Question

Solve:
$\int \frac{dx}{1-\tan x}$

Answer 1

$\begin{array}{c}I={\int }^{}\frac{dx}{1-\tan x}\\ ={\int }^{}\frac{dx}{1-\frac{2\tan x/2}{1+\tan x/2}}dx\\ ={\int }^{}\frac{{\sec }^{2}x/2}{{\left(\tan x/2-1\right)}^2}dx\\ Put\tan x/2-1=t\\ {\sec }^{2}x/2dx=2dt\\ =2{\int }^{}\frac{dt}{t^2}\\ =-\frac{2}{t}+C\\ I=-\frac{2}{\tan \frac{x}{2}-1}+c\end{array}$

Hence, this is the answer.