Question

Integrate: ${\int }_0^{\pi /4}(\sqrt{\tan x}+\sqrt{\cot x})dx$

Answer 1

$I={\int }_0^{\frac{\pi }{4}}(\sqrt{\tan x}+\sqrt{\cot x})dx$

$I={\int }_0^{\frac{\pi }{4}}[\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}]dx$

$I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{\sqrt{\sin x\times \cos x}}dx$

Let $\sin x-\cos x=t$

$(\cos x+\sin x)dx=dt$

$(\sin x-\cos x{)}^2=t^2$

$1-2\sin x.\sec x=t^2$

$\sin x.\cos x=\frac{1-t^2}{2}$

so $I={\int }_{-1}^0\frac{\sqrt{2}}{\sqrt{1-t^2}}dt$

$I=\sqrt{2}{\sin }^{-1}t{|}_{-1}^0$

$I=0-\sqrt{2}\times \frac{3\pi }{2}$

$I=\frac{-3\pi }{\sqrt{2}}$