Question

Find ${\int }_0^{\pi /4}\frac{dx}{{\cos }^{3}x\sqrt{2\sin 2x}}$

Answer 1

Taking ${\cos }^{4}x$ to the numerator it becomes ${\sec }^{4}x.$

$\Rightarrow \sin 2x=\frac{2\tan x}{1+{\tan }^{2}x}$ ${\sec }^{2}x=1+{\tan }^{2}x$

$\Rightarrow {\int }_0^{\pi /4}\frac{{\sec }^{4}x}{\sqrt{2\sin 2x}}dx={\int }_0^{\pi /4}\frac{{\sec }^{4}x(1+{\tan }^{2}x)}{\sqrt{\frac{4\tan 2}{1+{\tan }^{2}x}}}dx$

$\Rightarrow {\int }_0^{\pi /4}\frac{{\sec }^{4}x(1+{\tan }^{2}x)}{2\sqrt{\tan x}}dx$ Put $t=\tan x,dt={\sec }^{2}xdx$

$\Rightarrow {\int }_0^{\pi /4}\frac{\sqrt{1+t^2}}{2\sqrt{t}}dt={\int }_0^{\pi /4}\sqrt{\frac{1+t^2}{4t}}dt$

Solve by partial fraction.

As the degree of numerator is greater than the degree of denominator,

So they can be reduced by long division method.

$\Rightarrow 4t\times \frac{1}{4}t=t^2$

$\therefore {\int }_0^{\pi /4}\sqrt{\frac{1}{4}}t+\sqrt{\frac{1}{4t}}dt$

$=\frac{1}{2}{\int }_0^{\pi /4}\sqrt{t}+\frac{1}{\sqrt{t}}dt$

$=\frac{1}{2}{[\frac{t^{3/2}}{3/2}+\frac{t^{1/2}}{1/2}]}_0^{\pi /4}$

$=\frac{1}{2}[\frac{{2t}^{3/2}}{3}+{2t}^{1/2}]={[\frac{t^{3/2}}{3}+t^{1/2}]}_0^{\pi /4}$

$={[\frac{{\tan }^{3/2}x}{3}+\sqrt{\tan x}]}_0^{\pi /4}$

$=\frac{1}{3}+1=\frac{4}{3}.$

Hence, solved.