NCERT Solutions for Exercise 7.6 Class 12 Maths Chapter 7 - Integrals

Question 2: Integrate the functions $x \sin 3x$

Answer:

Given function is
$f(x)=x \sin 3x$
We will use integration by parts method

$\int x\sin 3x \, dx = x \int \sin 3x \, dx - \int \left( \frac{d(x)}{dx} \cdot \int \sin 3x \, dx \right) dx$

$\int x\sin 3x \, dx = x \cdot \left( \frac{-\cos 3x}{3} \right) - \int \left( 1 \cdot \frac{-\cos 3x}{3} \right) dx$

$\int x\sin 3x \, dx = -\frac{x\cos 3x}{3} + \frac{\sin 3x}{9} + C$
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

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

$\int x^2 e^x \, dx = x^2 e^x - \int (2x e^x) \, dx$
Again use integration by parts in $\int (2x.e^x)dx\\$

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

$\int 2x e^x \, dx = 2x e^x - \int 2 e^x \, dx$

$\int 2x e^x \, dx = 2x e^x - 2e^x$
Put this value in our equation
we will get,

$\int x^2 e^x \, dx = x^2 e^x - 2x e^x + 2e^x + C$

$\int x^2 e^x \, dx = e^x (x^2 - 2x + 2) + C$

Therefore, answer is $e^x(x^2-2x+2)+ C$

Question 4: Integrate the functions $x \log x$

Answer:

Given function is
$f(x)=x.\log x$
We will use integration by parts method
$\int x \log x \, dx = \log x \cdot \int x \, dx - \int \left( \frac{d(\log x)}{dx} \cdot \int x \, dx \right) dx$
$\int x \log x \, dx = \log x \cdot \frac{x^2}{2} - \int \left( \frac{1}{x} \cdot \frac{x^2}{2} \right) dx$
$\int x \log x \, dx = \log x \cdot \frac{x^2}{2} - \int \frac{x}{2} \, dx$
$\int x \log x \, dx = \log x \cdot \frac{x^2}{2} - \frac{x^2}{4} + C$

Therefore, the answer is $\frac{x^2}{2}\log x- \frac{x^2}{4}+ C$

Question 5: Integrate the functions $x \log 2x$

Answer:

Given function is
$f(x)=x.\log 2 x$
We will use integration by parts method

$\int x \log 2x \, dx = \log 2x \cdot \int x \, dx - \int \left( \frac{d(\log 2x)}{dx} \cdot \int x \, dx \right) dx$

$\int x \log 2x \, dx = \log 2x \cdot \frac{x^2}{2} - \int \left( \frac{2}{2x} \cdot \frac{x^2}{2} \right) dx$

$\int x \log 2x \, dx = \log 2x \cdot \frac{x^2}{2} - \int \frac{x}{2} \, dx$

$\int x \log 2x \, dx = \log 2x \cdot \frac{x^2}{2} - \frac{x^2}{4} + C$
Therefore, the answer is $\log 2x.\frac{x^2}{2}- \frac{x^2}{4}+ C$

Question 6: Integrate the functions $x^ 2 \log x$

Answer:

Given function is
$f(x)=x^2.\log x$
We will use integration by parts method

$\int x^2 \log x \, dx = \log x \cdot \int x^2 \, dx - \int \left( \frac{d(\log x)}{dx} \cdot \int x^2 \, dx \right) dx$

$\int x^2 \log x \, dx = \log x \cdot \frac{x^3}{3} - \int \left( \frac{1}{x} \cdot \frac{x^3}{3} \right) dx$

$\int x^2 \log x \, dx = \log x \cdot \frac{x^3}{3} - \int \frac{x^2}{3} \, dx$

$\int x^2 \log x \, dx = \log x \cdot \frac{x^3}{3} - \frac{x^3}{9} + C$
Therefore, the answer is $\log x.\frac{x^3}{3}- \frac{x^3}{9}+ C$

Question 7: Integrate the functions $x \sin ^{ -1} x$

Answer:

Given function is
$f(x)=x.\sin^{-1} x$
We will use integration by parts method

$\int x \sin^{-1}x \, dx = \sin^{-1}x \int x \, dx - \int \left( \frac{d(\sin^{-1}x)}{dx} \cdot \int x \, dx \right) dx$

$\int x \sin^{-1}x \, dx = \sin^{-1}x \cdot \frac{x^2}{2} - \int \left( \frac{1}{\sqrt{1 - x^2}} \cdot \frac{x^2}{2} \right) dx$
Now, we need to integrate $\int (\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$

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

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

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

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

$\int \frac{-x^2}{2\sqrt{1-x^2}} \, dx = \frac{x \sqrt{1-x^2}}{4} - \frac{\sin^{-1} x}{4} + C$

Put this value in our equation

Therefore, the answer is $\int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x\sqrt{1-x^2}}{4}$

Question 8: Integrate the functions $x \tan ^{-1} x$

Answer:

Given function is
$f(x)=x.\tan^{-1} x$
We will use integration by parts method

$\int x \tan^{-1} x \, dx = \tan^{-1} x \int x \, dx - \int \left( \frac{d(\tan^{-1} x)}{dx} \cdot \int x \, dx \right) dx$

$\int x \tan^{-1} x \, dx = \tan^{-1} x \cdot \frac{x^2}{2} - \int \left( \frac{1}{1 + x^2} \cdot \frac{x^2}{2} \right) dx$

$\int x \tan^{-1} x \, dx = \tan^{-1} x \cdot \frac{x^2}{2} - \frac{1}{2} \int \left( \frac{x^2 + 1}{1 + x^2} - \frac{1}{1 + x^2} \right) dx$

$\int x \tan^{-1} x \, dx = \tan^{-1} x \cdot \frac{x^2}{2} - \frac{1}{2} \int \left( 1 - \frac{1}{1 + x^2} \right) dx$

$\int x \tan^{-1} x \, dx = \tan^{-1} x \cdot \frac{x^2}{2} - \frac{1}{2} \left( x - \tan^{-1} x \right) + C$

$\int x \tan^{-1} x \, dx = \frac{\tan^{-1} x}{2} \left( 2x^2 + 1 \right) - \frac{x}{2} + C$
Put this value in our equation
$\int x\sin^{-1} xdx = \sin^{-1} x.\frac{x^2}{2}- \frac{x}{4\sqrt{1-x^2}}-\frac{\sin^{-1}x}{4}+C\\ \int x\sin^{-1} xdx =\frac{\sin^{-1}x}{4}(2x^2-1)-\frac{x}{4\sqrt{1-x^2}}$

Therefore, the answer is $\frac{\tan^{-1}x}{2}(2x^2+1)-\frac{x}{2}+C$

Question 9: Integrate the functions $x\cos ^{ -1} x$

Answer:

Given function is
$f(x)=x.\cos^{-1} x$
We will use integration by parts method

$\int x \cos^{-1} x \, dx = \cos^{-1} x \int x \, dx - \int \left( \frac{d(\cos^{-1} x)}{dx} \cdot \int x \, dx \right) dx$

$\int x \cos^{-1} x \, dx = \cos^{-1} x \cdot \frac{x^2}{2} - \int \left( \frac{-1}{\sqrt{1 - x^2}} \cdot \frac{x^2}{2} \right) dx$
Now, we need to integrate $\int (\frac{-1}{\sqrt{1-x^2}}.\frac{x^2}{2})dx\\$

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

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

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

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

$\int \frac{-x^2}{2\sqrt{1-x^2}} \, dx = \frac{x \sqrt{1 - x^2}}{4} - \frac{\cos^{-1} x}{4} + \frac{\cos^{-1} x}{2} + C$
Put this value in our equation
$\int x\cos^{-1} xdx = \cos^{-1} x.\frac{x^2}{2}-\left ( \frac{x\sqrt{1-x^2}}{4} -\frac{\cos^{-1}x}{4}+\frac{\cos^{-1}x}{2}+C \right )\\ \\ \int x\cos^{-1} xdx =\frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}$

Therefore, the answer is $\frac{\cos^{-1}x}{4}(2x^2-1)- \frac{x\sqrt{1-x^2}}{4}$

Question 10: Integrate the functions $( \sin ^{-1}x ) ^ 2$

Answer:

Given function is
$f(x)=( \sin ^{-1}x ) ^ 2$
we will use integration by parts method

$\int (\sin^{-1} x)^2 \, dx = (\sin^{-1} x)^2 \int 1 \, dx - \int \left( \frac{d((\sin^{-1} x)^2)}{dx} \cdot \int 1 \, dx \right) dx$

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

$\int (\sin^{-1} x)^2 \, dx = (\sin^{-1} x)^2 \cdot x + \left[ \sin^{-1} x \cdot \int \frac{-2x}{\sqrt{1 - x^2}} \, dx - \int \left( \frac{d(\sin^{-1} x)}{dx} \cdot \int \frac{-2x}{\sqrt{1 - x^2}} \, dx \right) \right]$

$ = (\sin^{-1} x)^2 \cdot x + \left[ \sin^{-1} x \cdot 2 \sqrt{1 - x^2} - \int \frac{1}{\sqrt{1 - x^2}} \cdot 2 \sqrt{1 - x^2} \, dx \right]$

$ = (\sin^{-1} x)^2 \cdot x + 2 \sin^{-1} x \sqrt{1 - x^2} - 2x + C$
Therefore, answer is $(\sin^{-1}x)^2.x + 2\sin^{-1}x\sqrt{1-x^2}-2x+C$

Question 11: Integrate the functions $\frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}$

Answer:

Consider $\int \frac{x \cos ^{-1}}{\sqrt { 1- x^2 }}dx =I$

So, we have then: $I = \frac{-1}{2}\int \frac{-2x}{\sqrt{1-x^2}}. \cos^{-1}x dx$

After taking $\cos ^{-1}x$ as a first function and $\left ( \frac{-2x}{\sqrt{1-x^2}} \right )$ as second function and integrating by parts, we get

$I =-\frac{1}{2}\left [ \cos^{-1}x\int\frac{-2x}{\sqrt{1-x^2}}dx - \int\left \{ \left ( \frac{d}{dx}\cos^{-1}x \right )\int \frac{-2x}{\sqrt{1-x^2}}dx \right \}dx \right ]$ $=-\frac{1}{2}\left [ \cos^{-1}x.2{\sqrt{1-x^2}} + \int \frac{-1}{\sqrt{1-x^2}}.2\sqrt{1-x^2}dx \right ]$

$=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-\int2dx \right ]$

$=\frac{-1}{2}\left [ 2\sqrt{1-x^2}\cos^{-1}x-2x \right ]+C$

Or, $- \left( \sqrt{1 - x^2} \cos^{-1}x + x \right) + C$

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

Answer:

Consider $x \sec ^2 x$

So, we have then: $I =\int x\sec^2 x dx$

After taking $x$ as a first function and $\sec^2x$ as second function and integrating by parts, we get

$I =x\int \sec^2 x dx -\int \left \{ \left ( \frac{d}{dx}x \right )\int \sec^2 x dx \right \}dx$

$= x\tan x -\int1.\tan x dx$

$= x\tan x +\log|\cos x | +C$

Question 13: Integrate the functions $\tan ^{-1} x$

Answer:

Consider $\tan ^{-1} x$

So, we have then: $I =\int 1.\tan^{-1}x dx$

After taking $\tan^{-1}x$ as a first function and $1$ as second function and integrating by parts, we get

$I = \tan^{-1}x \int 1dx -\int \left \{ \left ( \frac{d}{dx}\tan^{-1}x \right )\int1.dx \right \}dx$

$= \tan^{-1}x.x -\int \frac{1}{1+x^2}.xdx$

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

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

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

Question 14: Integrate the functions $x ( \log x )^ 2$

Answer:

Consider $x ( \log x )^ 2$

So, we have then: $I = \int x(\log x)^2 dx$

After taking $(\log x )^2$ as a first function and $x$ as second function and integrating by parts, we get

$I = (\log x )^2 \int xdx -\int \left \{ \left ( \frac{d}{dx} (\log x)^2 \right )\int x.dx \right \}dx$

$= (\log x)^2 .\frac{x^2}{2} - \int \frac{2\log x }{x}.\frac{x^2}{2} dx$

$= (\log x)^2 .\frac{x^2}{2} - \int x\log x dx$

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

Question 15: Integrate the functions $( x^2 + 1 ) \log x$

Answer:

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

So, we have then: $I = \int (x^2+1) \log x dx = \int x^2 \log x dx +\int \log x dx$

Let us take $I = I_{1} +I_{2}$ ....................(1)

Where, $I_{1} = \int x^2\log x dx$ and $I_{2} = \int \log x dx$

So, $I_{1} = \int x^2\log x dx$

After taking $\log x$ as a first function and $x^2$ as second function and integrating by parts, we get

$I = \log x \int x^2dx -\int \left \{ \left ( \frac{d}{dx} \log x \right )\int x^2.dx \right \}dx$

$= \log x .\frac{x^3}{3} - \int \frac{1}{x}.\frac{x^3}{3} dx$

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

$I_{2} = \int \log x dx$

After taking $\log x$ as a first function and $1$ as second function and integrating by parts, we get

$I_{2} = \log x \int 1.dx - \int \left \{ \left ( \frac{d}{dx}\log x \right ) \int 1.dx \right \}dx$

$= \log x .x -\int \frac{1}{x}. xdx$

$= x\log x -\int 1 dx$

$= x\log x -x +C_{2}$ ................(3)

Now, using the two equations (2) and (3) in (1) we get,

$I = \frac{x^3}{3}\log x -\frac{x^3}{9} +C_{1} +x\log x - x +C_{2}$

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

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

Question 16: Integrate the functions $e ^ x ( \sin x + \cos x )$

Answer:

Let suppose
$I =$ $e ^ x ( \sin x + \cos x )$
$f(x) = \sin x \Rightarrow f'(x) = \cos x$
we know that,
$I =\int e^x[f(x)+f'(x)]dx = e^x[f(x)]+C$
Thus, the solution of the given integral is given by

$\therefore I = e^x\sin x +C$

Question 17: Integrate the functions $\frac{x e ^x }{( 1+ x )^2}$

Answer:

$\frac{x e ^x }{( 1+ x )^2}$
Let suppose
$I = \int \frac{e^x(x)}{(1+x)^2}dx$
by rearranging the equation, we get
$\Rightarrow \int e^x[\frac{1}{1+x}-\frac{1}{(1+x)^2}]dx$
let
$f(x)=\frac{1}{1+x} \Rightarrow f'(x)= -\frac{1}{(1+x)^2}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$
therefore the solution of the given integral is

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

Question 18: Integrate the functions $e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )$

Answer:

Let
$I =e ^x \left ( \frac{1+ \sin x }{1+ \cos x } \right )$
substitute $1 =\sin ^2\frac{x}{2}+\cos^2\frac{x}{2}$ and $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$

$\\\Rightarrow e^x(\frac{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}})\\ =e^x(\frac{1}{2}\sec^2\frac{x}{2}+\tan\frac{x}{2})\\$
let
$f(x) =\tan\frac{x}{2} \Rightarrow f'(x)=\frac{1}{2}\sec^2\frac{x}{2}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$
Therefore the solution of the given integral is

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

Question 19: Integrate the functions $e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )$

Answer:

$e ^ x \left ( \frac{1 }{x} - \frac{1}{x^2}\right )$
It is known that
$\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$

let
$f(x)=\frac{1}{x}\Rightarrow f'(x)=-\frac{1}{x^2}$
Therefore the required solution of the given above integral is
$I = e^x.\frac{1}{x}+C$

Question 20: Integrate the functions $\frac{( x-3)e ^x }{( x-1)^3}$

Answer:

$\frac{( x-3)e ^x }{( x-1)^3}$
It is known that $\int e^x[f(x)+f'(x)]=e^x[f(x)]+C$

So, By adjusting the given equation, we get
$\int\frac{( x-3)e ^x }{( x-1)^3} =\int e^x(\frac{x-1-2}{(x-1)^3}) =\int e^x({\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3})}dx$

to let
$f(x)=\frac{1}{(x-1)^2}\Rightarrow f'(x)=-\frac{2}{(x-1)^3}$
Therefore the required solution of the given $I=\frac{e^x}{(x-1)^2}+C$ integral is

Question 21: Integrate the functions $e ^{ 2x } \sin x$

Answer:

Let
$I =e ^{ 2x } \sin x$
By using integrating by parts, we get

$\\ = \sin x \int e^{2x} \, dx - \int \left( \frac{d}{dx} \sin x \cdot \int e^{2x} \, dx \right) dx$

$= \frac{\sin x \cdot e^{2x}}{2} - \frac{1}{2} \int e^{2x} \cos x \, dx$

$= \frac{\sin x \cdot e^{2x}}{2} - \frac{1}{2} \left[ \cos x \int e^{2x} \, dx - \int \left( \frac{d}{dx} \cos x \cdot \int e^{2x} \, dx \right) dx \right]$

$= \frac{\sin x \cdot e^{2x}}{2} - \frac{1}{2} \left[ \cos x \cdot \frac{e^{2x}}{2} + \frac{1}{2} \int e^{2x} \sin x \, dx \right]$

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

$\Rightarrow \frac{5}{4} I = \frac{\sin x \cdot e^{2x}}{2} - \frac{1}{4} \cos x \cdot e^{2x}$

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

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

$\\ = 2 \left[ \theta \int \sec^2 \theta \, d\theta - \int \left( \frac{d}{d\theta} \theta \cdot \int \sec^2 \theta \, d\theta \right) d\theta \right]$

$ = 2 \left[ \theta \tan \theta - \int \tan \theta \, d\theta \right] + C$

$ = 2 \left[ \theta \tan \theta + \log \left| \cos \theta \right| \right] + C$

$ = 2x \tan^{-1} x + 2 \log \left( (1 + x^2)^{-1/2} \right)$

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