We have, $I=\int \frac{\sqrt{\cos 2 x}}{\sin x} d x$
Put $\cos 2 x=\frac{1-\tan ^2 x}{1+\tan ^2 x}$
$\because \quad I=\int \frac{\sqrt{1-\tan ^2 x}}{\sec x \sin x} d x$
Put $\quad 1-\tan ^2 x=t^2$
$$
\begin{aligned}
& \Rightarrow \quad-2 \tan x \sec ^2 x d x=2 t d t \\
& I=-\int \frac{t^2}{\left(1-t^2\right)\left(2-t^2\right)} d t=-\int \frac{t^2}{\left(t^2-1\right)\left(t^2-2\right.} d t \\
& I=-\int\left(\frac{2}{t^2-2}-\frac{1}{t^2-1}\right) d t=2 \int \frac{-1}{t^2-2} d t+\int \frac{d t}{t^2-1} \\
& I=\frac{-2}{2 \sqrt{2}} \log \left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right|+\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+c \\
& I=\frac{-1}{\sqrt{2}} \log \left|\frac{\sqrt{1-\tan ^2 x}-\sqrt{2}}{\sqrt{1-\tan ^2 x}+\sqrt{2}}\right|+\frac{1}{2} \log \left|\frac{\sqrt{1-\tan ^2 x}-1}{\sqrt{1-\tan ^2 x}+1}\right|+c \\
& I=\frac{1}{\sqrt{2}} \log \left|\frac{\sqrt{2}+\sqrt{1-\tan ^2 x}}{\sqrt{2}-\sqrt{1-\tan ^2 x}}\right|-\frac{1}{2} \log \left|\frac{1+\sqrt{1-\tan ^2 x}}{1-\sqrt{1-\tan ^2 x}}\right|+c
\end{aligned}
$$