How many ways to deal with the integral $\int \frac{d x}{1-\sin x \cos x}$? Multiplying both numerator and denominator of the integrand by $\sec^2 x$ yields
\begin{aligned}
& \int \frac{d x}{1-\sin x \cos x} \\
=& \int \frac{\sec ^{2} x}{\sec ^{2} x-\tan x} d x \\
=& \int \frac{d(\tan x)}{\tan ^{2} x-\tan x+1} \\
=& 2 \int \frac{d(2 \tan x-1)}{(2 \tan x-1)^{2}+(\sqrt{3})^{2}} \\
=& \frac{2}{\sqrt{3}} \arctan\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C .
\end{aligned}
Is there any simpler solution? Let me know if you have any, thank you for your attention.
Wish you enjoy the solution!

We could use the double-angle formula for sine,
$$\sin 2x = 2 \sin x \cos x$$
and then get
$$\newcommand{\II}{\mathcal{I}}
\newcommand{\dd}{\mathrm{d}}
\II := \int \frac{\dd x}{1 - \sin x \cos x} =  \int \frac{\dd x}{1 - \frac 1 2 \sin 2x} =2 \int \frac{\dd x}{2 - \sin 2x}$$
Then let $u = 2x$ to get
$$\II = \int \frac{\dd u}{2 - \sin u}$$
Weierstrass substitution ($t = \tan(u/2)$) then gives
$$\II = \int \frac{1}{2 - \frac{2t}{1+t^2}} \frac{2}{1+t^2} \, \dd t = \int \frac{\dd t}{t^2 - t + 1}$$
Integral $(13)$ from this PDF then gives the answer
$$\II 
= \frac{2}{\sqrt 3} \arctan \left( \frac{2t - 1}{\sqrt 3} \right) + C 
= \frac{2}{\sqrt 3} \arctan \left( \frac{2 \tan(x) - 1}{\sqrt 3} \right) + C $$
