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.

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.
