Chapter 7: Integrals

Ans: Say,  $I=\int{{{e}^{2x}}\sin xdx}$

Perform Integration by parts – $\int{uv}dx=u\int{vdx}-\int{\left( u'\int{vdx} \right)dx}$

With –$u=\sin x\text{   }v={{e}^{2x}}$

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

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

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

Perform Integration by parts for – $\int{\left( {{e}^{2x}}\cos x \right)dx}$

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

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

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

$=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}-\dfrac{1}{4}\left\{ \int{(\sin x){{e}^{2x}}dx} \right\}$

But, $I=\int{{{e}^{2x}}\sin xdx}$

$\Rightarrow I=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}-\dfrac{1}{4}I$

$\Rightarrow I+\dfrac{1}{4}I=\sin x\dfrac{{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow \dfrac{5}{4}I=\dfrac{{{e}^{2x}}\sin x}{2}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow \dfrac{5}{4}I=\dfrac{2{{e}^{2x}}\sin x}{4}-\dfrac{{{e}^{2x}}\cos x}{4}$

$\Rightarrow 5I={{e}^{2x}}(2\sin x-\cos x)$

Thus, we get – $I=\dfrac{{{e}^{2x}}}{5}(2\sin x-\cos x)+C$.

Ans: Say, $x=\tan \theta \text{ }\Rightarrow \text{dx=se}{{\text{c}}^{2}}\theta d\theta $

$\therefore {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta } \right)$

But, $\sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta }$

$\Rightarrow {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{3}}\theta } \right)=\therefore {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)={{\sin }^{-1}}\left( \sin 2\theta  \right)=2\theta $

Therefore, $\int{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)dx}=\int{2\theta \text{se}{{\text{c}}^{2}}\theta d\theta }$

$=2\int{\theta \text{se}{{\text{c}}^{2}}\theta d\theta }$

Perform Integration by parts – $\int{uv}dx=u\int{vdx}-\int{\left( u'\int{vdx} \right)dx}$

With –$u=\theta \text{   }v={{\sec }^{2}}\theta $

$2\int{\theta \text{se}{{\text{c}}^{2}}\theta d\theta }=2\left\{ \theta \int{{{\sec }^{2}}\theta d\theta }-\int{\left[ \left( \dfrac{d}{d\theta }\theta  \right)\int{{{\sec }^{2}}\theta d\theta } \right]d\theta } \right\}$

$=2\left\{ \theta \tan \theta -\int{\left[ \tan \theta  \right]d\theta } \right\}$

$=2\left\{ \theta \tan \theta -(-\log |\cos \theta |) \right\}+C$

$=2\left\{ \theta \tan \theta +\log |\cos \theta | \right\}+C$

Replace $\theta ={{\tan }^{-1}}x$

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

It is known that – ${{\tan }^{-1}}x={{\cos }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}}$

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

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

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

Here, $\log {{m}^{n}}=n\log m$

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

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

Thus, $\int{{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{3}}} \right)dx}=2x{{\tan }^{-1}}x-\log (1+{{x}^{2}})+C$

Answer: $\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx$ equals

1. $\dfrac{1}{3}{{e}^{{{x}^{3}}}}+C$

2. $\dfrac{1}{3}{{e}^{{{x}^{2}}}}+C$

3. $\dfrac{1}{2}{{e}^{{{x}^{3}}}}+C$

4. $\dfrac{1}{2}{{e}^{{{x}^{2}}}}+C$

Ans: Say, $I=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx$

Suppose, $t={{x}^{3}}\Rightarrow dt=3{{x}^{2}}dx$

Rewriting the equation – $I=\int{{{x}^{2}}{{e}^{{{x}^{3}}}}}dx=\dfrac{1}{3}\int{{{e}^{t}}}dt$

$\Rightarrow I=\dfrac{1}{3}\int{{{e}^{t}}}dt=\dfrac{1}{3}{{e}^{t}}+C$

Replacing $t={{x}^{3}}$

$\Rightarrow I=\dfrac{1}{3}{{e}^{{{x}^{3}}}}+C$

The correct option is A.

Answer: $\int{{{e}^{x}}\sec x(1+\tan x)}dx$

1. ${{e}^{x}}\cos x+C$

2. ${{e}^{x}}\sec x+C$

3. ${{e}^{x}}\sin x+C$

4. ${{e}^{x}}\tan x+C$

Ans: Say, $I=\int{{{e}^{x}}\sec x(1+\tan x)}dx$

$\Rightarrow I=\int{{{e}^{x}}(\sec x+\sec x\tan x)}dx$

Suppose, $f(x)=\sec x\Rightarrow f'(x)=\sec x\tan x$

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

$\Rightarrow I=\int{{{e}^{x}}(\sec x+\sec x\tan x)}dx={{e}^{x}}\sec x+C$

Thus, $I={{e}^{x}}\sec x+C$

The correct option is B.