TITLE: Interesting closed form for $\int_0^{\frac{\pi}{2}}\frac{1}{\left(\frac{1}{3}+\sin^2{\theta}\right)^{\frac{1}{3}}}\;d\theta$
QUESTION [25 upvotes]: Some time ago I used a formal approach to derive the following identity:

$$\int_0^{\frac{\pi}{2}}\frac{1}{\left(\frac{1}{3}+\sin^2{\theta}\right)^{\frac{1}{3}}}\;d\theta=\frac{3^{\frac{1}{12}}\pi\sqrt{2}}{AGM(1+\sqrt{3},\sqrt{8})}\tag{1}$$

where $AGM$ is the arithmetic-geometric mean. Wolfram Alpha does not tell me whether this is correct, but it does appear to be accurate to many decimal places. I have three questions:

Can anyone verify whether $(1)$ is in fact correct?
Is there a way of generalizing $(1)$ to integrals of the form $\int_0^{\frac{\pi}{2}}\left(a+\sin^2{\theta}\right)^{-\frac{1}{3}}\;d\theta$ or is this integral more special? My derivation (see below) appears to only work for $a=\frac{1}{3}$.
There is a superficial similarity between $(1)$ and elliptic integrals (e.g. the $AGM$ evaluation); is there a way to transform this integral into an elliptic integral that I have missed, or is it merely a coincidence that an integral of this form is the reciprocal of an $AGM$?


Derivation: I have put this here in case it helps to see where I am coming from; I apologize for its length. I began by using a multiple integration trick of squaring the integral and converting to polar coordinates to evaluate $\int_0^\infty e^{-x^6}dx=\frac{1}{6}\Gamma(\frac{1}{6})$ as follows:
$$\left[\int_0^\infty e^{-x^6}\;dx\right]^2=\int_0^\infty\int_0^{\frac{\pi}{2}}re^{-r^6(\cos^6\theta\;+\;\sin^6\theta)}\;d\theta\;dx={\int_0^\infty re^{-r^6}\int_0^{\frac{\pi}{2}}e^{3r^6\cos^2\theta\sin^2\theta}\;d\theta\;dx}$$
$$=\int_0^\infty re^{-r^6}\int_0^{\frac{\pi}{2}}e^{\frac{3r^6}{4}\sin^22\theta}\;d\theta\;dx={\int_0^\infty re^{-r^6}\int_0^{\frac{\pi}{2}}e^{\frac{3r^6}{4}\cos^2\theta}\;d\theta\;dx}$$
I then made use of the following formula (see here):
$$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}x^n=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}e^{4x\cos^2\theta}\;d\theta\tag{2}$$
Using $(2)$ and formally interchanging integration and summation we get:
$$\frac{\Gamma(\frac{1}{6})^2}{36}=\int_0^\infty re^{-r^6}\int_0^{\frac{\pi}{2}}e^{4\left(\frac{3r^6}{16}\right)\cos^2\theta}\;d\theta\;dx=\frac{\pi}{2}\int_0^\infty re^{-r^6}\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}\left(\frac{3r^6}{16}\right)^n\;dx$$
$$=\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}\left(\frac{3}{16}\right)^n \int_0^\infty r^{6n+1}e^{-r^6}\;dx=\frac{\pi}{12}\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}\left(\frac{3}{16}\right)^n \Gamma\left(n+\frac{1}{3}\right)$$
I then used Laplace transform identities and $(2)$, freely interchanging integrals and sums, to write:
$$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}\frac{\Gamma\left(n+\frac{1}{3}\right)}{s^{n+\frac{1}{3}}}=L\left[\sum_{n=0}^\infty \frac{(2n)!}{(n!)^3}t^{n-\frac{2}{3}}\right](s)={\frac{2}{\pi}L\left[t^{-\frac{2}{3}}\int_0^\frac{\pi}{2}e^{4t\cos^2\theta}\;d\theta\right](s)}={\frac{2}{\pi}\int_0^\frac{\pi}{2}L\left[t^{-\frac{2}{3}}e^{4t\cos^2\theta}\right](s)\;d\theta}={\frac{2}{\pi}\int_0^\frac{\pi}{2}\frac{\Gamma(\frac{1}{3})}{(s-4\cos^2\theta)^{\frac{1}{3}}}\;d\theta}$$
Accordingly, since $\frac{4}{3}-\cos^2\theta=\frac{1}{3}+\sin^2{\theta}$ we can deduce that:
$$\frac{\Gamma(\frac{1}{6})^2}{36}=\frac{\Gamma(\frac{1}{3})}{6}\left(\frac{4}{3}\right)^\frac{1}{3}\int_0^\frac{\pi}{2}\frac{1}{(\frac{1}{3}+\sin^2\theta)^{\frac{1}{3}}}\;d\theta$$
Reflection and duplication give $\Gamma(\frac{1}{6})=2^{-\frac{1}{3}}\sqrt{\frac{3}{\pi}}\Gamma(\frac{1}{3})^2$ and hence we have the following identity:

$$\int_0^{\frac{\pi}{2}}\frac{1}{\left(\frac{1}{3}+\sin^2{\theta}\right)^{\frac{1}{3}}}\;d\theta=\frac{3^\frac{1}{3}\Gamma(\frac{1}{3})^3}{2^\frac{7}{3}\pi}\tag{3}$$

while $(1)$ may be obtained by using the following identity (see here):
$$\Gamma\left(\frac{1}{6}\right)=\frac{2^\frac{14}{9}3^\frac{1}{3}\pi^\frac{5}{6}}{AGM(1+\sqrt{3},\sqrt{8})^\frac{2}{3}}$$
This completes the derivation; I cannot see how a method like this (especially with the conversion to polar coordinates) could be used to give results more general than $(1)$ and $(3)$.

REPLY [6 votes]: (Too long for a comment.)
Expanding on a concluding remark by Nemo, there is an interesting connection between the OP's,
$$\int_0^{\pi/2}\frac1{\sqrt[3]{\color{blue}{\alpha}+\sin^2 x}}dx=\frac{\pi}{2\,\sqrt[3]{\alpha+1}}\,_2F_1\Big(\frac12,\frac13;1;\,\frac1{\color{blue}{\alpha}+1}\Big)\tag1$$
and the integral involved in this post,
$$\int_0^{\infty}\frac1{\sqrt[3]{\color{blue}{1+2\alpha}+\cosh x}}dx=2^{2/3}\,3^{1/4}K(k_3)\;_2F_1\Big(\frac13,\frac13;\frac56;-\color{blue}{\alpha}\Big)\tag2$$
Equations $(1),(2)$ are valid for arbitrary $\alpha$. But if it is chosen such that,
$$\color{blue}{\alpha} =\alpha(\tau)= \frac1{4\sqrt{27}}\big(\lambda^3-\sqrt{27}\lambda^{-3}\big)^2$$
where,
$$\lambda =\frac{\eta\big(\frac{\tau+1}3\big)}{\eta(\tau)},\quad \tau=\frac{1+N\sqrt{-3}}2$$
then there is a beautifully simple relationship between $(1),(2)$ as,

$$\frac{N+1}{2^{1/3}\,3^{1/2}}\,\int_0^{\pi/2}\frac1{\sqrt[3]{\color{blue}{\alpha}+\sin^2 x}}dx=\int_0^{\infty}\frac1{\sqrt[3]{\color{blue}{1+2\alpha}+\cosh x}}dx\tag3$$

The closed-form for the RHS (via its hypergeometric equivalent) is already given in this answer, so that automatically leads to the form for the LHS as well. Some rational values are,
$$\alpha\big(\tfrac{1+3\sqrt{-3}}2\big)=\large\tfrac13$$
$$\alpha\big(\tfrac{1+5\sqrt{-3}}2\big)=4$$
$$\alpha\big(\tfrac{1+7\sqrt{-3}}2\big)=27$$
which will be algebraic for integer $N>1$. The first one was re-discovered by the OP and lead to the nice evaluations,
$$\int_0^{\pi/2}\frac1{\sqrt[3]{\tfrac13+\sin^2 x}}dx=3^{1/12}\,K(k_3)$$
$$\int_0^{\pi/2}\frac1{\sqrt[3]{4+\sin^2 x}}dx=\frac{3^{3/4}}{5^{5/6}}\,K(k_3)$$
$$\int_0^{\pi/2}\frac1{\sqrt[3]{27+\sin^2 x}}dx=\frac{3^{3/4}}7\,K(k_3)$$