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?

Using the tangent half-angle substitution
$$\int \frac{\sin (x)}{1-\sin (x) \cos (x)} \,d x=4\int \frac{t}{t^4+2 t^3+2 t^2-2 t+1}\,dt$$
The denominator has four roots
$$r_1=\frac{1}{2} \left(-1+\sqrt{3}-i \sqrt{2 \left(2-\sqrt{3}\right)}\right)\qquad r_2=\frac{1}{2} \left(-1+\sqrt{3}+i \sqrt{2 \left(2-\sqrt{3}\right)}\right)$$
$$r_3=\frac{1}{2} \left(-1-\sqrt{3}-i \sqrt{2 \left(2+\sqrt{3}\right)}\right)\qquad r_4=\frac{1}{2} \left(-1-\sqrt{3}+i \sqrt{2 \left(2+\sqrt{3}\right)}\right)$$ So, using partial fraction decomposition, four logarithms to be recombined.
