Chapter 7: Integrals

Ans: Given expression ${{\sin }^{-1}}\left( \cos x \right)$.

Let $\cos x=t$ 

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

$\Rightarrow -\sin xdx=dt$

$\Rightarrow dx=-\frac{dt}{\sin x}$

$\Rightarrow dx=-\frac{dt}{\sqrt{1-{{t}^{2}}}}$

Integration of given expression is

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\int{{{\sin }^{-1}}t\left( \frac{-dt}{\sqrt{1-{{t}^{2}}}} \right)}$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\int{\left( \frac{{{\sin }^{-1}}t}{\sqrt{1-{{t}^{2}}}} \right)dt}$

Let ${{\sin }^{-1}}t=u$ 

$\Rightarrow \frac{1}{\sqrt{1-{{t}^{2}}}}dt=du$ 

Integration becomes

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\int{4du}$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\frac{{{u}^{2}}}{2}+C$

Substitute ${{\sin }^{-1}}t=u$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\frac{{{\left( {{\sin }^{-1}}t \right)}^{2}}}{2}+C$

Substitute $\cos x=t$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\frac{{{\left[ {{\sin }^{-1}}\left( \cos x \right) \right]}^{2}}}{2}+C$ ……..(1)

We know that ${{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2}$ 

$\therefore {{\sin }^{-1}}\left( \cos x \right)=\frac{\pi }{2}-{{\cos }^{-1}}\left( \cos x \right)=\left( \frac{\pi }{2}-x \right)$ 

Substitute in eq. (1), we get

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\frac{-{{\left( \frac{\pi }{2}-x \right)}^{2}}}{2}+C$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\frac{1}{2}\left( \frac{{{\pi }^{2}}}{2}+{{x}^{2}}-\pi x \right)+C$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=-\frac{{{\pi }^{2}}}{4}-\frac{{{x}^{2}}}{2}+\frac{1}{2}\pi x+C$

$\Rightarrow \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\frac{\pi x}{2}-\frac{{{x}^{2}}}{2}+\left( C-\frac{{{\pi }^{2}}}{4} \right)$

$\therefore \int{{{\sin }^{-1}}\left( \cos x \right)dx}=\frac{\pi x}{2}-\frac{{{x}^{2}}}{2}+{{C}_{1}}$

Ans: Given expression $\frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}$.

Given expression can be written as

$\Rightarrow \frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\frac{1}{\sin \left( a-b \right)}\left[ \frac{\sin \left( a-b \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\frac{1}{\sin \left( a-b \right)}\left[ \frac{\sin \left[ \left( x-b \right)-\left( x-a \right) \right]}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\frac{1}{\sin \left( a-b \right)}\left[ \frac{\sin \left( x-b \right)\cos \left( x-a \right)-\cos \left( x-b \right)\sin \left( x-a \right)}{\cos \left( x-a \right)\cos \left( x-b \right)} \right]$

$\Rightarrow \frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}=\frac{1}{\sin \left( a-b \right)}\left[ \tan \left( x-b \right)-\tan \left( x-b \right) \right]$

Integration of given expression is

$\Rightarrow \int{\frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\frac{1}{\sin \left( a-b \right)}\left[ \tan \left( x-b \right)-\tan \left( x-b \right) \right]dx}$

$\Rightarrow \int{\frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\frac{1}{\sin \left( a-b \right)}\left[ -\log \left| \cos \left( x-b \right) \right|+\log \cos \left( x-a \right) \right]}$

$\Rightarrow \int{\frac{1}{\cos \left( x-a \right)\cos \left( x-b \right)}dx}=\int{\frac{1}{\sin \left( a-b \right)}\left[ \log \left| \frac{\cos \left( x-a \right)}{\cos \left( x-b \right)} \right| \right]}+C$

Ans: Given expression $\int{\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$.

Given expression can be written as

$\Rightarrow \int{\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{\frac{{{\sin }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}-\int{\frac{{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

$\Rightarrow \int{\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{{{\sec }^{2}}xdx}-\int{cose{{c}^{2}}xdx}$

$\therefore \int{\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\tan x+\cot x+C$

Therefore, option A is the correct answer.

Ans: Given expression $\int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}$.

Let ${{e}^{x}}x=t$ 

$\therefore \left( {{e}^{x}}.x+{{e}^{x}}.1 \right)dx=dt$ 

$\Rightarrow {{e}^{x}}\left( x+1 \right)dx=dt$

Integration of given expression is

$\Rightarrow \int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\int{\frac{dt}{{{\cos }^{2}}t}}$

\[\Rightarrow \int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\int{{{\sec }^{2}}t}dt\]

\[\Rightarrow \int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\tan t+C\]

Substitute ${{e}^{x}}x=t$,

\[\therefore \int{\frac{{{e}^{x}}\left( 1+x \right)}{{{\cos }^{2}}\left( {{e}^{x}}x \right)}dx}=\tan \left( {{e}^{x}}x \right)+C\]

Therefore, option B is the correct answer.