$$
\text { Let } \begin{aligned}
\mathrm{I} & =\int \frac{1}{\cos ^3 x \sqrt{\sin 2 x}} \mathrm{~d} x \\
& =\frac{1}{\sqrt{2}} \int \frac{\sec ^3 x}{\sqrt{\sin x \cos x}} \mathrm{~d} \\
& =\frac{1}{\sqrt{2}} \int \frac{\sec ^4 x}{\sqrt{\tan x}} \mathrm{~d} x
\end{aligned}
$$

Let $\tan x=\mathrm{t} \Rightarrow \sec ^2 x=\mathrm{dt}$
$$
\begin{aligned}
\therefore \quad I & =\frac{1}{\sqrt{2}} \int \frac{1+t^2}{\sqrt{t}} d t \\
& =\frac{1}{\sqrt{2}} \int t^{-\frac{1}{2}} d t+\frac{1}{\sqrt{2}} \int t^{\frac{3}{2}} d t \\
& =\frac{1}{\sqrt{2}}\left(\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right)+\frac{1}{\sqrt{2}}\left(\frac{t^{\frac{5}{2}}}{\frac{5}{2}}\right)+\mathrm{c} \\
& =\sqrt{2}\left(\sqrt{\tan x}+\frac{1}{5}(\tan x)^{\frac{5}{2}}\right)+\mathrm{c}
\end{aligned}
$$