The value of the definite integral ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{\tan x}dx$, is

$I={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{\tan x}dx \dots \dots \dots \dots \dots \left(1\right)$ ; $I={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sqrt{\cot x}dx \dots \dots \dots \dots \left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$, we get

$$\begin{gathered} 2I={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx=\sqrt{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\sin x +\cos x}{\sqrt{\sin 2x}}dx \\ =\sqrt{2}{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{\sin x +\cos x}{\sqrt{1-{\left(\sin x-\cos x\right)}^2}}dx \end{gathered}$$

    $=\sqrt{2}{\int }_{-1}^1\frac{dt}{\sqrt{1-t^2}}=2\sqrt{2}{\int }_0^1\frac{dt}{\sqrt{1-t^2}}=\sqrt{2}\mathrm{\pi }$   (where $\sin x-\cos x=t$)

$\therefore I=\frac{\mathrm{\pi }}{\sqrt{2}}$