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

We need to evaluate the integral $I = \int \frac{1-\cos x}{1+\cos x} dx$.
Using the trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$,we have:
$\frac{1-\cos x}{1+\cos x} = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = \tan^2 \frac{x}{2}$.
Using the identity $\tan^2 \theta = \sec^2 \theta - 1$,we can write:
$\tan^2 \frac{x}{2} = \sec^2 \frac{x}{2} - 1$.
Now,integrate the expression:
$I = \int (\sec^2 \frac{x}{2} - 1) dx = \int \sec^2 \frac{x}{2} dx - \int 1 dx$.
The integral of $\sec^2(ax)$ is $\frac{1}{a} \tan(ax) + C$.
Therefore,$I = \frac{\tan(x/2)}{1/2} - x + C = 2 \tan \frac{x}{2} - x + C$,where $C$ is an arbitrary constant.