Hyperelliptic generalization of Euler's formula Are there any hyperelliptic generalizations of the following formula, first proved by Euler in 1782,
$$\int\limits_0^1\frac{dx}{\sqrt{1-x^4}}\int\limits_0^1\frac{x^2\,dx}{\sqrt{1-x^4}}=\frac{\pi}{4}\,?$$
I found the formula in https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Almkvist-Berndt585-608.pdf (Gauss, Landen, Ramanujan, the Arithmetic-Geometric Mean, Ellipses, $\pi$, and the Ladies Diary, by Gert Almkvist and Bruce Berndt).
 A: I found the answer in  http://retro.seals.ch/digbib/view?pid=elemat-001:2000:55::180 (A Property of Euler's Elastic Curve, by  V.H. Moll,
P.A. Neill, J.L. Nowalsky and L. Solanilla) where a Euler-type proof is given that $$\int\limits_0^1\frac{dx}{\sqrt{1-x^{2n}}} \int\limits_0^1\frac{x^n\,dx}{\sqrt{1-x^{2n}}}=\frac{\pi}{2n}.$$
A simpler proof (suggested as an exercise 1.2.2 in Vladimir Tkachev's lectures "Elliptic functions: Introduction course" http://www.mai.liu.se/~vlatk48/papers/lect2-agm.pdf ) can be given by using Euler’s beta-function
$$B(\alpha,\beta)=\int\limits_0^1(1-t)^{\alpha-1}t^{\beta-1}\,dt=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}.$$
Indeed, after substitution $$x=t^\frac{1}{2n},$$ we get
$$\int\limits_0^1\frac{dx}{\sqrt{1-x^{2n}}}=\frac{1}{2n}B\left(\frac{1}{2},\frac{1}{2n}\right)=\frac{1}{2n}\,\frac{\Gamma\left(\frac{1}{2}\right)
\Gamma\left(\frac{1}{2n}\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{2n}\right)},$$
and $$\int\limits_0^1\frac{x^n\,dx}{\sqrt{1-x^{2n}}}=\frac{1}{2n}B\left(\frac{1}{2},\frac{1}{2n}+\frac{1}{2}\right)=
\frac{1}{2n}\,\frac{\Gamma\left(\frac{1}{2}\right)
\Gamma\left(\frac{1}{2n}+\frac{1}{2}\right)}{\Gamma\left(1+\frac{1}{2n}\right)}.$$
But $$\Gamma\left(1+\frac{1}{2n}\right)=\frac{1}{2n}\Gamma\left(\frac{1}{2n}\right)$$ and we get
$$\int\limits_0^1\frac{dx}{\sqrt{1-x^{2n}}} \int\limits_0^1\frac{x^n\,dx}{\sqrt{1-x^{2n}}}=\frac{\Gamma\left(\frac{1}{2}\right)^2}{2n}=\frac{\pi}{2n}.$$
