Question

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

Answer 1

$1-\sin 2x={(\sin x-\cos x)}^2\Rightarrow \sin 2x=1-{(\sin x-\cos x)}^2\int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx=\int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx=\int \frac{\cos x+\sin x}{\sqrt{1-{(\sin x-\cos x)}^2}}dxLet\sin x-\cos x=t(\cos x+\sin x)dx=dt\int \frac{dt}{\sqrt{1-t^2}}={\sin }^{-1}t+c={\sin }^{-1}(\sin x-\cos x)+c$