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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.
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.
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
Integration by substitution |
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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: |
Also, read,
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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.
Integration is used to find the area, centre of mass etc. It has great application in Physics also.
Integration along with Application of Derivatives, holds good weightage in the examination. Aroung 20% marks questions are asked from these 2 chapters.
Initial concepts are quite easy to grasp but in later exercises, significant hard work is required to understand the concepts in a holistic manner.
Topics like integrals of root functions and trigonometric functions are discussed in this chapter.
There are 39 questions in exercise 7.2 Class 12 Maths
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.
It can help in solving questions faster. Hence integrals of basic functions can be memorised.
Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
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I hope this information helps you.
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If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
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