Question

Integrate the following

$\int \frac{(sinx+cosx)}{\sqrt{sin2x}}dx$

Answer 1

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

Let $\sin x-\cos x=t\Rightarrow (\sin x+\cos x)dx=dt$

Putting these substituting we have

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