Question

Integrate the function $\frac{1}{1-\tan x}$

Answer 1

Let $I=\int \frac{1}{1-\tan x}dx$
$=\int \frac{\cos x}{\cos x-\sin x}dx$
$=\frac{1}{2}\in \frac{2\cos x}{\cos x-\sin x}dx$
$=\frac{1}{2}\int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{(\cos x-\sin x)}dx$
$=\frac{1}{2}\int 1dx+\frac{1}{2}\int \frac{\cos x+\sin x}{\cos x-\sin x}dx$
$=\frac{x}{2}+\frac{1}{2}\int \frac{\cos x+\sin x}{\cos x-\sin x}dx$
Put $\cos x-\sin x=t\Rightarrow (-\sin x-\cos x)dx=dt$
$\therefore I=\frac{x}{2}+\frac{1}{2}\int \frac{-\left(dt\right)}{t}$
$=\frac{x}{2}-\frac{1}{2}\log |t|+C$
$=\frac{x}{2}-\frac{1}{2}\log |\cos x-\sin x|+C$