# NCERT Solutions for Exercise 7.2 Class 12 Maths Chapter 7 - Integrals

NCERT solutions for Class 12 Maths chapter 7 exercise 7.2 is one of the exercises of chapter Integrals. In the Mathematics NCERT Book Class 12 chapter 7 first exercise we learnt basic concepts like finding out integrals of basic functions but in this exercise, advanced level questions are asked which strengthen the concepts of students. Exercise 7.2 Class 12 Maths increases the knowledge of Integrals which can further help in physics also.

NCERT solutions for exercise 7.2 Class 12 Maths chapter 7 should not be missed by any serious aspirant because it is observed in the past that some questions in JEE and NEET have been asked from this exercise. NCERT solutions for Class 12 Maths chapter 7 exercise 7.2 is provided below in detail made by subject matter experts. The NCERT chapter Integrals consists of below exercises also .

• Integrals Exercise 7.1

• Integrals Exercise 7.3

• Integrals Exercise 7.4

• Integrals Exercise 7.5

• Integrals Exercise 7.6

• Integrals Exercise 7.7

• Integrals Exercise 7.8

• Integrals Exercise 7.9

• Integrals Exercise 7.10

• Integrals Exercise 7.11

• Integrals Miscellaneous Exercise

## Integrals Class 12 Chapter 7Exercise 7.2

Question:1 Integrate the functions

Given to integrate function,

Let us assume

we get,

now back substituting the value of

as is positive we can write

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

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

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

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

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

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

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

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

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

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

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

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

$= \log|t| +C$

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

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

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

Let us assume $\cos x =t$

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

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

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

$= \cos t +C$

Back substituting the value of t we get,

$= \cos (\cos x ) +C$

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

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

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

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

we get, $2adx =dt$

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

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

Now, by back substituting the value of t,

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

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

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

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

we get, $adx =dt$

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

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

Now, by back substituting the value of t,

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

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

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

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

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

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

$\therefore dx =dt$

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

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

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

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

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

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

Back substituting the value of t in the above equation.

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

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

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

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

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

$\therefore 4xdx =dt$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Can be written in the form:

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

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

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

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

$= 2\log|t| +C$

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

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

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

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

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

$\therefore dx=dt$

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

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

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

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

, where C is any constant value.

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

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

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

Assume the $x^3-1 = t$

$\therefore 3x^2dx=dt$

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

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

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

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

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

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

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

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

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

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

$\therefore 9x^2dx=dt$

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

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

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

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

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

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

Assume the $\log x =t$

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

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

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

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

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

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

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

$\therefore -8xdx =dt$

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

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

Now back substituting the value of t ;

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

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

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

Assume the $2x+3 =t$

$\therefore 2dx =dt$

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

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

Now back substituting the value of t ;

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

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

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

Assume the $x^2=t$

$\therefore 2xdx =dt$

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

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

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

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

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

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

Given,

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

Let's do the following substitution

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

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

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

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

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

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

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

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

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

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

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

$= \log |t| +C$

Now, back substituting the value of t,

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

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

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

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

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

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

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

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

Now, back substituting the value of t,

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

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

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

Assume the $2x-3 =t$

$\therefore 2dx =dt$

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

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

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

Now, back substituting the value of t,

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

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

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

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

Assume the $7-4x=t$

$\therefore -4dx =dt$

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

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

Now, back substituted the value of t.

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

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

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

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

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

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

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

Now, back substituted the value of t.

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

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

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

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

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

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

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

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

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

Now, back substituted the value of t.

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

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

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

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

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

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

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

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

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

Now, back substituted the value of t.

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

where C is any constant value.

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

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

Assume the $\sqrt x =t$

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

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

$= 2\sin t +C$

Now, back substituted the value of t.

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

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

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

Assume the $\sin 2x = t$

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

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

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

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

Now, back substituted the value of t.

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

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

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

Assume the $1+\sin x =t$

$\therefore \cos x dx = dt$

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

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

$= 2\sqrt t +C$

Now, back substituted the value of t.

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

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

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

Assume the $\log \sin x =t$

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

$\cot x dx =dt$

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

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

Now, back substituted the value of t.

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

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

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

Assume the $1+\cos x =t$

$\therefore -\sin x dx =dt$

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

$= -\log|t| +C$

Now, back substituted the value of t.

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

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

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

Assume the $1+\cos x =t$

$\therefore -\sin x dx =dt$

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

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

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

Now, back substituted the value of t.

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

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

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

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

Now solving the assumed integral;

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

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

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

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

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

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

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

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

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

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

Now, back substituting the value of t,

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

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

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

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

Now solving the assumed integral;

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

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

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

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

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

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

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

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

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

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

Now, back substituting the value of t,

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

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

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

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

Now solving the assumed integral;

Multiplying numerator and denominator by $\cos x$ ;

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

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

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

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

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

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

$=2\sqrt t +C$

Now, back substituting the value of t,

$= 2\sqrt{\tan x } +C$

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

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

Assume that $1+\log x =t$

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

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

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

Now, back substituting the value of t,

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

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

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

Simplifying to solve easier;

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

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

Assume that $x+\log x =t$

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

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

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

Now, back substituting the value of t,

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

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

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

Assume that $x^4 =t$

$\therefore 4x^3 dx =dt$

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

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

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

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

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

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

Now back substitute the value of u,

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

and then back substituting the value of t,

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

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

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

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

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

Now differentiating both sides we get

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

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

$= \log t +C$

Back substituting the value of t,

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

Therefore the correct answer is D.

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

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

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

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

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

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

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

$=\tan x -\cot x +C$

Therefore, the correct answer is B.

## More About NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2

The NCERT Class 12 Maths chapter Integrals gives the complete idea of Integrals and few of its applications. Exercise 7.2 Class 12 Maths provides solutions to 39 main questions and their sub-questions. Topics like integration of square root functions, exponential functions etc. are discussed. NCERT solutions for Class 12 Maths chapter 7 exercise 7.2 is recommended to students to perform better in the examination.

Also Read| Integrals Class 12 Notes

## Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2

• The NCERT syllabus Class 12th Maths chapter 7.2 exercise is made by experienced subject matter experts.

• Practicing exercise 7.2 Class 12 Maths is going to be beneficial for sure to perform better in the examination.

• These Class 12 Maths chapter 7 exercise 7.2 solutions are asked word to word in the board examinations.

• NCERT Solutions for Class 12 Maths chapter 7 exercise 7.2 can be helpful in physics topics also

## Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 7

• NCERT Solutions for Class 12 Maths Chapter 7

## NCERT Solutions Subject Wise

• NCERT Solutions Class 12 Chemistry

• NCERT Solutions for Class 12 Physics

• NCERT Solutions for Class 12 Biology

• NCERT Solutions for Class 12 Mathematics

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 12 Maths

• NCERT Exemplar Class 12 Physics

• NCERT Exemplar Class 12 Chemistry

• NCERT Exemplar Class 12 Biology

Happy learning!!!