How to show that $\int_0^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}dx=\frac{\pi^2}{4}$? I am trying to show that
$$\int_0^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}dx=\frac{\pi^2}{4}$$
Context: I was working on another question ("Attempt $2$") and miscopied an integral, so I was trying to evaluate the integral in my question here. Anyway, now I'm intrigued by this integral, because my computer strongly suggests that it has a closed form, $\frac{\pi^2}{4}$.
My attempt: I have tried substituting $u=\cos{x}$ or $u=\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}$, and the half-angle tangent substitution, but they do not seem to work.
(Ideally, there is an elementary solution, but any solution would be appreciated.)

One can show that $\sin(x)/\sqrt{5/4-\cos(x)}$ increases from $0$ to $1$ for $x$ going from $0$ to $\pi/3$ and decreases from $1$ to $0$ for $x$ going from $\pi/3$ to $\pi$. Then the integral can be separated into
\begin{align}
 I&=\int_0^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}\,dx\\
 &=\int_0^{\pi/3} \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}\,dx+\int_{\pi/3}^\pi \arcsin{\left(\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}\right)}\,dx
\end{align}
We define
\begin{equation}
 u=\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}
\end{equation}
or $u^2(5/4-\cos{x})=1-\cos^2x$.
Solving this equation for $\cos x$,

*

*for $0\le x\le\pi/3$, we have $\cos x=(u^2+\sqrt{u^4-5u^2+4})/2$

*for $\pi/3\le x\le\pi$, we have $\cos x=(u^2-\sqrt{u^4-5u^2+4})/2$.

Substituting $u$ into the integral, it comes
\begin{align}
 I&=\int_0^1\arcsin(u)\frac{\sqrt{5-2 u^{2}-2 \sqrt{u^{4}-5 u^{2}+4}}}{\sqrt{u^{4}-5 u^{2}+4}}\,du+
 \int_0^1\arcsin(u)\frac{\sqrt{5-2 u^{2}+2 \sqrt{u^{4}-5 u^{2}+4}}}{\sqrt{u^{4}-5 u^{2}+4}}\,du\\
 &=\int_0^1\frac{\arcsin(u)}{\sqrt{u^{4}-5 u^{2}+4}}
 f(u)\,du
\end{align}
where $f(u)=\sqrt{5-2 u^{2}-2 \sqrt{u^{4}-5 u^{2}+4}}+\sqrt{5-2 u^{2}+2 \sqrt{u^{4}-5 u^{2}+4}}$.
Now, it is possible to simplify:
\begin{align}
 \left[f(u)\right]^2&=10-4u^2+2\sqrt{5-2 u^{2}-2 \sqrt{u^{4}-5 u^{2}+4}}\sqrt{5-2 u^{2}+2 \sqrt{u^{4}-5 u^{2}+4}}\\
 &=4(4-u^2)
\end{align}
and
\begin{equation}
 \sqrt{\frac{4(4-u^2)}{u^{4}-5 u^{2}+4}}=\frac{2}{\sqrt{1-u^2}}
\end{equation}
Then
\begin{align}
 I&=2\int_0^1\frac{\arcsin(u)}{\sqrt{1-u^2}}\,du\\
&=\left.\arcsin^2(x)\right|_0^1\\
 &=\frac{\pi^2}{4}
\end{align}
as expected.
