Question

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

Answer 1

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

$I=\int \frac{{\sec }^{2}xdx}{(\tan x+1)(1+{\tan }^{2}x)}$

Let $t=\tan x$

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

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

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

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

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

$I=\frac{1}{2}log(\tan x+1)-\frac{1}{4}log({\tan }^{2}x+1)+\frac{1}{2}{\tan }^{-1}\left(\tan x\right)+c$

$I=\frac{1}{2}log(\tan x+1)-\frac{1}{2}logsecx)+\frac{1}{2}x+c$

$I=\frac{1}{2}log\left(\frac{\tan x+1}{\sec x}\right)+\frac{x}{2}+c$

$I=\frac{1}{2}\left[log\right(\sin x+\cos x\left)\right]+\frac{x}{2}+c$.