Evaluate the following integrals:

Let I $=\int \tan ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right) d x$

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

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

In the numerator, we have $\sin 2 x=2 \sin x \cos x$

Therefore, we can write the integral as

$\mathrm{I}=\int \tan ^{-1}\left(\frac{2 \sin \mathrm{x} \cos \mathrm{x}}{2 \cos ^{2} \mathrm{x}}\right) \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int \tan ^{-1}\left(\frac{\sin \mathrm{x}}{\cos \mathrm{x}}\right) \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int \tan ^{-1}(\tan \mathrm{x}) \mathrm{dx}$

$\Rightarrow \mathrm{I}=\int \mathrm{xdx}$

Recall $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow I=\frac{X^{1+1}}{1+1}+c$

$\therefore \mathrm{I}=\frac{\mathrm{x}^{2}}{2}+\mathrm{c}$

Thus, $\int \tan ^{-1}\left(\frac{\sin 2 \mathrm{x}}{1+\cos 2 \mathrm{x}}\right) \mathrm{dx}=\frac{\mathrm{x}^{2}}{2}+\mathrm{c}$