Prove $\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx=\frac{\sqrt2}{8}\left( \frac{5\pi^{3}}{12}-\pi^2\ln2-\pi \ln^22 \right)$ I came across the follwing integral:
$$\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx=\frac{\sqrt2}{8}\left( \frac{5\pi^{3}}{12}-\pi^2\ln2-\pi \ln^22 \right)$$
I try to do it like the following:
Consider
$$I(a,b)=\int_0^{\pi} \frac{\cos ax}{\sin^b x}\ dx=2\int_0^{\pi/2} \frac{\cos 2ax}{\sin^b 2x}\ dx$$
Then
$$I''(a,b)=-8\int_0^{\pi/2}x^2 \frac{\cos 2ax}{\sin^b 2x}\ dx$$
Let $a=1/2,b=1/2$
$$I''(1/2,1/2)=-8\int_0^{\pi/2}x^2 \frac{\cos x}{\sqrt{\sin 2x}}\ dx=-\frac{8}{\sqrt2}\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx $$
Back to $I(a,b),\quad I(a,b) $can be expressed by beta fuction by Wolfram Mathematica.
$$I(a,b)=\int_0^{\pi} \frac{\cos ax}{\sin^b x}\ dx=\frac{\pi \cdot 2^b\cdot\cos (\pi a/2)\cdot \Gamma(1-b) }{\Gamma(a/2-b/2+1) \cdot \Gamma(-a/2-b/2+1)} dx$$
Then we can get $I''(a,b)$ approach from other way.
Finally,we can get $\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx$,but it seems a little complex. 
For a similar integral: 
$$\int_0^{\pi/2}x\cdot\tan^p x \ dx=\frac{\pi}{4\sin (p\pi/2)}\left(\Psi\left(\frac{1}{2}\right)-\Psi\left(\frac{1-p}{2}\right) \right)$$
The above integral can be solved by method of parametric development.Let $p=-\frac{1}{2}$,We can get 
$$\int_0^{\pi/2}x\sqrt{\cot x} \ dx=\frac{\pi\left(\pi-2\ln 2\right)}{4\sqrt2}$$
But to this one, it seems to be difficult with method of parametric development.  Could you suggest any ideas how to prove this?
 A: A slightly easier form for $I(a, \frac{1}{2})$ is
$$I(a, \frac{1}{2}) = 2^{a-\frac{3}{2}} \beta\left(\frac{1+2a}{4}\right) (1 + \sin \pi a + \cos \pi a), $$
where $\beta(p) = \Gamma(p)^2/\Gamma(2p)$ is the central beta function. (This can be easily obtained by applying the Euler reflection formula and the Legendre duplication formula to OP's representation.) Now from the relation
$$ \frac{d}{dp} \log \beta(p) \bigg|_{p=\frac{1}{2}} = 2\psi(1/2) - 2\psi(1) = 2 \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+\frac{1}{2}} \right) = -4 \log 2 $$
and
$$ \frac{d^2}{dp^2} \log \beta(p) \bigg|_{p=\frac{1}{2}} = 2\psi^{(1)}(1/2) - 4\psi^{(1)}(1) = \frac{\pi^2}{3}, $$
we can compute $\frac{\partial^2 I}{\partial a^2}(\frac{1}{2}, \frac{1}{2})$ as follows:
\begin{align*}
&\frac{\partial^2 I}{\partial a^2} (1/2, 1/2) \\
&= \frac{1}{4} \beta''(1/2) - \left(\frac{\pi}{2}-\log 2\right) \beta'(1/2) - \left(\frac{\pi^2}{2} + \pi\log 2-\log^2 2\right) \beta(1/2) \\
&= \frac{1}{4} \pi \left( 16\log^2 2 + \frac{\pi^2}{3} \right) \\
&\qquad - \left(\frac{\pi}{2}-\log 2\right) (-4\pi \log 2)
 - \left(\frac{\pi^2}{2} + \pi\log 2-\log^2 2\right) \pi \\
&= -\left(\frac{5\pi^3}{12} - \pi^2\log2 - \pi \log^2 2 \right). 
\end{align*}

I am also trying a more direct approach. Using contour integral, we can check that
$$ \int_{0}^{\frac{\pi}{2}} x^2 \sqrt{\cot x} \, \mathrm{d}x
= \frac{1}{3\sqrt{2}} \left( \frac{\pi^3}{4} - \frac{3\pi}{2} \int_{0}^{1} \frac{\operatorname{artanh}^2 t}{\sqrt{t}} \, \mathrm{d}t \right). $$
So I am tackling the last integral, but have no good news at this point.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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It's quite clear the OP already evaluated the integral
$\ds{\color{#f00}{\int_{0}^{\pi/2}x^{2}\root{\cot\pars{x}}\,\dd x}}$. However, the OP is $\texttt{still looking for a simple evaluation}$ which start from $\ds{\,\mrm{I}\pars{a,b}\ \pars{~\mbox{see below}~}}$. Hereafter we presente a 'relatively simple' evaluation of it.

$\ds{\mrm{I}\pars{a,b} =
{\pi\,2^{b}\,\cos\pars{\pi a/2}\Gamma(1 - b) \over
\Gamma\pars{a/2 - b/2 + 1}\Gamma\pars{-a/2 - b/2 + 1}}\,,\qquad\Gamma\pars{\half} = \root{\pi}}$

\begin{align}
\mrm{I}\pars{a,\half} & =
\root{2}\pi^{3/2}\,\,
{\cos\pars{\pi a/2} \over \Gamma\pars{3/4 + a/2}\Gamma\pars{3/4 - a/2}}
\end{align}

Note that
$$
\color{#f00}{\int_{0}^{\pi/2}x^{2}\root{\cot\pars{x}}\,\dd x} =
-\,{\root{2} \over 8}\braces{2\bracks{\epsilon^{2}}\mrm{I}\pars{\half + \epsilon,\half}}
$$
where
\begin{align}
\mrm{I}\pars{\half + \epsilon,\half} & =
\pi^{3/2}\,\,{\cos\pars{\pi\epsilon/2} - \sin\pars{\pi\epsilon/2} \over
\Gamma\pars{1 + \epsilon/2}\Gamma\pars{1/2 - \epsilon/2}} =
\pi^{3/2}\,\,{\cos\pars{\pi\epsilon/2} - \sin\pars{\pi\epsilon/2} \over
\pars{\epsilon/2}!\pars{-1/2 - \epsilon/2}!}
\\[5mm] & =
\pi^{3/2}\,{1 \over \pars{-1/2}!}{-1/2 \choose \epsilon/2}
\bracks{\cos\pars{\pi\epsilon/2} - \sin\pars{\pi\epsilon/2}}
\\[5mm] & =
\pi\,{-1/2 \choose \epsilon/2}
\bracks{\cos\pars{\pi\epsilon \over 2} - \sin\pars{\pi\epsilon \over 2}}
\end{align}


We just need the binomial and 'the inside brackets term' expansion up to $\ds{\epsilon^{2}}$. The binomial expansion, up to order $\ds{\epsilon^{2}}$, is simple but laborious: It is simplified by using the Digamma value $\ds{\Psi\pars{1/2} = -\gamma - 2\ln\pars{2}}$. $\ds{\gamma}$: Euler-Mascheroni Constant. 

$$
\left\{\begin{array}{rcl}
\ds{-1/2 \choose \epsilon/2} & \ds{=} &
\ds{1 - \ln\pars{2}\,\epsilon +
\bracks{\half\,\ln^{2}\pars{2} - {\pi^{2} \over 12}}\epsilon^{2} + \,\mrm{O}\pars{\epsilon^{3}}}
\\[3mm]
\ds{\cos\pars{\pi\epsilon \over 2} - \sin\pars{\pi\epsilon \over 2}} & \ds{=} & \ds{1 - {\pi \over 2}\,\epsilon - {\pi^{2} \over 8}\,\epsilon^{2} + \,\mrm{O}\pars{\epsilon^{3}}}
\end{array}\right.
$$

\begin{align}
&\color{#f00}{\int_{0}^{\pi/2}x^{2}\root{\cot\pars{x}}\,\dd x}
\\[5mm] = &\
-\,{\root{2} \over 8}\,\times 2\times \pi\braces{%
1\times\pars{-\,{\pi^{2} \over 8}} +
\bracks{-\ln\pars{2}}\pars{-\,{\pi \over 2}} +
\bracks{\half\,\ln^{2}\pars{2} - {\pi^{2} \over 12}}\times 1}
\\[5mm] & =
\color{#f00}{{\root{2} \over 8}\bracks{{5\pi^{3} \over 12} -
\pi^{2}\ln\pars{2} - \pi\ln^{2}\pars{2}}} \approx 0.8077
\end{align}
