Question

$\int \frac{dx}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}=$

• A. $\frac{-2}{\sin \alpha }\sqrt{\frac{\sin (x+\alpha )}{\sin x}}+c$
• B. $\frac{-1}{sin\alpha }\sqrt{\frac{sinx}{(sinx+\alpha )}}+c$
• C. $\frac{2}{\sin \alpha }\sqrt{\frac{\sin (x+\alpha )}{\sin x}}+c$
• D. $\frac{-2}{\sin \alpha }\sqrt{\frac{\sin x}{\sin (x+\alpha )}}+c$

Answer 1

Answer: (A)

$\frac{-2}{\sin \alpha }\sqrt{\frac{\sin (x+\alpha )}{\sin x}}+c$

$I=\int \frac{dx}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}$
$=\int \frac{\sqrt{\sin x}}{{\sin }^{2}x\sqrt{\sin (x+\alpha )}}$
Put $\frac{\sin (x+\alpha )}{\sin x}=t$
$\Rightarrow -\frac{\sin \alpha }{{\sin }^{2}x}=\frac{dt}{dx}$
So, $I=-\frac{1}{\sin \alpha }\int \frac{1}{\sqrt{t}}dt$
$=-\frac{2}{\sin \alpha }{t}^{1/2}+C$
$I=\frac{-2}{\sin \alpha }\sqrt{\frac{\sin (x+\alpha )}{\sin x}}+c$