integration - Evaluating $intfrac{cos^2x}{1+tan x},dx$

I'd like to evaluate the following integral:

$$\int \frac{\cos^2 x}{1+\tan x}dx$$

I tried integration by substitution, but I was not able to proceed.

Answer

$\displaystyle \int \frac{\cos^2 x}{1+\tan x}dx = \int\frac{\cos^3 x}{\sin x+\cos x}dx = \frac{1}{2}\int\frac{(\cos^3 x+\sin ^3 x)+(\cos^3 x-\sin ^3 x)}{\cos x+\sin x}dx$

$\displaystyle = \frac{1}{4}\int (2-\sin 2x)dx+\frac{1}{4}\int\frac{(2+\sin 2x)(\cos x-\sin x)}{\cos x+\sin x}dx$

put $\cos x+\sin x= t$ and $1+\sin 2x = t^2$ in second integral