Evaluate the following integrals:

$\operatorname{Let} I=\int \frac{1-\cos 2 x}{1+\cos 2 x} d x$

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

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

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

Therefore, we can write the integral as

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

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

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

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

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

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

$\therefore I=\tan x-x+c$

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