Definite integration involving square root function

Question

How to integrate this definite integral: $$\int_{0}^{\pi/2} \big(\sqrt{\cos x}+ \sqrt{\cot x}\,\big)\,\mathrm dx$$

Answer 1

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:

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

Answer 2

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}