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