Just because it has a finite answer

$\begin{aligned} I & = \int_0^\frac \pi 2 \sqrt{\tan x} \ dx \\ & = \int_0^\frac \pi 2 \sin^\frac 12 x \cos^{-\frac 12} x \ dx & \small \blue{\text{Beta function }B(m, n) = \int_0^\frac \pi 2 2\sin^{2m-1} x \cos^{2n-1}x \ dx} \\ & = \frac 12 B \left(\frac 34, \frac 14 \right) & \small \blue{B(m,n) = \frac {\Gamma(m)\Gamma(n)}{\Gamma(m+n)}\text{, where }\Gamma (\cdot) \text{ denotes the gamma function.}} \\ & = \frac \blue{\Gamma \left(\frac 14\right)\Gamma \left(\frac 34\right)}{2\red{\Gamma (1)}} & \small \blue{\text{Euler's reflection formula: } \Gamma(z)\Gamma(1-z)=\frac \pi{\sin (\pi z)}} \\ & = \frac \blue \pi{2\blue{\sin \frac \pi4} \cdot \red{0!}} & \small \red{\text{Note that }\Gamma(n) = (n-1)!} \\ & = \frac \pi{\sqrt 2} \end{aligned}$

Therefore $m = \boxed 2$ .

References :

• Beta function
• Gamma function