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.
We want to prove that: $$\frac{I}{J}=\frac{\int_0^\pi x^3\ln(\sin x)dx} {\int_0^\pi x^2\ln\left(\sqrt 2\sin x\right)dx}=\frac{3\pi}2$$
First, let's take the $I$ integral and perform the $x\to \pi-x$ substitution:
$$I=\int_0^\pi x^3\ln(\sin x)dx\overset{x\to\pi-x}=\int_0^\pi (\pi^3-3\pi^2x+3\pi x^2-x^3)\ln(\sin x)dx$$
In the $J$ integral we had an additional $\ln \sqrt 2$ for the $x^2$ term, thus we can also add it here:
$$\Rightarrow I=\int_0^\pi (\pi^3-3\pi^2x+3\pi x^2-x^3)\ln(\sin x)dx+ 3\pi(\underbrace{\ln \sqrt 2-\ln \sqrt 2}_{=0})\int_0^\pi x^2 dx$$
$$=\pi^3 \underbrace{\int_0^\pi \ln(\sin x)dx}_{=\mathcal L}-3\pi^2 \underbrace{\int_0^\pi x\ln(\sin x)dx}_{=\mathcal K}+3\pi J-I-{\pi^4}\ln \sqrt 2$$
$$\Rightarrow 2I=\left(\pi^3-\frac{3\pi^3}{2}\right)\int_0^\pi \ln(\sin x)dx+3\pi J-{\pi^4}\ln \sqrt 2$$
$$\require{cancel} 2I=\cancel{\frac{\pi^3}{2}\cdot 2\pi \ln \sqrt 2}+3\pi J-\cancel{\pi^4 \ln \sqrt 2}\Rightarrow I=\frac{3\pi}2J$$
Note that above we used:
$$\mathcal K=\int_0^\pi x\ln(\sin x)dx\overset{x\to \pi-x}=\int_0^\pi (\pi-x)\ln(\sin x)dx$$
$$\Rightarrow \mathcal K=\frac{\pi}{2}\underbrace{\int_0^\pi \ln(\sin x)dx}_{=\mathcal L}=\frac{\pi}{2}\mathcal L$$
$$\mathcal L=\int_0^\pi \ln(\sin x)dx=\int_0^\frac{\pi}{2} \ln(\sin x)dx+\int_0^\frac{\pi}{2} \ln(\cos x)dx$$
$$=\int_0^\frac{\pi}{2} \ln\left(\sin x\cos x\right)=\int_0^\frac{\pi}{2} \ln(\sin 2x)dx-\int_0^\frac{\pi}{2} \ln 2dx$$
$$=\frac12 \underbrace{\int_0^\pi \ln(\sin x) dx}_{=\mathcal L}-\ln\sqrt 2 \int_0^{\pi} dx\Rightarrow \mathcal L=-2\pi \ln\sqrt 2$$
Best Answer
Here's an answer that does not rely on special functions. As in Mokrane's original approach, we combine the two integrals to obtain a double integral (the same method is also used in this answer to a similar question).
We have $$ P \equiv \int \limits_0^{\pi/2} \sqrt{\sin(x)} \, \mathrm{d} x \int \limits_0^{\pi/2} \frac{\mathrm{d} y}{\sqrt{\sin(y)}} = \int \limits_0^{\pi/2} \int \limits_0^{\pi/2} \sqrt{\frac{\sin(x)}{\sin(y)}} \, \mathrm{d} x \, \mathrm{d} y \ . $$ Letting $u = \sin(x)$ and $v = \sin(y)$, we find \begin{align} P &= \int \limits_0^1 \int \limits_0^1 \frac{\sqrt{\frac{u}{v}}}{\sqrt{1-u^2}\sqrt{1-v^2}} \, \mathrm{d} u \, \mathrm{d} v = \frac{1}{2} \int \limits_0^1 \int \limits_0^1 \frac{\sqrt{\frac{u}{v}} + \sqrt{\frac{v}{u}}}{\sqrt{1-u^2}\sqrt{1-v^2}} \, \mathrm{d} u \, \mathrm{d} v \\ &= \int \limits_0^1 \int \limits_0^v \frac{\sqrt{\frac{u}{v}} + \sqrt{\frac{v}{u}}}{\sqrt{1-u^2}\sqrt{1-v^2}} \, \mathrm{d} u \, \mathrm{d} v \, , \end{align} where we exploited the symmetry of the integrand to simplify the result. Now the substitution $u = v t^2$ in the inner integral yields \begin{align} P &= 2 \int \limits_0^1 \int \limits_0^1 \frac{(1 + t^2) v}{\sqrt{1 - t^4 v^2}\sqrt{1-v^2}} \, \mathrm{d} t \, \mathrm{d} v = 2 \int \limits_0^1 (1+t^2) \int \limits_0^1 \frac{v}{\sqrt{1 - t^4 v^2}\sqrt{1-v^2}} \, \mathrm{d} v \, \mathrm{d} t \\ &\overset{(*)}{=} 2 \int \limits_0^1 (1+t^2) \frac{\operatorname{artanh}(t^2)}{t^2} \, \mathrm{d} t = 2 \int \limits_0^1 \left(1 + \frac{1}{t^2}\right) \operatorname{artanh}(t^2) \, \mathrm{d} t \overset{\text{IBP}}{=} 2 \int \limits_0^1 \left(\frac{1}{t} - t\right) \frac{2 t}{1-t^4} \, \mathrm{d} t \\ &= 4 \int \limits_0^1 \frac{\mathrm{d} t}{1+t^2} = 4 \arctan(1) = \pi \, . \end{align}
Proof of $(*)$:
The Euler substitution $w = \sqrt{\frac{1-v^2}{a^{-2}-v^2}}$ yields $$ \int \limits_0^1 \frac{v}{\sqrt{1-a^2 v^2}\sqrt{1-v^2}} \, \mathrm{d} v = \frac{1}{a} \int \limits_0^a \frac{\mathrm{d}w}{1-w^2} = \frac{\operatorname{artanh}(a)}{a} $$ for $a \in (0,1)$.