Question

Find $\left(\begin{array}{c}{\int }_0^{\pi /4}\frac{dx}{{\cos }^{3}x\sqrt{2\sin 2x}}\end{array}\right)$.

Answer 1

${\int }_0^{\pi /4}\frac{dx}{{\cos }^{3}x\sqrt{2sin2x}}={\int }_0^{\pi /4}\frac{dx}{{\cos }^{3}x\sqrt{4sinx\cos x}}={\int }_0^{\pi /4}\frac{dx}{{\cos }^{4}x\sqrt{4\tan x}}=\frac{1}{2}{\int }_0^{\pi /4}\frac{{\sec }^{4}x}{\sqrt{\tan x}}dx={\int }_0^{\pi /4}\frac{{\sec }^{2}x(1+{\tan }^{2}x)}{\sqrt{\tan x}}dxt=\tan x\Rightarrow dt={\sec }^{2}xdx\Rightarrow \frac{1}{2}{\int }_0^{\pi /4}\frac{{\sec }^{2}x(1+{\tan }^{2}x)}{\sqrt{\tan x}}dx=\frac{1}{2}{\int }_0^1(t^{-\frac{1}{2}}+t^{\frac{3}{2}})dt={[t^{\frac{1}{2}}+\frac{t^{\frac{5}{2}}}{5}]}_0^1=\frac{6}{5}$