Write a value

We know that

$1+\sin 2 x=\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x$

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

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

$y=\int \frac{(\sin x-\cos x)}{(\sin x+\cos x)} d x$

Let, $\sin x+\cos x=t$

Differentiating both sides with respect to $x$

$\frac{d t}{d x}=\cos x-\sin x$

$\Rightarrow-\mathrm{dt}=(\sin x-\cos x) d x$

$y=\int \frac{-1}{t} d t$

Use formula $\int \frac{1}{t}=\log t$

$y=-\log t+c$

Again, put $t=\sin x+\cos x$

$y=-\log (\sin x+\cos x)+c$