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

Toolbox:
• If an integral function $f(x)=t,$ then $f'(x)dx=dt$ then $\int f(x)dx=\int tdt$
• $\int x^n dx=\frac{x^{n+1}}{n+1}+c$
• $\sin(A+B)=\sin A\cos B+\cos A\sin B$
Step 1:
Let $I=\int\large\frac{1}{\sqrt {\sin^3x\sin(x+\alpha)}}$
Let us consider $\sin^3x\sin(x+\alpha)$
$\sin^3x\sin(x+\alpha)=\sin^3x(\sin x\cos \alpha+\cos x\sin\alpha)$
$\qquad\qquad\;\;\;\;\quad=\sin^4x\cos\alpha+\sin^3x\cos x\sin\alpha$
$\qquad\qquad\;\;\;\;\quad=\sin^4x(\cos\alpha+\cot x\sin\alpha)$
Step 2:
Put $\cos\alpha+\cot x\sin\alpha=t$
On differentiating we get,