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

Let $I = \int \frac{1}{1-\tan x} dx$.
$= \int \frac{1}{1-\frac{\sin x}{\cos x}} dx = \int \frac{\cos x}{\cos x-\sin x} dx$.
Multiply and divide by $2$:
$= \frac{1}{2} \int \frac{2 \cos x}{\cos x-\sin x} dx$.
Rewrite the numerator:
$= \frac{1}{2} \int \frac{(\cos x-\sin x)+(\cos x+\sin x)}{\cos x-\sin x} dx$.
$= \frac{1}{2} \int 1 dx + \frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} dx$.
$= \frac{x}{2} + \frac{1}{2} \int \frac{\cos x+\sin x}{\cos x-\sin x} dx$.
Let $t = \cos x - \sin x$,then $dt = -(\sin x + \cos x) dx$,so $(\cos x + \sin x) dx = -dt$.
$= \frac{x}{2} - \frac{1}{2} \int \frac{1}{t} dt$.
$= \frac{x}{2} - \frac{1}{2} \ln |\cos x - \sin x| + C$,where $C$ is the constant of integration.