Let $I=\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} \cdot d x$
$\begin{aligned} & I=\int \frac{\sin x-\cos x}{\sqrt{1+2 \sin x \cdot \cos x-1}} d x \\ & I=\int \frac{\sin x-\cos x}{\sqrt{(\cos x+\sin x)^2-1}} d x\end{aligned}$
Let $\sin x+\cos x=t$
$\begin{array}{lr}\Rightarrow & \cos x-\sin x=\frac{d t}{d x} \\ \Rightarrow & -(\sin x-\cos x) d x=d t \\ \Rightarrow & (\sin x-\cos x) d x=-d t\end{array}$
$\therefore I=-\int \frac{d t}{\sqrt{t^2-1}}$
$\left[\because \int \frac{1}{\sqrt{x^2-a^2}} d x=\log \left|x+\sqrt{x^2-a^2}\right|+C\right]$
$\begin{aligned} & =-\log \left|t+\sqrt{t^2-1}\right| \\ & =-\log \left|\sin x+\cos x+\sqrt{(\sin x+\cos x)^2}-1\right|+C \\ & I=-\log |\sin x+\cos x+\sqrt{\sin 2 x}|+C\end{aligned}$