There is numerical evidence that $I=\int_0^{\pi/3}\arccos(2\sin^2 x-\cos x)\mathrm dx=\frac{\pi^2}{5}$. How can this be proved?
I was trying to answer another question, and I got it down to this integral. I tried integration by parts, substitution and Maclaurin series, but it seems that this integral requires more advanced techniques. Wolfram does not evaluate the indefinite integral.
Here is the graph of $y=\arccos(2\sin^2 x-\cos x)$ for $0\le x\le \frac{\pi}{3}$. It intersects the axes at $(\frac{\pi}{3},0)$ and $(0,\pi)$.
Not sure if this helps, but apparently we also have $\int_0^{\pi/3}\arccos^{\color{red}{2}}(2\sin^2 x-\cos x)\mathrm dx=\frac{19\pi^3}{135}$.
Best Answer
$$\int_0^\frac{\pi}{3}\arccos(2\sin^2 x-\cos x)dx\overset{\large \tan\frac{x}{2}\to x}=2\int_0^\frac{1}{\sqrt 3}\frac{\arccos\left(\frac{-1+8x^2+x^4}{(1+x^2)^2}\right)}{1+x^2}dx$$
$$=4\int_0^\frac{1}{\sqrt 3} \frac{\arctan\left(\frac{1}{x}\sqrt{\frac{1-3x^2}{5+x^2}}\right)}{1+x^2}dx\overset{\large \frac{1}{x}\to\sqrt x}=2\int_3^\infty \frac{\arctan \left(\sqrt x\sqrt{\frac{x-3}{5x+1}}\right)}{\sqrt{x}(1+x)}dx$$
$$\overset{\large \frac{x-3}{5x+1}\to x}=8\int_0^\frac{1}{5} \frac{1}{(1-x)(1-5x)} \frac{\arctan\left(\sqrt x \sqrt{\frac{3+x}{1-5x}}\right)}{\sqrt{\frac{3+x}{1-5x}}} dx$$
$$=8\int_0^\frac{1}{5} \frac{1}{(1-x)(1-5x)} \left(\int_0^{\sqrt x} \frac{1}{1+\frac{3+x}{1-5x}y^2}dy\right)dx$$
$$=8\int_0^\frac{1}{\sqrt 5}\int_{y^2}^\frac{1}{5} \frac{1}{1-x} \frac{1}{(1+3y^2)-(5-y^2)x}dxdy=2\int_0^\frac{1}{\sqrt 5} \frac{\ln\left(\frac{1-y^2}{4y^2}\right)}{1-y^2}dy$$
$$\overset{y\to\frac{1-y}{1+y}}=\int_{\large\frac{1}{\phi^2}}^1 \frac{\ln\left(\large \frac{y}{(1-y)^2}\right)}{y}dy=2 \operatorname{Li}_2(1)-2\operatorname{Li}_2\left(\frac{1}{\phi^2}\right)-2\ln^2\phi=\boxed{\frac{\pi^2}{5}}$$
$$\operatorname{Li}_2(1)=\frac{\pi^2}{6},\quad \operatorname{Li}_2\left(\frac{1}{\phi^2}\right) = \frac{\pi^2}{15}-\ln^2\phi,\quad \phi=\frac{1+\sqrt 5}{2}$$