Question

Integrate $\int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}$

Answer 1

Let $I=\int \frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$

Let $\sin x-\cos x=t$

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

Further $t^2={(\sin x-\cos x)}^2={\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x$

$\Rightarrow t^2=1-2\sin x\cos x$

$\Rightarrow 2\sin x\cos x=\sin 2x=1-t^2$

$\therefore I=\int \frac{dt}{\sqrt{1-t^2}}dx$

$={\sin }^{-1}t+c$ where $c$ is the constant of integration.

$={\sin }^{-1}(\sin x-\cos x)+c$

Since $\sin x-\cos x=\sqrt{2}(\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x)$

$=\sqrt{2}(\cos \frac{\pi }{4}\sin x-\sin \frac{\pi }{4}\cos x)+c$

$\therefore I=\sqrt{2}\sin (x-\frac{\pi }{4})+c$