Definite integration involving square root function How to integrate this definite integral:
$$\int_{0}^{\pi/2}  \big(\sqrt{\cos x}+ \sqrt{\cot x}\,\big)\,\mathrm dx$$
 A: Using the properties of Euler's Beta function and the residue theorem we have:
$$\begin{eqnarray*} \int_{0}^{\pi/2}\sqrt{\cos x}\,dx &=& \int_{0}^{\pi/2}\sqrt{\sin{x}}\,dx = \int_{0}^{1}\frac{\sqrt{t}}{\sqrt{1-t^2}}\,dt =\frac{1}{2}\int_{0}^{1}u^{-1/4}(1-u)^{-1/2}\\&=&\frac{1}{2}B(3/4,1/2)=\color{red}{\sqrt{\frac{2}{\pi}}\,\Gamma^2\left(\frac{3}{4}\right)}\tag{1}\end{eqnarray*}$$
and:
$$\begin{eqnarray*}\int_{0}^{\pi/2}\sqrt{\cot x}\,dx=\int_{0}^{+\infty}\frac{\sqrt{t}}{1+t^2}\,dt = \int_{-\infty}^{+\infty}\frac{u^2}{1+u^4}\,du = \color{blue}{\frac{\pi}{\sqrt{2}}},\tag{2}\end{eqnarray*}$$
so:

$$ I = \int_{0}^{\pi/2}\left(\sqrt{\cos x}+\sqrt{\cot x}\right)\,dx = \color{blue}{\frac{\pi}{\sqrt{2}}}+\color{red}{\sqrt{\frac{2}{\pi}}\,\Gamma^2\left(\frac{3}{4}\right)}.\tag{3}$$

A: We have
$$\operatorname{B}(x,y) =2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,\mathrm{d}\theta,\qquad \mathrm{Re}(x)>0,\ \mathrm{Re}(y)>0$$
where $\operatorname{B}(x,y)$ is the beta function. Hence
\begin{align}
\int_{0}^{\pi/2}  \big(\sqrt{\cos x}+ \sqrt{\cot x}\,\big)\,\mathrm dx&=\int_{0}^{\pi/2}  \left(\sqrt{\cos x}+ \sqrt{\frac{\cos x}{\sin x}}\,\right)\,\mathrm dx\\[7pt]
&=\frac{1}{2}\left[\operatorname{B}\left(\frac{1}{2},\frac{3}{4}\right)+\operatorname{B}\left(\frac{1}{4},\frac{3}{4}\right)\right]\\[7pt]
\end{align}
