Chapter 7: Integrals

Ans: The given expression is, $\dfrac{{{x}^{2}}+x+1}{{{(x+1)}^{2}}\left( x+2 \right)}$. 

Now consider it as shown below,

$\therefore \dfrac{{{x}^{2}}+x+1}{{{(x+1)}^{2}}\left( x+2 \right)}=\dfrac{A}{\left( x+1 \right)}+\dfrac{B}{\left( x+1 \right)}+\dfrac{C}{\left( x+2 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow {{x}^{2}}+x+1=A\left( x+1 \right)\left( x+2 \right)+B\left( x+2 \right)+C{{\left( x+1 \right)}^{2}}$

$\Rightarrow {{x}^{2}}+x+1=A\left( {{x}^{2}}+3x+2 \right)+B\left( x+2 \right)+C\left( {{x}^{2}}+2x+1 \right)$

$\Rightarrow {{x}^{2}}+x+1=\left( A+C \right){{x}^{2}}+x\left( 3A+B+2C \right)+\left( 2A+2B+C \right)$

On equating the coefficients of ${{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+C=1\,\,\,...\left( 2 \right)$ 

$3A+B+2C=1\,\,\,...\left( 3 \right)$

$2A+2B+C=1\,\,\,...\left( 4 \right)$

And on solving these equations, the values of $A,\,B,\,C$ can be obtained as,

$A=-2,\,B=1,\,C=3$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$\int{\dfrac{{{x}^{2}}+x+1}{{{(x+1)}^{2}}\left( x+2 \right)}dx}=\int{\left\{ \dfrac{-2}{\left( x+1 \right)}+\dfrac{1}{\left( x+1 \right)}+\dfrac{3}{\left( x+2 \right)} \right\}dx}$ 

$=-2\int{\dfrac{1}{x+1}dx}+\int{\dfrac{1}{{{\left( x+1 \right)}^{2}}}dx}+3\int{\dfrac{1}{x+2}dx}$

$=-2\log \left| x+1 \right|+3\log \left| x+2 \right|-\dfrac{1}{(x+1)}+C$, where $C$ is any arbitrary constant.

Ans: The given expression is, ${{\tan }^{-1}}\sqrt{\dfrac{1-x}{1+x}}$. 

Assume, $I=\int{{{\tan }^{-1}}\sqrt{\dfrac{1-x}{1+x}}dx}$ 

Now, consider $x=\cos \theta $ 

$\therefore dx=-\sin \theta d\theta $

$\therefore I=\int{{{\tan }^{-1}}\sqrt{\dfrac{1-\cos \theta }{1+\cos \theta }}\left( -\sin \theta  \right)d\theta }=-\int{{{\tan }^{-1}}\sqrt{\dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{2{{\cos }^{2}}\dfrac{\theta }{2}}}\sin \theta d\theta }=-\int{{{\tan }^{-1}}\tan \dfrac{\theta }{2}\sin \theta d\theta }$ $\Rightarrow I=-\dfrac{1}{2}\int{\theta \sin \theta d\theta }=-\dfrac{1}{2}\left[ \theta \left( -\cos \theta  \right)-\int{1.(-\cos \theta )d\theta } \right]=-\dfrac{1}{2}\left[ \theta \left( \cos \theta  \right)+\sin \theta  \right]$$\Rightarrow I=\dfrac{x}{2}{{\cos }^{-1}}x-\dfrac{1}{2}\sqrt{1-{{x}^{2}}}+C=\dfrac{1}{2}\left( x{{\cos }^{-1}}x-\sqrt{1-{{x}^{2}}} \right)+C$, where $C$ is any arbitrary constant.

Ans: The given expression is, $\dfrac{\sqrt{{{x}^{2}}+1}\left[ \log \left( {{x}^{2}}+1 \right)-2\log x \right]}{{{x}^{4}}}$. 

Assume, $I=\int{\dfrac{\sqrt{{{x}^{2}}+1}\left[ \log \left( {{x}^{2}}+1 \right)-2\log x \right]}{{{x}^{4}}}dx}$ $\Rightarrow I=\dfrac{\sqrt{{{x}^{2}}+1}}{{{x}^{4}}}\left[ \log \left( {{x}^{2}}+1 \right)-\log {{x}^{2}} \right]=\dfrac{\sqrt{{{x}^{2}}+1}}{{{x}^{4}}}\left[ \log \left( \dfrac{{{x}^{2}}+1}{{{x}^{2}}} \right) \right]=\dfrac{1}{{{x}^{3}}}\sqrt{\dfrac{{{x}^{2}}+1}{{{x}^{2}}}}\left[ \log \left( 1+\dfrac{1}{{{x}^{2}}} \right) \right]$Consider $1+\dfrac{1}{{{x}^{2}}}=t$ 

$\therefore \dfrac{-2}{{{x}^{3}}}dx=dt$

Now, integrate the given expression as shown below $\therefore I=\int{\dfrac{\sqrt{{{x}^{2}}+1}\left[ \log \left( {{x}^{2}}+1 \right)-2\log x \right]}{{{x}^{4}}}dx}=\int{\dfrac{1}{{{x}^{3}}}\sqrt{\dfrac{{{x}^{2}}+1}{{{x}^{2}}}}\left[ \log \left( 1+\dfrac{1}{{{x}^{2}}} \right) \right]dx}=\dfrac{-1}{2}\int{{{t}^{\dfrac{1}{2}}}\log tdt}+C$Using integration by parts, it can be obtained that,

$I=\dfrac{-1}{2}\left[ \log t.\int{{{t}^{\dfrac{1}{2}}}dt-\left\{ \left( \dfrac{d}{dt}\log t \right)\int{{{t}^{\dfrac{1}{2}}}dt} \right\}dt} \right]=\dfrac{-1}{2}\left[ \log t.\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-\int{\dfrac{1}{t}.\dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}dt} \right]$ 

$\Rightarrow I=\dfrac{-1}{2}\left[ \dfrac{2}{3}{{t}^{\dfrac{3}{2}}}\log t-\dfrac{2}{3}\int{{{t}^{\dfrac{1}{2}}}dt} \right]=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\log t+\dfrac{2}{9}{{t}^{\dfrac{3}{2}}}=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\left[ \log t-\dfrac{2}{3} \right]$

$\Rightarrow I=\dfrac{-1}{2}\left[ \dfrac{2}{3}{{t}^{\dfrac{3}{2}}}\log t-\dfrac{2}{3}\int{{{t}^{\dfrac{1}{2}}}dt} \right]=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\log t+\dfrac{2}{9}{{t}^{\dfrac{3}{2}}}=\dfrac{-1}{3}{{t}^{\dfrac{3}{2}}}\left[ \log t-\dfrac{2}{3} \right]$

$\Rightarrow I=\dfrac{-1}{2}\left[ 1+\dfrac{1}{{{x}^{2}}} \right]\left( \log \left( 1+\dfrac{1}{{{x}^{2}}} \right)-\dfrac{2}{3} \right)+C$, where $C$ is any arbitrary constant. 

Evaluate the definite integrals in Exercises 24 to 31.

Ans: The given expression is, $\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{1-\sin x}{1-\cos x} \right)dx}$. 

Assume, $I=\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{1-\sin x}{1-\cos x} \right)dx}$ 

$\Rightarrow I=\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{1-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{+2{{\sin }^{2}}\dfrac{x}{2}} \right)dx=}\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( \dfrac{\cos e{{c}^{2}}\dfrac{x}{2}}{2}-\cot \dfrac{x}{2} \right)dx}$

Now, substitute $f\left( x \right)=-\cot \dfrac{x}{2}$ 

$\therefore f'\left( x \right)=-\left( -\dfrac{1}{2}\cos e{{c}^{2}}\dfrac{x}{2} \right)dx=\dfrac{1}{2}\cos e{{c}^{2}}\dfrac{x}{2}dx$

$\therefore I=\int_{\dfrac{\pi }{2}}^{\pi }{{{e}^{x}}\left( f(x)+f'(x) \right)dx}=\left[ {{e}^{x}}f(x) \right]_{\dfrac{\pi }{2}}^{\pi }=\left[ {{e}^{x}}\cot \dfrac{x}{2} \right]_{\dfrac{\pi }{2}}^{\pi }$

$\Rightarrow I=\left[ {{e}^{\pi }}\cot \dfrac{\pi }{2}-{{e}^{\dfrac{\pi }{2}}}\cot \dfrac{\pi }{4} \right]=\left[ 0-{{e}^{\dfrac{\pi }{2}}} \right]=-{{e}^{\dfrac{\pi }{2}}}$

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x\cos x}{{{\sin }^{4}}x+{{\cos }^{4}}x}dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\dfrac{\sin x\cos x}{{{\cos }^{4}}x}}{\dfrac{{{\sin }^{4}}x+{{\cos }^{4}}x}{{{\cos }^{4}}x}}dx}=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\tan x{{\sec }^{2}}x}{1+{{\tan }^{4}}x}dx}$

Now, substitute ${{\tan }^{2}}x=t$ 

$\therefore 2\tan x{{\sec }^{2}}xdx=dt$

And also when $x=0,\,t=0$ and when $x=\dfrac{\pi }{4},\,t=1$.

$\therefore I=\dfrac{1}{2}\int_{0}^{1}{\dfrac{dt}{1+{{t}^{2}}}}=\dfrac{1}{2}\left[ {{\tan }^{-1}}t \right]_{0}^{1}=\dfrac{1}{2}\left[ {{\tan }^{-1}}\left( 1 \right)-{{\tan }^{-1}}\left( 0 \right) \right]=\dfrac{1}{2}\left( \dfrac{\pi }{4} \right)=\dfrac{\pi }{8}$

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x+4{{\sin }^{2}}x}dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x+4{{\sin }^{2}}x}dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x+4\left( 1-{{\cos }^{2}}x \right)}dx}=\dfrac{-1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4-4-3{{\cos }^{2}}x}{-3{{\cos }^{2}}x+4}dx}$

$\Rightarrow I=-\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4-3{{\cos }^{2}}x}{4-3{{\cos }^{2}}x}dx}+\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4}{4-3{{\cos }^{2}}x}dx}=-\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{dx}+\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4se{{c}^{2}}x}{4{{\sec }^{2}}x-3}dx}$

$\Rightarrow I=\dfrac{-1}{3}\left[ x \right]_{0}^{\dfrac{\pi }{2}}+\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{4se{{c}^{2}}x}{4\left( 1+{{\tan }^{2}}x \right)-3}dx}=\dfrac{-\pi }{6}+\dfrac{2}{3}\int_{0}^{\dfrac{\pi }{2}}{\dfrac{2se{{c}^{2}}x}{\left( 1+4{{\tan }^{2}}x \right)}dx}\,\,\,...\left( 1 \right)$

Observe, $\int_{0}^{\dfrac{\pi }{2}}{\dfrac{2se{{c}^{2}}x}{\left( 1+4{{\tan }^{2}}x \right)}dx}$

Now, substitute $2\tan x=t$ 

$\therefore 2{{\sec }^{2}}xdx=dt$

And also when $x=0,\,t=0$ and when $x=\dfrac{\pi }{2},\,t=\infty $. 

$\therefore \int_{0}^{\dfrac{\pi }{2}}{\dfrac{2se{{c}^{2}}x}{\left( 1+4{{\tan }^{2}}x \right)}dx}=\int_{0}^{\infty }{\dfrac{dt}{\left( 1+{{t}^{2}} \right)}dx}=\left[ {{\tan }^{-1}}\left( t \right) \right]_{0}^{\infty }=\left[ {{\tan }^{-1}}\left( \infty  \right)-{{\tan }^{-1}}\left( 0 \right) \right]=\dfrac{\pi }{2}$Henceforth from equation $\left( 1 \right)$, it can be obtained that, 

$I=-\dfrac{\pi }{6}+\dfrac{2}{3}\left( \dfrac{\pi }{2} \right)=-\dfrac{\pi }{6}+\dfrac{2\pi }{6}=\dfrac{\pi }{6}$

Ans: The given expression is, $\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$. 

Assume, $I=\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{\sin 2x}}dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{-\left( -1+1-2\sin x\cos x \right)}}dx}=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{1-\left( {{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x \right)}}dx}$

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x+\cos x}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}dx}$

Now, substitute $\left( \sin x-\cos x \right)=t$ 

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

And also when $x=\dfrac{\pi }{6},\,t=\left( \dfrac{1-\sqrt{3}}{2} \right)$ and when $x=\dfrac{\pi }{3},\,t=\left( \dfrac{\sqrt{3}-1}{2} \right)$. 

$\therefore I=\int_{\dfrac{1-\sqrt{3}}{2}}^{\dfrac{\sqrt{3}-1}{2}}{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}=\int_{-\left( \dfrac{-1+\sqrt{3}}{2} \right)}^{\dfrac{\sqrt{3}-1}{2}}{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}$

As $\dfrac{1}{\sqrt{1-{{\left( -t \right)}^{2}}}}=\dfrac{1}{\sqrt{1-{{t}^{2}}}}$, it can be thus obtained that $\dfrac{1}{\sqrt{1-{{t}^{2}}}}$ is an even function, 

$\therefore \int_{-a}^{a}{f\left( x \right)dx}=2\int_{0}^{a}{f\left( x \right)dx}$ 

$\therefore I=2\int_{0}^{\dfrac{\sqrt{3}-1}{2}}{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}=\left[ 2{{\sin }^{-1}}t \right]_{0}^{\dfrac{\sqrt{3}-1}{2}}=2{{\sin }^{-1}}\left( \dfrac{\sqrt{3}-1}{2} \right)$ $x=\dfrac{\pi }{6},\,t=\left( \dfrac{1-\sqrt{3}}{2} \right)$ and when $x=\dfrac{\pi }{3},\,t=\left( \dfrac{\sqrt{3}-1}{2} \right)$. 

Ans: The given expression is, $\int_{0}^{1}{\dfrac{dx}{\sqrt{1+x}-\sqrt{x}}}$. 

Assume, $I=\int_{0}^{1}{\dfrac{dx}{\sqrt{1+x}-\sqrt{x}}}$ 

$\Rightarrow I=\int_{0}^{1}{\dfrac{1}{\sqrt{1+x}-\sqrt{x}}\times \dfrac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}dx}=\int_{0}^{1}{\dfrac{\sqrt{1+x}+\sqrt{x}}{1+x-x}dx}$

$\Rightarrow I=\int_{0}^{1}{\sqrt{1+x}dx}+\int_{0}^{1}{\sqrt{x}dx}=\dfrac{2}{3}\left[ {{\left( 1+x \right)}^{\dfrac{2}{3}}} \right]_{0}^{1}+\dfrac{2}{3}\left[ {{\left( x \right)}^{\dfrac{3}{2}}} \right]_{0}^{1}=\dfrac{2}{3}\left[ {{\left( 2 \right)}^{\dfrac{2}{3}}}-1 \right]+\dfrac{2}{3}=\dfrac{4\sqrt{2}}{3}$

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x+\cos x}{9+16\sin 2x}dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{4}}{\dfrac{\sin x+\cos x}{9+16\sin 2x}dx}$ 

Now, substitute $\sin x-\cos x=t$ 

$\therefore \left( \cos x+\sin x \right)dx=dt$

And also when $x=0,\,t=-1$ and when $x=\dfrac{\pi }{4},\,t=0$.

$\therefore {{\left( \sin x-\cos x \right)}^{2}}={{t}^{2}}$

$\Rightarrow 1-2\sin x\cos x={{t}^{2}}$

$\Rightarrow 1-\sin 2x={{t}^{2}}$

$\Rightarrow \sin 2x=1-{{t}^{2}}$

$\therefore I=\int_{-1}^{0}{\dfrac{dt}{9+16\left( 1-{{t}^{2}} \right)}}=\int_{-1}^{0}{\dfrac{dt}{25-16{{t}^{2}}}}=\int_{-1}^{0}{\dfrac{dt}{{{\left( 5 \right)}^{2}}-{{\left( 4t \right)}^{2}}}}$

$\Rightarrow I=\dfrac{1}{4}\left[ \dfrac{1}{2\left( 5 \right)}\log \left| \dfrac{5+4t}{5-4t} \right| \right]_{-1}^{0}=\dfrac{1}{40}\left[ \log \left| 1 \right|-\log \left| \dfrac{1}{9} \right| \right]=\dfrac{1}{40}\log \left| 9 \right|$

Ans: The given expression is, $\int_{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}$. 

Assume, $I=\int_{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}$ 

$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{\sin 2x{{\tan }^{-1}}\left( \sin x \right)dx}=\int_{0}^{\dfrac{\pi }{2}}{2\sin x\cos x{{\tan }^{-1}}\left( \sin x \right)dx}$

Now, substitute $\sin x=t$ 

$\therefore \cos xdx=dt$

And also when $x=0,\,t=0$ and when $x=\dfrac{\pi }{2},\,t=1$. 

$\therefore I=2\int_{0}^{1}{t{{\tan }^{-1}}\left( t \right)dt}$

Observe, $\int{t{{\tan }^{-1}}\left( t \right)dt}$

$\therefore \int{t{{\tan }^{-1}}\left( t \right)dt}={{\tan }^{-1}}t\int{tdt}-\int{\left\{ \dfrac{d\left( {{\tan }^{-1}}t \right)}{dt}\int{tdt} \right\}dt}={{\tan }^{-1}}t.\dfrac{{{t}^{2}}}{2}-\int{\dfrac{1}{1+{{t}^{2}}}.\dfrac{{{t}^{2}}}{2}dt}$

$=\dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\int{\dfrac{{{t}^{2}}+1-1}{1+{{t}^{2}}}dt}=\dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\int{1dt}+\int{\dfrac{1}{1+{{t}^{2}}}dt}=\dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\dfrac{1}{2}t+\dfrac{1}{2}{{\tan }^{-1}}t$

$\therefore \int_{0}^{1}{t.{{\tan }^{-1}}tdt}=\left[ \dfrac{{{t}^{2}}{{\tan }^{-1}}t}{2}-\dfrac{1}{2}t+\dfrac{1}{2}{{\tan }^{-1}}t \right]_{0}^{1}=\dfrac{1}{2}\left[ \dfrac{\pi }{4}-1+\dfrac{\pi }{4} \right]=\dfrac{\pi }{4}-\dfrac{1}{2}$

Henceforth from equation $\left( 1 \right)$, it can be obtained that, 

$I=2\left[ \dfrac{\pi }{2}-\dfrac{1}{2} \right]=\dfrac{\pi }{2}-1$