<s>[INST] 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.
 [/INST] see my nice answer:
$$\int \frac{2\ dx}{(\cos x-\sin x)^2}=\int \frac{2\ dx}{\cos^2 x\left(1-\frac{\sin x}{\cos x}\right)^2}$$  $$=2\int \frac{\sec^2 x\ dx}{\left(1-\tan x\right)^2}$$
$$=-2\int \frac{d(1-\tan x)}{\left(1-\tan x\right)^2}$$
$$=-2 \frac{-1}{\left(1-\tan x\right)}+C$$$$=\frac{2\cos x}{\cos x-\sin x}+C$$
</s>