Evaluate:

Let $I=\int \frac{(1-\cos x)}{(1+\cos x)} d x$

$=\int \frac{(1-\cos x)}{(1+\cos x)} d x$

$=\int \frac{\left(2 \sin ^{2} \frac{x}{2}\right)}{2 \cos ^{2} \frac{x}{2}}$

$=\int \tan ^{2} \frac{x}{2} d x$

$=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x$

$=\frac{\left(\tan \frac{x}{2}\right)}{\frac{1}{2}}-x$

Hence, $\mathrm{I}=2 \tan \frac{\mathrm{x}}{2}-\mathrm{x}+\mathrm{C}$