Question

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

Answer 1

$I=\int \frac{1}{1+tanx}dxI=\int \frac{1}{1+\frac{\sin x}{\cos x}}dx=\frac{1}{2}\int \frac{(\cos x+\sin x)+(\cos x-\sin x)}{\cos x+\sin x}dx=\frac{1}{2}\int dx+\frac{1}{2}\int \frac{\cos x-\sin x}{\cos x+\sin x}\cos x+\sin x=t(-\sin x+\cos x)dx=dt=\frac{1}{2}\int dx+\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}x+\frac{1}{2}log\left|t\right|+C=\frac{x}{2}+\frac{1}{2}log|\sin x+\cos x|+C$

Hence, this is the answer.