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 $a \in (0,1)$.
