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

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