Integral $\int_0^{\frac{\pi}{2}} x^2 \sqrt{\sin x} \, dx$ Greetings I am trying to find a closed form for:  $$I=\int_0^{\frac{\pi}{2}} x^2 \sqrt{\sin x}\,dx$$ If we rewrite the integral as $$I=\int_0^\infty x^2 \sqrt{\frac{1}{\sqrt{1+\cot^2 x}}}\,dx$$ now with $$\cot x =t $$ $$I=\int_0^{\infty} \operatorname{arccot}^2 (x)(1+x^2)^{-\frac{5}{4}}dx$$ and with https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms $$I=\frac{1}{4i}\int_0^{\infty}\log^2\left(\frac{z-i}{z+i}\right)(1+x^2)^{-\frac{5}{4}} \, dx$$ Now for the $\log$ I thought to expand into power series but since the radius of converge is abit smaller, this fails. Also integrating by parts or combining the initial integral with $\int_0^{\frac{\pi}{2}} x^2 \sqrt{\cos x}\,dx$ wasn't much of a help, could you help me evaluate this integral ?

The substitution $\sin(x) = \sqrt{t}$ leads to the expression
$$I = \frac{1}{2} \int \limits_0^1 \frac{t^{-1/4} \arcsin^2 (\sqrt{t})}{\sqrt{1-t}} \, \mathrm{d} t \, . $$
Now you can use the power series for $\arcsin^2$ (see for example this question) and integrate term by term (monotone convergence). Using the beta function you will find
\begin{align}
I &= \frac{1}{4} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \int \limits_0^1 t^{n-\frac{1}{4}} (1-t)^{-\frac{1}{2}} \, \mathrm{d} t \\
&= \frac{1}{4} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \operatorname{B}\left(n+\frac{3}{4},\frac{1}{2}\right) \\
&= \frac{\sqrt{\pi}}{4} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \frac{\Gamma\left(n+\frac{3}{4}\right)}{\Gamma\left(n+\frac{5}{4}\right)} \\
&= \frac{\sqrt{\pi} \, \Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \frac{\prod_{k=1}^n (4k-1)}{\prod_{l=1}^{n+1} (4l-3)} \\
&= \frac{\pi \sqrt{2 \pi}}{\Gamma\left(\frac{1}{4}\right)^2} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (4n+1) (2n-1)!!} \prod \limits_{k=1}^n \frac{4k-1}{4k-3} \, .
\end{align}
Mathematica gives the following expression in terms of a hypergeometric function:
$$ I = \frac{6 \pi \sqrt{2 \pi}}{5 \Gamma\left(\frac{1}{4}\right)^2} \,  {}_4 \! \operatorname{F}_3 \left(1,1,1,\frac{7}{4};\frac{3}{2},2,\frac{9}{4};1\right) \approx 1.208656578687 \, .$$
Inverse symbolic calculators do not seem to give any expression for this number, so this might be as good as it gets. 
