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

We have the integral $I = \int \frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}} dx$.
Using the identity $\sin(x+\alpha) = \sin x \cos \alpha + \cos x \sin \alpha$,we get:
$I = \int \frac{1}{\sqrt{\sin ^{3} x (\sin x \cos \alpha + \cos x \sin \alpha)}} dx$
$= \int \frac{1}{\sqrt{\sin ^{4} x \cos \alpha + \sin ^{3} x \cos x \sin \alpha}} dx$
$= \int \frac{1}{\sin ^{2} x \sqrt{\cos \alpha + \cot x \sin \alpha}} dx$
$= \int \frac{\csc^{2} x}{\sqrt{\cos \alpha + \cot x \sin \alpha}} dx$.
Let $t = \cos \alpha + \cot x \sin \alpha$. Then $dt = -\csc^{2} x \sin \alpha dx$,which implies $\csc^{2} x dx = -\frac{dt}{\sin \alpha}$.
Substituting these into the integral:
$I = \int \frac{-dt}{\sin \alpha \sqrt{t}} = -\frac{1}{\sin \alpha} \int t^{-1/2} dt$
$= -\frac{1}{\sin \alpha} [2 \sqrt{t}] + C$
$= -\frac{2}{\sin \alpha} \sqrt{\cos \alpha + \cot x \sin \alpha} + C$
$= -\frac{2}{\sin \alpha} \sqrt{\cos \alpha + \frac{\cos x \sin \alpha}{\sin x}} + C$
$= -\frac{2}{\sin \alpha} \sqrt{\frac{\sin x \cos \alpha + \cos x \sin \alpha}{\sin x}} + C$
$= -\frac{2}{\sin \alpha} \sqrt{\frac{\sin (x+\alpha)}{\sin x}} + C$.