Question

Prove that: ${\int }_0^{\pi /4}\frac{dx}{{\cos }^{3}x\sqrt{2\sin 2x}}=\frac{6}{5}$

Answer 1

$\begin{array}{c}{\int }_0^{\pi /4}\frac{dx}{{\cos }^{3}x\sqrt{2\sin 2x}}=\frac{6}{5}\\ ={\int }_0^{\pi /4}\frac{se{c}^{3}xdx}{\sqrt{4\sin x\cos x}}=\frac{6}{5}\\ ={\int }_0^{\pi /4}\frac{se{c}^{3}xdx}{2\cos x\sqrt{\tan x}}=\frac{6}{5}\\ =\frac{1}{2}{\int }_0^{\pi /4}\frac{se{c}^{4}xdx}{\sqrt{\tan x}}=\frac{6}{5}\\ put,y=\tan x\\ dy={\sec }^{2}xdx\\ \frac{1}{2}\underset{0}{\overset{1}{\int }}\frac{\left(1+y^2\right)}{\sqrt{y}}dy\\ =\frac{1}{2}\left({\left[2\sqrt{y}\right]}_0^1+{\left[\frac{y^{5/2}}{5/2}\right]}_0^1\right)\\ =\frac{1}{2}\left(2+\frac{2}{5}\right)=\frac{6}{5}\end{array}$