$\int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$
$\int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$
We can write above integral as
$=\int \frac{\sin x+\cos x}{\sqrt{1-1+\sin 2 x}} d x$ [Adding and subtracting 1 in denominator]
$=\int \frac{\sin x+\cos x}{\sqrt{1-(1-\sin 2 x)}} d x$
$=\int \frac{\sin x+\cos x}{\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)}} d x \because \sin ^{2} x+\cos ^{2} x=1$ and
$\sin 2 x=2 \sin x \cos x$
$=\int \frac{(\sin x+\cos x)}{\sqrt{1-(\sin x-\cos x)^{2}}} d x \because \sin ^{2} x+\cos ^{2} x-2 \sin x \cos x=(\sin x-\cos x)^{2}$
Put $\sin x-\cos x=t$
Differentiating w.r.t $x$ we get,
$(\cos x+\sin x) d x=d t$
Putting values we get,
$=\int \frac{(\sin x+\cos x)}{\sqrt{1-(\sin x-\cos x)^{2}}} d x=\int \frac{d t}{\sqrt{1-t^{2}}}$
$=\int \frac{d t}{\sqrt{1-t^{2}}}=\sin ^{-1} t+C$
Putting value of $t$ we get,
$\therefore \int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x=\sin ^{-1}(\sin x-\cos x)+C$