Question

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

Answer 1

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

$=\int \frac{\cos x}{\sin x+\cos x}dx$

$=\frac{1}{2}\int \frac{2\cos x}{\sin x+\cos x}dx$

$=\frac{1}{2}\left[\int \frac{\cos x+\sin x}{\sin x+\cos x}dx\right]$$+\frac{1}{2}\left[\int \frac{\cos x-\sin x}{\sin x+\cos x}dx\right]$

$=\frac{1}{2}\left[\int dx\right]$$+\frac{1}{2}\left[\int \frac{d(\cos x+\sin x)}{\sin x+\cos x}\right]$

$=\frac{x}{2}+\frac{1}{2}\log |(\sin x+\cos x)|+c$ [Where $c$ being integrating constant.]