Question

Solve $\int \frac{dx}{\sin x\sqrt{\sin (2x+\alpha )}}$

Answer 1

$\int \frac{dx}{\sin x\sqrt{\sin (2x+\alpha )}}=\int \frac{dx}{\sin x\sqrt{\sin 2x\cos \alpha +\sin \alpha \cos 2\alpha }}$

$=\int \frac{dx}{\sin x\sqrt{2\sin x\cos x\cos \alpha +\sin \alpha ({\cos }^{2}x-{\sin }^{2}x)}}$

$=\int \frac{dx}{{\sin }^{2}x\sqrt{2\cot x\cos \alpha +\sin \alpha ({\cot }^{2}x-1)}}$

$=\frac{-1}{\sqrt{\sin \alpha }}\int \frac{-{\csc }^{2}xdx}{\sqrt{{\cot }^{2}x+2\cot \alpha \cot x-1}}$ Let $t=\cot x;dt=-{\csc }^{2}xdx$

$=\frac{-1}{\sqrt{\sin \alpha }}\int \frac{dt}{\sqrt{t^2+2\cot \alpha t-1}}$

$=\frac{-1}{\sqrt{\sin \alpha }}\int \frac{dt}{\sqrt{{(t+\cot \alpha )}^2-(1+{\cot }^2\alpha )}}$

$=\frac{-1}{\sqrt{\sin \alpha }}\int \frac{dt}{\sqrt{{(t+\cot \alpha )}^2-{\csc }^2\alpha }}$

We know that $\int \frac{1}{\sqrt{x^2-a^2}}dx=\ln |x+\sqrt{x^2-a^2}|+C$

$=\frac{-1}{\sqrt{\sin \alpha }}\ln |(t+\cot \alpha )+\sqrt{{(t+\cot \alpha )}^2-{\csc }^2\alpha }|+C$

$=\frac{-1}{\sqrt{\sin \alpha }}\ln |(\cot x+\cot \alpha )+\sqrt{{\cot }^{2}x+2\cot \alpha \cot x-1}|+C$