integral exercise 1/(1+tanx)

Exercise:

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

Solution: we know that $\tan x=\frac{\sin x}{\cos x}\Leftrightarrow 1+\tan x=\frac{\cos x+\sin x}{\cos x}\Leftrightarrow \frac{1}{1+\tan x}=\frac{\cos x}{\cos x+\sin x}$

So $\int{\frac{dx}{1+\tan x}=\int{\frac{\cos x}{\cos x+\sin x}dx}=\frac{1}{2}\int{\frac{2\cos x+\sin x-\sin x}{\cos x+\sin x}}\,dx}$

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

$=\frac{1}{2}x+\frac{1}{2}\int{\frac{d\left( \cos x+\sin x \right)}{\cos x+\sin x}}=\frac{1}{2}\left( x+\ln \left| \cos x+\sin x \right| \right)+c$