Solving $\int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$ 
$$\int_0^\frac{\pi}{2} \frac{\sqrt[3]{\sin x}}{4-\sin^2 x}dx$$

I tried much of elementary methods to solve above integral but is not advancing.
Any methods from elementary to advanced are appreciated.
 A: Hint. One may write
$$
\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}=\frac14\sum_{n=0}^\infty\frac1{4^n}\left(\sin x\right)^{2n+\frac13}
$$ then one is allowed to perform a termwise integration
$$
\int_0^{\Large \frac{\pi}2}\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}\,dx=\sum_{n=0}^\infty\frac1{4^{n+1}}\int_0^{\Large \frac{\pi}2}\left(\sin x\right)^{2n+\frac13}\,dx
$$using the classic Euler evaluation
$$
\int_0^{\Large \frac{\pi}2}\left(\sin x\right)^s\,dx=\frac{\sqrt{\pi} \,\Gamma \left(\frac{s+1}{2}\right)}{2\, \Gamma \left(\frac{s}{2}+1\right)},\qquad s>0,
$$ obtaining

$$
\int_0^{\Large \frac{\pi}2}\frac{\sqrt[3]{\sin x}}{4-\sin^2 x}\,dx=\frac{3 \sqrt{\pi} \,\, _2F_1\left(\frac{2}{3},1;\frac{7}{6};\frac{1}{4}\right)\, \Gamma \left(\frac{2}{3}\right)}{4\, \Gamma \left(\frac{1}{6}\right)}.
$$

A path to the simplification $\dfrac{\pi}{2^{2/3} 3^{3/2}}$ would be interesting.
A: Here is a way to simplify @Olivier Oloa's answer for the integral that is written in terms of the Gauss hypergeometric function.
Starting from (4) from a previous answer I gave here, it was shown that
$$_2F_1 \left (1, \frac{2}{3}; \frac{7}{6}; \frac{1}{4} \right ) = \frac{2^{4/3}}{\sqrt{\pi}} \Gamma \left (\frac{4}{3} \right ) \Gamma \left (\frac{7}{6} \right ).$$
Thus
\begin{align}
I &= \frac{3 \sqrt{\pi}}{2^2} \cdot \frac{2^{4/3}}{\sqrt{\pi}} \cdot \frac{\Gamma (\frac{4}{3}) \Gamma (\frac{7}{6}) \Gamma (\frac{2}{3})}{\Gamma (\frac{1}{6})}\\
&= \frac{1}{2^{2/3} 6} \Gamma \left (1 - \frac{1}{3} \right ) \Gamma \left (\frac{1}{3} \right )\\
&= \frac{1}{2^{2/3} 6} \cdot \frac{\pi}{\sin (\frac{\pi}{3})}\\
&= \frac{\pi}{2^{2/3} 3^{3/2}},
\end{align}
giving the required simplification. 
Actually, the reason for this coincidence is, as I have just discovered, unsurprising. As @Zacky correctly observes in the comments section of the linked question, the integral considered here is, to within a numerical factor, just that considered in the linked question, namely
$$\int_0^1 \frac{x^{2/3}}{\sqrt[3]{1 - x} (1 - x + x^2)} \, dx = 2^{5/3} \int_0^{\frac{\pi}{2}} \frac{\sqrt[3]{\sin x}}{4 - \sin^2 x} \, dx.$$ 
