I came across the following indefinite integral
$$ \int \frac{2dx}{(\cos(x) - \sin(x))^2} $$
and was able to solve it by doing the following:
First I wrote
$$\begin{align*}
\int \frac{2dx}{(\cos(x) - \sin(x))^2} &= \int \frac{2dx}{1 - 2\cos(x)\sin(x)} \\
&= \int \frac{2dx}{1 - \sin(2x)} \\
\end{align*}$$

Then setting $2x = z$,
$$\begin{align*}
&= \int \frac{dz}{1-\sin(z)}\\
&= \int \frac{1+\sin(z)}{\cos^2(z)}dz \\
&= \int \sec^2(z) + \tan(z)\sec(z) dz \\
&= \tan(z) + \sec(z) \\
&= \tan(2x) + \sec(2x).
\end{align*}$$

The solutions to the problem were given as $\tan(x + \pi/2)$ or $\frac{\cos(x) + \sin(x)}{\cos(x) - \sin(x)}$. I checked that these solutions are in face equivalent to my solution of $\tan(2x) + \sec(2x)$. 

My question is, **are there other ways to calculate this integral that more "directly" produce these solutions?** Actually, any elegant calculation methods in general would be interesting.