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?

Multiplying both numerator and denominator of the integrand by $1+\sin x \cos x$ transforms the integral into
$$
\begin{aligned}
I &=\int \frac{\sin x(1+\sin x \cos x)}{1-\sin ^{2} x \cos ^{2} x} d x \\
&=\int \frac{\sin x d x}{1-\sin ^{2} x \cos ^{2} x}+\int \frac{\sin ^{2} x \cos x}{1-\sin ^{2} x \cos ^{2} x} d x \\
&=-  \underbrace{\int \frac{d c}{1-\left(1-c^{2}\right) c^{2}}}_{J}+  \underbrace{\int \frac{s^{2} d s}{1-s^{2}\left(1-s^{2}\right)}}_{K}
\end{aligned}
$$
where $c=\cos x$ and $s=\sin x.$
$$
\begin{aligned}
J&=-\int \frac{d c}{c^{4}-c^{2}+1}\\
&=-\int \frac{\frac{1}{c^{2}}}{c^{2}+\frac{1}{c^{2}}-1} d c
\end{aligned}
$$
Playing a little trick on the integrand gives
$$
\begin{aligned}
J &=-\frac{1}{2} \int \frac{\left(1+\frac{1}{c^{2}}\right)+\left(1-\frac{1}{c^{2}}\right)}{c^{2}+\frac{1}{c^{2}}-1} d c \\
&=-\frac{1}{2} \int \frac{d\left(c-\frac{1}{c}\right)}{\left(c-\frac{1}{c}\right)^{2}+1}+\frac{1}{2} \int \frac{d\left(c+\frac{1}{c}\right)}{\left(c+\frac{1}{c}\right)^{2}-3} \\
&=-\frac{1}{2} \tan ^{-1}\left(c-\frac{1}{c}\right)+\frac{1}{4 \sqrt{3}} \ln \left|\frac{c+\frac{1}{c}-\sqrt{3}}{c+\frac{1}{c}+\sqrt{3}}\right|+C_{1}
\end{aligned}
$$
Similarly,
$$
\begin{aligned}
K &=\int \frac{1}{s^{2}+\frac{1}{s^{2}}-1} d s \\
&=\frac{1}{2} \int \frac{\left(1+\frac{1}{s^{2}}\right)-\left(1-\frac{1}{s^{2}}\right)}{s^{2}+\frac{1}{s^{2}}-1} d s \\
&=\frac{1}{2} \tan ^{-1}\left(s-\frac{1}{s}\right)+\frac{1}{4 \sqrt{3}} \ln \left|\frac{s+\frac{1}{s}+\sqrt{3}}{s+\frac{1}{s}-\sqrt{3}}\right|+C_{2}
\end{aligned}
$$
Now we can conclude that
$$
\begin{aligned}
\therefore I=&-\frac{1}{2} \tan ^{-1}(\cos x-\sec x)+\frac{1}{2} \tan ^{-1}(\sin x-\csc x) \\
&+\frac{1}{4 \sqrt{3}} \ln \left|\frac{\cos ^{2} x-\sqrt{3} \cos x+1}{\cos ^{2} x+\sqrt{3} \cos x+1}\right|+\frac{1}{4 \sqrt{3}} \ln \left|\frac{\sin ^{2} x+\sqrt{3} \sin x+1}{\sin ^{2} x-\sqrt{3} \sin x+1}\right|+C
\end{aligned}
$$
