Find the integral of the function $\frac{\cos x}{1+\cos x}$.

We need to evaluate the integral $I = \int \frac{\cos x}{1+\cos x} dx$.
First,we simplify the integrand:
$\frac{\cos x}{1+\cos x} = \frac{\cos x + 1 - 1}{1+\cos x} = 1 - \frac{1}{1+\cos x}$.
Using the identity $1+\cos x = 2\cos^2 \frac{x}{2}$,we get:
$1 - \frac{1}{2\cos^2 \frac{x}{2}} = 1 - \frac{1}{2}\sec^2 \frac{x}{2}$.
Now,integrate term by term:
$I = \int (1 - \frac{1}{2}\sec^2 \frac{x}{2}) dx = \int 1 dx - \frac{1}{2} \int \sec^2 \frac{x}{2} dx$.
$= x - \frac{1}{2} \cdot \frac{\tan \frac{x}{2}}{\frac{1}{2}} + C$.
$= x - \tan \frac{x}{2} + C$,where $C$ is an arbitrary constant.