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

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

Edited By Komal Miglani | Updated on May 08, 2025 02:33 PM IST | #CBSE Class 12th
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In the world of calculus, integrals teach us the power of accumulation—a minor change at a time adds up to something big. NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.2 are significant in the learning process of integrals. This exercise helps students get accustomed to one of the important integration methods: Substitution. Integration by substitution is like the reverse chain rule in calculus. These NCERT solutions follow the latest CBSE guidelines.

This Story also Contains
  1. Class 12 Maths Chapter 7 Exercise 7.2 Solutions: Download PDF
  2. Integrals Class 12 Chapter 7 Exercise 7.2
  3. Topics covered in Chapter 1 Integrals: Exercise 7.2
  4. NCERT Solutions Subject Wise
  5. Subject-Wise NCERT Exemplar Solutions

The main purpose of 12th class Maths exercise 7.2 of NCERT is to help students understand how to simplify complex integrals by changing variables to make the expression easier to integrate. Careers360 experts supplied in-depth explanations and structured answers to ensure no step went unexplained.

Class 12 Maths Chapter 7 Exercise 7.2 Solutions: Download PDF

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Integrals Class 12 Chapter 7 Exercise 7.2

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

Answer:

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

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

we get, $2x\,dx = dt$

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

$= \log|t| + C$

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

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

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

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

Answer:

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

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

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

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

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

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

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

Answer:

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

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

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

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

$= \log|t| +C$

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

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

Answer:

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

Let us assume $\cos x =t$

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

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

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

$= \cos t +C$

Back substituting the value of t we get,

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

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

Answer:

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

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

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

we get, $2adx =dt$

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

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

Now, by back substituting the value of t,

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

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

Answer:

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

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

we get, $adx =dt$

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

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

Now, by back substituting the value of t,

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

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

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

Answer:

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

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

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

$\therefore dx =dt$

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

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

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

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

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

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

Back substituting the value of t in the above equation.

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

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

Answer:

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

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

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

$\therefore 4xdx =dt$

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

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

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

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

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

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 function $\frac{x }{ \sqrt{ x +4} }$ , 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 function $( x ^3 - 1 ) ^{1/3} x ^ 5$

Answer:

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

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

Assume the $x^3-1 = t$

$\therefore 3x^2dx=dt$

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

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

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

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

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

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

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

Answer:

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

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

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

$\therefore 9x^2dx=dt$

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

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

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

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

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

Answer:

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

Assume the $\log x =t$

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

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

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

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

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

Answer:

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

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

$\therefore -8xdx =dt$

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

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

Now back substituting the value of t ;

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

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

Answer:

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

Assume the $2x+3 =t$

$\therefore 2dx =dt$

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

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

Now back substituting the value of t ;

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

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

Answer:

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

Assume the $x^2=t$

$\therefore 2xdx =dt$

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

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

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

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

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

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

Answer:

Given,

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

Let's do the following substitution

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

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

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

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

Answer:

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

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

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

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

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

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

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

$= \log |t| +C$

Now, back substituting the value of t,

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

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

Answer:

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

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

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

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

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

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

Now, back substituting the value of t,

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

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

Answer:

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

Assume the $2x-3 =t$

$\therefore 2dx =dt$

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

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

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

Now, back substituting the value of t,

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

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

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

Answer

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

Assume the $7-4x=t$

$\therefore -4dx =dt$

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

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

Now, back substituted the value of t.

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

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

Answer:

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

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

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

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

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

Now, back substituted the value of t.

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

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

Answer:

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

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

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

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

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

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

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

Now, back substituted the value of t.

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

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

Answer:

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

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

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

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

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

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

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

Now, back substituted the value of t.

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

where C is any constant value.

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

Answer:

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

Assume the $\sqrt x =t$

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

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

$= 2\sin t +C$

Now, back substituted the value of t.

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

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

Answer:

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

Assume the $\sin 2x = t$

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

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

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

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

Now, back substituted the value of t.

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

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

Answer:

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

Assume the $1+\sin x =t$

$\therefore \cos x dx = dt$

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

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

$= 2\sqrt t +C$

Now, back substituted the value of t.

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

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

Answer:

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

Assume the $\log \sin x =t$

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

$\cot x dx =dt$

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

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

Now, back substituted the value of t.

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

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

Answer:

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

Assume the $1+\cos x =t$

$\therefore -\sin x dx =dt$

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

$= -\log|t| +C$

Now, back substituted the value of t.

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

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

Answer:

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

Assume the $1+\cos x =t$

$\therefore -\sin x dx =dt$

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

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

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

Now, back substituted the value of t.

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

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

Answer:

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

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

Now solving the assumed integral;

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

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

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

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

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

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

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

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

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

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

Now, back substituting the value of t,

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

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

Answer:

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

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

Now solving the assumed integral;

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

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

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

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

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

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

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

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

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

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

Now, back substituting the value of t,

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

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

Answer:

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

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

Now solving the assumed integral;

Multiplying numerator and denominator by $\cos x$ ;

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

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

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

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

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

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

$=2\sqrt t +C$

Now, back substituting the value of t,

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

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

Answer:

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

Assume that $1+\log x =t$

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

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

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

Now, back substituting the value of t,

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

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

Answer:

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

Simplifying to solve easier;

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

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

Assume that $x+\log x =t$

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

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

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

Now, back substituting the value of t,

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

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

Answer:

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

Assume that $x^4 =t$

$\therefore 4x^3 dx =dt$

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

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

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

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

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

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

Now back substitute the value of u,

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

and then back substituting the value of t,

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

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

$ (A) \ 10^x - x^{10} + C $

$ (B) \ 10^x + x^{10} + C $

$ (C) \ (10^x - x^{10})^{-1} + C $

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

Answer:

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

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

Now differentiating both sides we get

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

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

$= \log t +C$

Back substituting the value of t,

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

Therefore the correct answer is $\log(10^x + x^{10}) + C$.

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

$(A) \tan x + \cot x + C$

$(B) \tan x - \cot x + C$

$(C) \tan x \cot x + C$

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

Answer:

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

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

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

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

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

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

Therefore, the correct answer is $\tan x - \cot x + C$.

Also, read

Topics covered in Chapter 1 Integrals: Exercise 7.2

Integration by substitution

It is a method used to simplify integration by changing the variable. It helps solve integrals involving composite functions by making the substitution for part of the expression.

If $I=\int f(g(x)) \cdot g^{\prime}(x) d x$, let $u=g(x)$ so that $d u=g^{\prime}(x) d x$, then:
$I=\int f(u) d u$


Also, read,

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Frequently Asked Questions (FAQs)

1. Define Integration ?

Integration is the process of finding the antiderivative of a function which gives the area of the curve formed by a function. It is useful in finding the area, centre of mass etc. 

2. What is the use of Integration?

Integration is used to find the area, centre of mass etc. It has great application in Physics also. 

3. What is the weightage of Integrals in Board examination ?

Integration along with Application of Derivatives, holds good weightage in the examination. Aroung 20% marks questions are asked from these 2 chapters. 

4. How difficult is the chapter Integrals ?

Initial concepts are quite easy to grasp but in later exercises, significant hard work is required to understand the concepts in a holistic manner. 

5. What are the topics covered in Exercise 7.2 Class 12 Maths ?

Topics like integrals of root functions and trigonometric functions are discussed in this chapter.

6. How many questions are there in the exercise 7.2 Class 12 Maths ?

There are 39 questions in exercise 7.2 Class 12 Maths

7. What is the importance of chapter integration from Boards perspective?

Integration is a topic which has varied applications, it is used heavily in application of Integrals, physics as well as sometimes in chemistry (quantum chemistry). Hence it has a significant role in Board examination.

8. Should we memorise values of basic integral functions?

It can help in solving questions faster. Hence integrals of basic functions can be memorised.

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I hope this information helps you.







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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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