Chapter 7: Integrals

Ans:  Let $I=\int{\dfrac{x{{\cos }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx$

Multiplying and dividing by 2

$I=\dfrac{-1}{2}\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}\cdot {{\cos }^{-1}}xdx$

Consider  $\text{u}={{\cos }^{-1}}\text{x}$ and $\text{v}=\left( \dfrac{-2x}{\sqrt{1-{{x}^{2}}}} \right)$and integrating by parts, to obtain

$I=\dfrac{-1}{2}\left[ {{\cos }^{-1}}x\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\cos }^{-1}}x \right)\int{\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}}dx \right\}}dx \right]$

$=\dfrac{-1}{2}\left[ {{\cos }^{-1}}x\cdot 2\sqrt{1-{{x}^{2}}}-\int{\dfrac{-1}{\sqrt{1-{{x}^{2}}}}}\cdot 2\sqrt{1-{{x}^{2}}}dx \right]=\dfrac{-1}{2}\left[ 2\sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+\int{2}dx \right]$

Simplifying,

$=\dfrac{-1}{2}\left[ 2\sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+2x \right]+C$

$=-\left[ \sqrt{1-{{x}^{2}}}{{\cos }^{-1}}x+x \right]+C$

Ans: Let $I=\int{x}{{\sec }^{2}}xdx$

Consider$\text{u}=\text{x}$ and $\text{v}={{\sec }^{2}}\text{x}$ and integrating by parts, to obtain

$I=x\int{{{\sec }^{2}}}xdx-\int{\left\{ \left\{ \dfrac{d}{dx}x \right\}\int{{{\sec }^{2}}}xdx \right\}}dx$

$=x\tan x-\int{1}\cdot \tan xdx$

$=x\tan x+\log |\cos x|+C$

Ans: Let $I=\int{1}\cdot {{\tan }^{-1}}xdx$

Consider $\text{u}={{\tan }^{-1}}\text{x}$ and $\text{v}=1$ and integrating by parts, to obtain

$I={{\tan }^{-1}}x\int{1}dx-\int{\left\{ \left( \dfrac{d}{dx}{{\tan }^{-1}}x \right)\int{1}.dx \right\}}dx={{\tan }^{-1}}xx-\int{\dfrac{1}{1+{{x}^{2}}}}xd$

$=x{{\tan }^{-1}}x-\dfrac{1}{2}\int{\dfrac{2x}{1+{{x}^{2}}}}dx$

$=x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left| 1+{{x}^{2}} \right|+C$ 

$=x{{\tan }^{-1}}x-\dfrac{1}{2}\log \left( 1+{{x}^{2}} \right)+C$

Ans:  $I=\int{x}{{(\log x)}^{2}}dx$

Consider$u={{(\log x)}^{2}}$ and $v=1$ and integrating by parts, to obtain

$I={{(\log )}^{2}}\int{x}dx-\int{\left[ \left\{ {{\left( \dfrac{d}{dx}\log x \right)}^{2}} \right\}\int{x}dx \right]}dx$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \int{2}\log x\cdot \dfrac{1}{x}\cdot \dfrac{{{x}^{2}}}{2}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\int{x}\log xdx$

Again using integration by parts, to obtain

$I=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \log x\int{x}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{x}dx \right\}}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\left[ \dfrac{{{x}^{2}}}{2}-\log x-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{2}}}{2}dx \right]$

$=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\dfrac{{{x}^{2}}}{2}\log x+\dfrac{1}{2}\int{x}dx=\dfrac{{{x}^{2}}}{2}{{(\log x)}^{2}}-\dfrac{{{x}^{2}}}{2}\log x+\dfrac{{{x}^{2}}}{4}+C$

Ans: Let $I=\int{\left( {{x}^{2}}+1 \right)}\log xdx=\int{{{x}^{2}}}\log xdx+\int{\log }xdx$

Let $\text{I}={{\text{I}}_{1}}+{{\text{I}}_{2}}\ldots (1)$

Where,${{I}_{1}}=\int{{{x}^{2}}}\log xdx\,\,\,\,\,\,and\,\,\,{{\text{I}}_{2}}=\int{\log }xdx$

${{I}_{1}}=\int{{{x}^{2}}}\log xdx$ 

Consider$\text{u}=\log \text{x}$ and $v=x^2$ and integrating by parts, to obtain

${{I}_{1}}=\log x-\int{{{x}^{2}}}dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{{{x}^{2}}}dx \right\}}dx$

$=\log x\cdot \dfrac{{{x}^{3}}}{3}-\int{\dfrac{1}{x}}\cdot \dfrac{{{x}^{3}}}{3}dx=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{1}{3}\int{{{x}^{2}}}dx$

$=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+{{C}_{1}}\quad \ldots (2)$

${{I}_{2}}=\int{\log }xdx$

Consider $\text{u}=\log \text{x}$ and $\text{v}=1$and integrating by parts, to obtain

${{I}_{2}}=\log x\int{1}.dx-\int{\left\{ \left( \dfrac{d}{dx}\log x \right)\int{1}.dx \right\}}$

$=\log x\cdot x-\int{\dfrac{1}{x}}xdx$

$=x\log x-x..\left( 3 \right)$

Using equations (2) and (3) in (1), we get

$I=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+{{C}_{1}}+x\log x-x+{{C}_{2}}$

$=\dfrac{{{x}^{3}}}{3}\log x-\dfrac{{{x}^{3}}}{9}+x\log x-x+\left( {{C}_{1}}+{{C}_{2}} \right)$

$=\left( \dfrac{{{x}^{3}}}{3}+x \right)\log x-\dfrac{{{x}^{3}}}{9}-x+C$

Ans: Consider$I=\int{{{e}^{x}}}(\sin x+\cos x)dx$

Consider$f(x)=\sin x$

${{f}^{\prime }}(x)=\cos x$

$I=\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx$

Since, $\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx={{e}^{x}}f(x)+C$

$\therefore I={{e}^{x}}\sin x+C$

Ans:  Consider $I=\int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\int{{{e}^{x}}}\left\{ \dfrac{x}{{{(1+x)}^{2}}} \right\}dx$

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

Here, $f(x)=\dfrac{1}{1+x}\quad {{f}^{\prime }}(x)=\dfrac{-1}{{{(1+x)}^{2}}}$

$\Rightarrow \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx$

Since, $\int{{{e}^{x}}}\left\{ f(x)+{{f}^{\prime }}(x) \right\}dx={{e}^{x}}f(x)+C$

$\therefore \int{\dfrac{x{{e}^{x}}}{{{(1+x)}^{2}}}}dx=\dfrac{{{e}^{x}}}{1+x}+C$

Ans: First simplify –${{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)$

It is known that – 

$1+\sin x={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$

$1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$

$\therefore {{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)={{e}^{x}}\left( \dfrac{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} \right)$

$={{e}^{x}}\left( \dfrac{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}{2{{\cos }^{2}}\dfrac{x}{2}} \right)$

$=\dfrac{1}{2}{{e}^{x}}\left( \dfrac{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}{{{\cos }^{2}}\dfrac{x}{2}} \right)$

$=\dfrac{1}{2}{{e}^{x}}{{\left( \dfrac{\sin \dfrac{x}{2}+\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{2}}$

$=\dfrac{1}{2}{{e}^{x}}{{\left( \dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}+\dfrac{\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)}^{2}}$

$=\dfrac{1}{2}{{e}^{x}}{{\left( \tan \dfrac{x}{2}+1 \right)}^{2}}$

$=\dfrac{1}{2}{{e}^{x}}\left( {{\tan }^{2}}\dfrac{x}{2}+1+2\tan \dfrac{x}{2} \right)$

But, $1+{{\tan }^{2}}\dfrac{x}{2}={{\sec }^{2}}\dfrac{x}{2}$

$=\dfrac{1}{2}{{e}^{x}}\left( {{\sec }^{2}}\dfrac{x}{2}+2\tan \dfrac{x}{2} \right)$

$={{e}^{x}}\left( \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\tan \dfrac{x}{2} \right)$

$\Rightarrow {{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)={{e}^{x}}\left( \dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}+\tan \dfrac{x}{2} \right)$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

If we say, $f(x)=\tan \dfrac{x}{2}\Rightarrow f'(x)=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}$

Thus, we get – $\int{{{e}^{x}}\left( \dfrac{1+\sin x}{1+\cos x} \right)}dx={{e}^{x}}\tan \dfrac{x}{2}+C$

Ans: Say, $I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}$

Suppose, $f(x)=\dfrac{1}{x}\Rightarrow f'(x)=-\dfrac{1}{{{x}^{2}}}$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

Thus, we get – $I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}=\dfrac{{{e}^{x}}}{x}+C$

Ans: $\int{{{e}^{x}}\dfrac{(x-3)}{{{(x-1)}^{3}}}dx=\int{{{e}^{x}}\left[ \dfrac{(x-1-2)}{{{(x-1)}^{3}}} \right]dx}}$

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

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

Suppose, $f(x)=\dfrac{1}{{{(x-1)}^{2}}}\Rightarrow f'(x)=-\dfrac{2}{{{(x-1)}^{3}}}$

It is known that, $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx=}{{e}^{x}}f(x)+C$

Thus, $\int{{{e}^{x}}\dfrac{(x-3)}{{{(x-1)}^{3}}}dx=\dfrac{{{e}^{x}}}{{{(x-1)}^{2}}}+C}$