Question

Integrate:

$\int \frac{1}{1+\tan x}dx$

Answer 1

Consider the given integral.

$I=\int \frac{1}{1+\tan x}dx$

Put,

$t=\tan x$

$dt={\sec }^{2}xdx$

$dt=(1+{\tan }^{2}x)dx$

Therefore,

$I=\int \frac{1}{(t+1)(t^2+1)}dt$

$I=\frac{1}{2}\int \frac{1}{(t+1)}dt-\frac{1}{2}\int \frac{t-1}{(t^2+1)}dt$

$I=\frac{1}{2}\int \frac{1}{(t+1)}dt-\frac{1}{2}\int \frac{t}{(t^2+1)}dt+\frac{1}{2}\int \frac{1}{(t^2+1)}dt$

$I=\frac{1}{2}\int \frac{1}{(t+1)}dt-\frac{1}{4}\int \frac{2t}{(t^2+1)}dt+\frac{1}{2}\int \frac{1}{(t^2+1)}dt$

$I=\frac{1}{2}\ln (t+1)-\frac{1}{4}\ln (t^2+1)+\frac{1}{2}{\tan }^{-1}t+C$

On putting the value of $t$, we get

$I=\frac{1}{2}\ln (\tan x+1)-\frac{1}{4}\ln (ta{n}^{2}x+1)+\frac{1}{2}{\tan }^{-1}\left(\tan x\right)+C$

$I=\frac{1}{2}\ln (\tan x+1)-\frac{1}{4}\ln ({\tan }^{2}x+1)+\frac{x}{2}+C$

Hence, this is the required value of the integral.