Evaluate the following integrals:

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

We have $\cos x=\cos \left(2 \times \frac{x}{2}\right)$

We know $\cos 2 \theta=1-2 \sin ^{2} \theta=2 \cos ^{2} \theta-1$

Hence, in the numerator, we can write $1-\cos x=2 \sin ^{2} \frac{x}{2}$

In the denominator, we can write $1+\cos x=2 \cos ^{2} \frac{x}{2}$

Therefore, we can write the integral as

$I=\int \frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} d x$

$\Rightarrow I=\int \frac{\sin ^{2} \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x$

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

$\Rightarrow \mathrm{I}=\int\left(\sec ^{2} \frac{\mathrm{x}}{2}-1\right) \mathrm{dx}\left[\because \sec ^{2} \theta-\tan ^{2} \theta=1\right]$

$\Rightarrow I=\int \sec ^{2} \frac{x}{2} d x-\int d x$

Recall $\int \sec ^{2} x d x=\tan x+c$ and $\int d x=x+c$

$\Rightarrow \mathrm{I}=\frac{\tan \frac{\mathrm{x}}{2}}{\frac{1}{2}}-\mathrm{x}+\mathrm{c}$

$\therefore I=2 \tan \frac{x}{2}-x+c$

Thus, $\int \frac{1-\cos x}{1+\cos x} d x=2 \tan \frac{x}{2}-x+c$