${\displaystyle \int \frac{dx}{{\cos }^{3}x\sqrt{\sin 2x}}}$ is equal to

${\displaystyle \int \frac{dx}{{\cos }^{3}x\sqrt{\sin 2x}}={\displaystyle \int \frac{dx}{{\cos }^{3}x\sqrt{\frac{2\tan x}{1+{\tan }^{2}x}}}}}$

                          $={\displaystyle \int \frac{{\sec }^{4}x}{\sqrt{2\tan x}}dx=\frac{1}{\sqrt{2}}{\displaystyle \int \frac{\left(1+z^4\right).2zdz}{z}}}$

                                       $\left[Putting\tan x=z^2\Rightarrow {\sec }^{2}xdx=2zdz\right]$

                          $=\sqrt{2}\left(z+\frac{z^5}{5}\right)+C=\sqrt{2}\left(\sqrt{\tan x}+\frac{1}{5}{\tan }^{5/2}x\right)+C.$