TITLE: Calculating $\int_0^{\pi/2} \sqrt{\cot x} + \sqrt{\cos x} dx$
QUESTION [5 upvotes]: How should I solve the following integral:
$$\int_0^{\pi/2} (\sqrt{\cot x} + \sqrt{\cos x} )\,\mathrm dx$$

REPLY [3 votes]: Substituting $u=\sin^2(x)$ and $1-u=\cos^2(x)$, we get the same answer as Alexis, but go a bit further.
$$
\begin{align}
\int_0^{\pi/2}\left(\sqrt{\cot(x)}+\sqrt{\cos(x)}\right)\,\mathrm{d}x
&=\int_0^{\pi/2}\sqrt{\cot(x)}\,\mathrm{d}x+\int_0^{\pi/2}\sqrt{\cos(x)}\,\mathrm{d}x\\
&=\int_0^{\pi/2}\sqrt{\frac1{\sin(x)\cos(x)}}\,\mathrm{d}\sin(x)+\int_0^{\pi/2}\sqrt{\frac1{\cos(x)}}\,\mathrm{d}\sin(x)\\
&=\frac12\int_0^1u^{-3/4}(1-u)^{-1/4}\,\mathrm{d}u+\frac12\int_0^1u^{-1/2}(1-u)^{-1/4}\,\mathrm{d}u\\
&=\frac12\frac{\Gamma(1/4)\Gamma(3/4)}{\Gamma(1)}+\frac12\frac{\Gamma(1/2)\Gamma(3/4)}{1/4\Gamma(1/4)}\\
&=\frac\pi{\sqrt2}+\frac{(2\pi)^{3/2}}{\Gamma(1/4)^2}\\[6pt]
&\doteq3.4195817038147753309
\end{align}
$$