Question

$\int \frac{1}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}dx,\alpha \ne n\pi ,n\in Z$ is equal to

• A. $-2\text{cosec}\alpha (\cos \alpha -\tan x\sin \alpha {)}^{1/2}+C$
• B. $-2(\cos \alpha +\cot x\sin \alpha {)}^{1/2}+C$
• C. $-2\text{cosec}\alpha (\cos \alpha +\cot x\sin \alpha {)}^{1/2}+C$
• D. $-2\text{cosec}\alpha (\sin \alpha +\cot x\cos \alpha {)}^{1/2}+C$

Answer 1

Answer: (C)

${\sin }^{3}x\sin (x+\alpha )={\sin }^{3}x(\sin x\cos \alpha +\cos x\sin \alpha )$
$={\sin }^{4}x(\cos \alpha +\cot x\sin \alpha )$
$I=\int \frac{1}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}dx$
$=\int \frac{1}{{\sin }^{2}x\sqrt{\cos \alpha +\cot x\sin \alpha }}dx$
$=\int \frac{{\text{cosec}}^{2}x}{\sqrt{\cos \alpha +\cot x\sin \alpha }}dx$
Putting $\cos \alpha +\cot x\sin \alpha =t$
and $-{\text{cosec}}^{2}x\sin \alpha dx=dt$, we have
$I=\int -\frac{1}{\sin \alpha \sqrt{t}}dt=-\frac{1}{\sin \alpha }\int t^{-1/2}dt$
$=-\frac{1}{\sin \alpha }\left(\frac{t^{1/2}}{1/2}\right)+C$
$\Rightarrow I=-2\text{cosec}\alpha \sqrt{t}+C$
$=-2\text{cosec}\alpha (\cos \alpha +\cot x\sin \alpha {)}^{1/2}+C$