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

Solve the indefinite Integration $\int \frac{1}{\sqrt{\sin^3x\sin(x+\alpha)}}dx$

$$\int \frac{1}{\sqrt{\sin^3x\sin(x+\alpha)}}dx=\int \frac{1}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx\\=\frac{1}{\sin\alpha}\int \frac{\sin(x+\alpha-x)}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx=\frac{1}{\sin\alpha}\int\frac{\sin(x+\alpha)\cos x-\cos(x+\alpha)\sin{x}}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx\\ =\frac{1}{\sin\alpha}\int\frac{\sin(x+\alpha)\cos x}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx-\frac{1}{\sin\alpha}\int\frac{\cos(x+\alpha)\sin{x}}{\sin x\sqrt{\sin x\sin(x+\alpha)}}dx$$

Is it possible to proceed further and complete the integration or what is the right substitution to find the solution ?

Note: I'm looking for a simple way to solve this, unlike here Finding indefinite integral $\int{dx\over \sqrt{\sin^3 x+\sin (x+\alpha)}}$.

Divide by $\sin^2 x$ on numerator and denominator and use the identity $\sin(A+B) = \sin A \cos B+\cos A \sin B$ to get to following:

$$\int\frac{\csc^2 x \, dx}{\sqrt{\cos a + \sin a\cot x}}$$

Noting that $\cot'(x) = -\csc^2 x$, it is straightforward now. The final antiderivative is:

$$-\frac{2 \sqrt{\cos a + \sin a \cot x}}{\sin a}+c$$

Hint:

$$\sin^3x\sin(x+a)=\sin^4x(\cos a+\sin a\cot x)$$

Now $\dfrac1{\sqrt{\sin^4x}}=|\csc^2x|$

Set $\cos a+\sin a\cot x=u$