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

We are going to evaluate the integral by auxiliary angle. $$
\begin{aligned}
\int \frac{2 d x}{(\cos x-\sin x)^{2}} &=\int \frac{2 d x}{\left[\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)\right]^{2}} \\
&=\int \sec ^{2}\left(x+\frac{\pi}{4}\right) d x \\
&=\tan \left(x+\frac{\pi}{4}\right)+C \\ (\textrm{ OR }&=\frac{2 \sin x}{\cos x-\sin x}+C’)
\end{aligned}
$$
