beyond NCERT for JEE?

NCERT Solutions for Class 12 Maths: Chapter Wise

Students can access all the Maths solutions from the NCERT book from the links below.

Also, read,

• NCERT Exemplar Class 12 Maths Solutions Chapter 7 Integrals
• NCERT Notes Class 12 Maths Chapter 7 Integrals

NCERT Exemplar solutions for class 12 - subject-wise

Given below are the subject-wise links for the NCERT exemplar solutions of class 12:

• NCERT Exemplar Class 12 Chemistry Solutions
• NCERT Exemplar Class 12 Mathematics Solutions
• NCERT Exemplar Class 12 Biology Solutions
• NCERT Exemplar Class 12 Physics Solution

NCERT solutions for class 12 - subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

• NCERT solutions for class 12 mathematics
• NCERT solutions class 12 chemistry
• NCERT solutions for class 12 physics
• NCERT solutions for class 12 biology

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and NCERT syllabus for class 12:

• NCERT Books Class 12 Maths
• NCERT Syllabus Class 12 Maths
• NCERT Books Class 12
• NCERT Syllabus Class