<s>[INST] 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!
 [/INST] another method:
$\sin(x) = \frac{tan(x)}{sec(x)}$ and $\cos(x) = \frac{1}{sec(x)}$
now you have to do u-substitution and complete the square.
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