Explanation

(a)
Let $I = \int_{}^{}\frac{dx}{{\cos^{3}x}\sqrt{2\sin{2x}}}$
$= \int_{}^{}{\frac{dx}{\cos^{3}{x\sqrt{4\sin{x\cos x}}}} = \frac{1}{2}\int_{}^{}\frac{dx}{\cos^{7/2}{x\sin^{1/2}x}}}$
$= \frac{1}{2}\int_{}^{}{\frac{\sec^{4}x}{\sqrt{\tan x}}dx = \frac{1}{2}\int_{}^{}{\frac{(1 + \tan^{2}{x)\sec^{2}x}}{\sqrt{\tan x}}dx}}$
Put $\tan{x = t\ \Rightarrow \sec^{2}{x\ dx = dt}}$
$\therefore I = \frac{1}{2}\int_{}^{}{\frac{1 + t^{2}}{\sqrt{t}}dt}$
$= \frac{1}{2}\int_{}^{}{t^{- 1/2}dt + \frac{1}{2}\int_{}^{}{t^{3/2}dt = t^{1/2} + \frac{t^{5/2}}{5} + c}}$
$= \sqrt{\tan x} + \frac{1}{5}\tan^{5/2}x + c$