Question

Integrate the function $\frac{1}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}$

Answer 1

$\frac{1}{\sqrt{{\sin }^{3}x\sin (x+\alpha )}}=\frac{1}{\sqrt{{\sin }^{3}x(\sin x\cos \alpha +\cos x\sin \alpha )}}$
$=\frac{1}{\sqrt{{\sin }^{4}x\cos \alpha +{\sin }^{2}x\cos x\sin \alpha }}$
$=\frac{1}{{\sin }^{2}x\sqrt{\cos \alpha +\cot x\sin \alpha }}$
$=\frac{co{\sec }^{2}x}{\sqrt{\cos \alpha +\cot x\sin \alpha }}$
Let $\cos \alpha +\cot xsin\alpha =t\Rightarrow -co{\sec }^{2}x\sin \alpha dx=dt$
$\therefore \int \frac{1}{{\sin }^{3}x\sin (x+\alpha )}dx=\int \frac{co{\sec }^{2}x}{\sqrt{\cos \alpha +\cot x\sin \alpha }}dx$
$=\frac{-1}{\sin \alpha }\int \frac{dt}{\sqrt{t}}$
$=\frac{-1}{\sin \alpha }\left[2\sqrt{t}\right]+C$
$=\frac{-1}{\sin \alpha }\left[2\sqrt{\cos \alpha +\cos x\sin \alpha }\right]+C$
$=\frac{-2}{\sin \alpha }\sqrt{\cos \alpha +\frac{\cos x\sin \alpha }{\sin x}}+C$
$=\frac{-2}{\sin \alpha }\sqrt{\frac{\sin x\cos \alpha +\cos x\sin \alpha }{\sin x}}+C$
$=\frac{-2}{sin\alpha }\sqrt{\frac{\sin (x+\alpha )}{\sin x}}+C$