The solution to the following integral
$$\int_0^{\tfrac{\pi }{2}} \cos ^{-1}\frac{1}{2 \cos x+1} \, \mathrm dx$$
is equal to the numerical value of $\frac{\pi^2}{6}$. It is verifiable numerically, or with a computer algebra system (for example, Mathematica using NIntegrate).
Could this solution be calculated or determined analytically?
\begin{align}&\int_0^{\frac{\pi }{2}} \cos ^{-1}\frac{1}{2 \cos x+1} \overset{t=\tan\frac x2} {dx}\\ =&\ 4\int_0^1 \frac{\tan^{-1}\sqrt{\frac{1-t^2}2}}{1+t^2}dt =4\int_0^1\int_0^1\frac{\sqrt{\frac{1-t^2}2}}{(1+t^2)(1+ \frac{1-t^2}2 s^2)} ds\ dt\\ =& \ 4\int_0^1\int_0^1\frac{1}{(1+s^2) \sqrt \frac{1-t^2}2}\bigg(\frac{1}{1+t^2} -\frac1{2+ (1-t^2)s^2}\bigg)dt\ ds\\ =& \ 2\pi\int_0^1 \frac{1}{1+s^2}\bigg(1-\frac1{\sqrt{2+s^2}}\bigg) ds =2\pi\left(\frac\pi4-\frac\pi{6} \right)=\frac{\pi^2}6 \end{align}
With a few steps we can recover the integral discussed here:
$$\begin{align*} I &= \int_0^\tfrac\pi2 \operatorname{arcsec}\left(1+2\cos x\right) \, dx \\ &= \int_0^\tfrac\pi2 \frac{x \sin x}{\sqrt{\cos x}\sqrt{1+\cos x}(1+2\cos x)} \, dx & \text{by parts} \\ &= \int_0^\tfrac\pi4 \frac{4\sqrt2 \, x \sin(2x)}{\sqrt{1-2\sin^2x}\left(3-4\sin^2x\right)} \, dx & x\to2x \\ &= \int_0^1 \frac{4y \arcsin\frac y{\sqrt2}}{\sqrt{1-y^2}\sqrt{2-y^2}\left(3-2y^2\right)} \, dy & \sin x=\frac y{\sqrt2} \\ &= \int_0^1 \frac{4\sqrt2 \, z\arctan z}{\sqrt{1-z^2} \left(3-z^2\right)} & y=\frac{\sqrt2\,z}{\sqrt{1+z^2}} \\ &= 4\cdot\frac{\pi^2}{24} = \boxed{\frac{\pi^2}6} \end{align*}$$
