# 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.)

## 1 Answer

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 $$$$u=\frac{\sin{x}}{\sqrt{5/4-\cos{x}}}$$$$ or $$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 $$$$\sqrt{\frac{4(4-u^2)}{u^{4}-5 u^{2}+4}}=\frac{2}{\sqrt{1-u^2}}$$$$ 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.