How many method are there to handle the integral $\int \frac{\sin x}{1-\sin x \cos x} d x$ $$
\begin{aligned}
\int \frac{\sin x}{1-\sin x \cos x} d x =& \frac{1}{2} \int \frac{(\sin x+\cos x)+(\sin x-\cos x)}{1-\sin x \cos x} d x \\
=& \int \frac{d(\sin x-\cos x)}{2-2 \sin x \cos x}-\int \frac{d(\sin x+\cos x)}{2-2 \sin x \cos x} \\
=& \int \frac{d(\sin x-\cos x)}{1+(\sin x-\cos x)^{2}}-\int \frac{d(\sin x+\cos x)}{3-(\sin x+\cos x)^{2}} \\
=& \tan ^{-1}(\sin x-\cos x)+\frac{1}{2 \sqrt{3}} \ln \left| \frac{\sin x+\cos x-\sqrt{3}}{\sin x+\cos x+\sqrt{3}} \right|+C
\end{aligned}
$$
Is there any other method?

Letting $y=\frac{\pi}{2}-x$ converts the integral into
$$
I=\int \frac{\cos y}{1-\sin y \cos y} d y
$$
Using the Weierstrass substitution, $t =\tan \frac{y}{2} $, we simplify and get
$$
I=-2 \int \frac{t^{2}-1}{t^{4}-2 t^{3}+2 t^{2}-2 t+1} d t
$$
Playing a little trick on the $I$ gives
$$
I =-2 \int \frac{1-\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}-2\left(t+\frac{1}{t}\right)+2} d t =-2 \int \frac{d\left(t+\frac{1}{t}\right)}{t^{2}+\frac{1}{t^{2}}-2\left(t+\frac{1}{t}\right)+2}
$$
Letting $T=t+\frac{1}{t}$ yields
$$
\begin{aligned}
I &=-2 \int \frac{d T}{T^{2}-2 T} \\
&=-2 \int\left(\frac{1}{T-2}-\frac{1}{T}\right) d T \\
&=2 \ln \left|\frac{T}{T-2}\right|+C \\
&=2 \ln \left|\frac{t+\frac{1}{t}}{t+\frac{1}{t}-2}\right|+C\\&= 2 \ln \left|\frac{  \tan^2 \frac{y}{2}+1}{\tan ^{2} \frac{y}{2}-2 \tan \frac{y}{2}+1}\right|+C\\&= 2 \ln \left|\frac{\tan ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)+1}{\tan ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)-2 \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)+1}\right|+C
\end{aligned}
$$
