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.
 A: HINT:
$$1-\sin2x=1-\cos2\left(\dfrac\pi4-x\right)=2\sin^2\left(\dfrac\pi4-x\right)=\dfrac2{\csc^2\left(\dfrac\pi4-x\right)}$$
As $\csc(-A)=-\csc A,$
$$\csc^2\left(\dfrac\pi4-x\right)=\csc^2\left(x-\dfrac\pi4\right)$$
$$\int\csc^2y\ dy=-\cot y+K$$

Alternatively,
$$1-\sin2x=\dfrac{(1-\tan x)^2}{1+\tan^2x}$$
A: $$\int\frac{dx}{(\cos x-\sin x)^2}=\int\frac{dx}{(\sqrt2\cos(x+\frac\pi4))^2}$$ where you recognize the derivative of a tangent.

A very general and useful method is the use of the exponential representation
$$\cos x:=\frac{e^{ix}+e^{-ix}}2,\\\sin x:=\frac{e^{ix}-e^{-ix}}{2i}$$ together with the change of variable $z:=e^{ix}$ such that $dx=dz/iz$.
In your case
$$\int\frac{dx}{(\cos x-\sin x)^2}
=\int \frac{4\,dz}{(z+z^{-1}+iz-iz^{-1})^2iz}
=\int \frac{2\,d(z^2)}{((1+i)z^2+(1-i))^2i}$$ which is elementary.
A: see my nice answer:
$$\int \frac{2\ dx}{(\cos x-\sin x)^2}=\int \frac{2\ dx}{\cos^2 x\left(1-\frac{\sin x}{\cos x}\right)^2}$$  $$=2\int \frac{\sec^2 x\ dx}{\left(1-\tan x\right)^2}$$
$$=-2\int \frac{d(1-\tan x)}{\left(1-\tan x\right)^2}$$
$$=-2 \frac{-1}{\left(1-\tan x\right)}+C$$$$=\frac{2\cos x}{\cos x-\sin x}+C$$
A: We are going to evaluate the integral by auxiliary angle. $$
\begin{aligned}
\int \frac{2 d x}{(\cos x-\sin x)^{2}} &=\int \frac{2 d x}{\left[\sqrt{2} \cos \left(x+\frac{\pi}{4}\right)\right]^{2}} \\
&=\int \sec ^{2}\left(x+\frac{\pi}{4}\right) d x \\
&=\tan \left(x+\frac{\pi}{4}\right)+C \\ (\textrm{ OR }&=\frac{2 \sin x}{\cos x-\sin x}+C’)
\end{aligned}
$$
