Calculating $\int_0^{\pi/2} \sqrt{\cot x} + \sqrt{\cos x} dx$ How should I solve the following integral:
$$\int_0^{\pi/2} (\sqrt{\cot x} + \sqrt{\cos x} )\,\mathrm dx$$
 A: I think can use 
\begin{align}I&= \int_0^{\frac{\pi}{2}}\left(\sqrt{\cot (x)}+\sqrt{\cos (x)}\right)\,\mathrm dx\\&=\int _0^1\frac{1}{\cos(x)}\left(\sqrt{\cot (x)}+\sqrt{\cos (x)}\right)\, \mathrm d(\sin(x))\\&=\int _0^1\frac{1}{\sqrt{1-a^2}}\left(\sqrt{\frac{\sqrt{1-a^2}}{a}}+\sqrt{\sqrt{1-a^2}}\right)\,\mathrm da\\ &=\int _0^1 a^{-\frac{1}{2}}(1-a^2)^{-\frac{1}{4}}\,\mathrm da+\int _0^1 (1-a^2)^{-\frac{1}{4}}\,\mathrm da\\& =\frac{1}{2}\int _0^1 (a^2)^{-\frac{3}{4}}(1-a^2)^{-\frac{1}{4}}\,\mathrm d(a^2)+\frac{1}{2}\int _0^1 (a^2)^{-\frac{1}{2}}(1-a^2)^{-\frac{1}{4}}\,\mathrm d(a^2)\\ &=\frac{1}{2}B \left(\frac{1}{4}, \frac{3}{4}\right)+\frac{1}{2}B\left(\frac{1}{2},\frac{3}{4}\right)\end{align} with $B$ is beta function. Hope it can help.
A: Substituting $u=\sin^2(x)$ and $1-u=\cos^2(x)$, we get the same answer as Alexis, but go a bit further.
$$
\begin{align}
\int_0^{\pi/2}\left(\sqrt{\cot(x)}+\sqrt{\cos(x)}\right)\,\mathrm{d}x
&=\int_0^{\pi/2}\sqrt{\cot(x)}\,\mathrm{d}x+\int_0^{\pi/2}\sqrt{\cos(x)}\,\mathrm{d}x\\
&=\int_0^{\pi/2}\sqrt{\frac1{\sin(x)\cos(x)}}\,\mathrm{d}\sin(x)+\int_0^{\pi/2}\sqrt{\frac1{\cos(x)}}\,\mathrm{d}\sin(x)\\
&=\frac12\int_0^1u^{-3/4}(1-u)^{-1/4}\,\mathrm{d}u+\frac12\int_0^1u^{-1/2}(1-u)^{-1/4}\,\mathrm{d}u\\
&=\frac12\frac{\Gamma(1/4)\Gamma(3/4)}{\Gamma(1)}+\frac12\frac{\Gamma(1/2)\Gamma(3/4)}{1/4\Gamma(1/4)}\\
&=\frac\pi{\sqrt2}+\frac{(2\pi)^{3/2}}{\Gamma(1/4)^2}\\[6pt]
&\doteq3.4195817038147753309
\end{align}
$$
