Evaluate the following integrals:

In the given equation $\cos 2 x=\cos ^{2} x-\sin ^{2} x$

Also we know $\cos ^{2} x+\sin ^{2} x=1$

$\therefore$ Substituting the values in the above equation we get

$\Rightarrow \int \frac{1}{\sqrt{\sin ^{2} x+\cos ^{2} x-\left(-\sin ^{2} x+\cos ^{2} x\right)}} d x$

$\Rightarrow \int \frac{1}{\sqrt{\sin ^{2} x+\cos ^{2} x+\sin ^{2} x-\cos ^{2} x}} d x$

$\Rightarrow \int \frac{1}{\sqrt{2 \sin ^{2} x}} \mathrm{dx}$

$\Rightarrow \int \frac{1}{\sqrt{2} \sin x} \mathrm{dx}$

$\Rightarrow \int \frac{\csc x}{\sqrt{2}} \mathrm{dx}$

$\Rightarrow \frac{1}{\sqrt{2}} \int \csc x d x$

$\Rightarrow \frac{1}{\sqrt{2}} \log \left|\frac{\tan x}{2}\right|+c$