Integral $\int_0^{\frac{\pi}{2}} x^2 \sqrt{\sin x} \, dx$

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.

The original integral in the question unfortunately lacks a closed form. You can rewrite it in terms of the hypergeometric function, but that's the best you can get. However I did a bit of digging and found an explicit expression for the integral below

$$\mathfrak{I}=\int\limits_0^{\pi/2}\mathrm dx\, x^2\sqrt{\cos x}$$

Perhaps this may help in some way. First, let $x=\arcsin\sqrt t$ and our integral transforms into

$$\mathfrak{I}=\frac 12\int\limits_0^1\mathrm dt\,t^{-1/2}(1-t)^{-1/4}\arcsin^2\sqrt t$$

The integrand is very similar to the beta function, so we can substitute $\arcsin(\cdot)$ with it's infinite series counterpart and rewrite the integral in terms of the gamma function. Recalling that

$$\arcsin^2z=\sum\limits_{n\geq1}\frac {2^{2n-1}z^{2n}}{n^2\binom {2n}n}$$

We get

$$\begin{align*}\mathfrak{I} & =\frac 14\sum\limits_{n\geq1}\frac {4^n}{n^2\binom {2n}n}\int\limits_0^1\mathrm dt\, t^{n-1/2}(1-t)^{-1/4}\\ & =\frac 14\sum\limits_{n\geq1}\frac {4^n\Gamma^2(n+1)}{n^2\Gamma(2n+1)}\frac {\Gamma\left(n+\frac 12\right)\Gamma\left(\frac 34\right)}{\Gamma\left(n+\frac 54\right)}\end{align*}$$

Using the two equations$$\Gamma\left(n+\frac 12\right)=\frac {\Gamma(2n+1)\sqrt\pi}{4^n\Gamma(n+1)}$$$$\Gamma\left(\frac 14\right)\Gamma\left(\frac 34\right)=\pi\sqrt2$$

The infinite sum simplifies into something much less daunting

$$\mathfrak{I}=\frac {\pi\sqrt{2\pi}}{4\Gamma\left(\frac 14\right)}\sum\limits_{n\geq1}\frac {\Gamma(n)}{n\Gamma\left(n+\frac 54\right)}$$

The gamma functions can again be rewritten as the beta function and converted back to an integral to get

$$\begin{align*}\mathfrak{I} & =\frac {\pi\sqrt{2\pi}}{\Gamma^2\left(\frac 14\right)}\sum\limits_{n\geq1}\frac 1n\operatorname{B}\left(n,\frac 54\right)\\ & =\frac {\pi\sqrt{2\pi}}{\Gamma^2\left(\frac 14\right)}\int\limits_0^1\mathrm dt\,\frac {(1-t)^{1/4}}t\sum\limits_{n\geq1}\frac {t^n}n\\ & =-\frac {\pi\sqrt{2\pi}}{\Gamma^2\left(\frac 14\right)}\int\limits_0^1\mathrm dt\,\frac {(1-t)^{1/4}\log(1-t)}t\end{align*}$$

The resulting integral is rather easy to compute by first using the substitution $1-t\mapsto t^4$ and then using partial fraction decomposition to arrive at

$$\int\limits_0^1\mathrm dt\,\frac {(1-t)^{1/4}\log(1-t)}t=16-8G-\pi^2$$

where $G$ is Catalan's constant. Therefore, it's easy to see that the final result, in color, is

$$\mathfrak{I}=\int\limits_0^{\pi/2}\mathrm dx\, x^2\sqrt{\sin x}\color{blue}{=\frac {\pi\sqrt{2\pi}}{\Gamma^2\left(\frac 14\right)}\biggr[\pi^2+8G-16\biggr]}$$

If we put that together into the original integral, another possible solution, given by ComplexYetTrivial, is $$\int\limits_0^{\pi/2}\mathrm dx\, x^2\sqrt{\sin x}\color{red}{=\frac{\pi\sqrt{2\pi}}{\Gamma^2\left(\frac 14\right)}\biggr[\frac {3\pi^2}2+8G-16\biggr]-\frac {4\pi}{15}{}_3F_2\left[\begin{array}{c}1,\frac 54,\frac 54\\\frac 74,\frac 94\end{array}\right]}$$