Prove

$\frac{1-\cos x}{1+\cos x}=\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}$    $\left[2 \sin ^{2} \frac{x}{2}=1-\cos x\right.$ and $\left.2 \cos ^{2} \frac{x}{2}=1+\cos x\right]$

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

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

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

$=\left[\frac{\tan \frac{x}{2}}{\frac{1}{2}}-x\right]+\mathrm{C}$

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