I want to evaluate the following integral in a closed form involving elementary functions:
$$I = \displaystyle \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sqrt[3]{\frac{1}{3} + \sin^2 x}} = \frac{\sqrt[3]{3}\Gamma^3\left(\frac{1}{3}\right)}{\sqrt[3]{2^7}\pi} $$
The integral converges numerically to the solution proposed.
Here is my try:
From $ \cos^2 x + \sin^2 x = 1$
$$ I = \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sqrt[3]{\frac{1}{3} + \sin^2 x}} = \int_{0}^{\frac{\pi}{2}} \frac{dx}{\sqrt[3]{\frac{4}{3} - \cos^2 x}} = \frac{\sqrt[3]{3}}{2^{\frac{2}{3}}}\int_{0}^{\frac{\pi}{2}} \frac{dx}{\sqrt[3]{1 - \frac{3}{4}\cos^2 x}} $$
Since $\displaystyle 1> \left|\frac{3}{4}\cos^2(x) \right|$ we can expand the integrand with the generalized binomial theorem:
$$ I = \frac{\sqrt[3]{3}}{2^{\frac{2}{3}}}\int_{0}^{\frac{\pi}{2}} \sum_{j=0}^{\infty} \binom{-\frac{1}{3}}{j}(-1)^j\left(\frac{3}{4}\right)^j\cos^{2j} x dx = \frac{\sqrt[3]{3}}{2^{\frac{2}{3}}}\sum_{j=0}^{\infty} \binom{-\frac{1}{3}}{j}(-1)^j\left(\frac{3}{4}\right)^j\int_{0}^{\frac{\pi}{2}} \cos^{2j} x dx$$
The integral
$$ \int_{0}^{\frac{\pi}{2}} \cos^{2j} x dx = \frac{\sqrt{\pi}}{2} \frac{\Gamma\left(j+\frac{1}{2}\right)}{\Gamma(j+1)}$$
is a particular case of one of the integral representations of the complete beta function
Therefore
$$ \frac{\sqrt[3]{3}}{2^{\frac{2}{3}}}\sum_{j=0}^{\infty} \binom{-\frac{1}{3}}{j}(-1)^j\left(\frac{3}{4}\right)^j\int_{0}^{\frac{\pi}{2}} \cos^{2j} x dx = \frac{\sqrt[3]{3}}{2^{\frac{2}{3}}}\sum_{j=0}^{\infty} \binom{-\frac{1}{3}}{j}(-1)^j\left(\frac{3}{4}\right)^j\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(j+\frac{1}{2}\right)}{\Gamma(j+1)}$$
Using
$$ \binom{v}{m} = \frac{(v-m+1)_{m}}{m!}$$ $$ (-x)_{n} = (-1)^n(x-n+1)_{n}$$ $$ (x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}$$
where $(x)_{n}$ is the rising factorial
we have
$$ I = \frac{\sqrt[3]{3}\pi}{2^{\frac{5}{3}}}\sum_{j=0}^{\infty} \frac{\left(\frac{1}{3}\right)_{j}\left(\frac{1}{2}\right)_{j}}{(1)_{j}} \frac{\left(\frac{3}{4}\right)^j}{j!} = \frac{\sqrt[3]{3}\pi}{2^{\frac{5}{3}}} F\left(\frac{1}{3},\frac{1}{2};1;\frac{3}{4} \right)$$
where $F(a,b;c;z)$ is the Gaussian hypergeometric function.
This solution converges numerically. However, I cannot find a closed form for $\displaystyle F\left(\frac{1}{3},\frac{1}{2};1;\frac{3}{4} \right)$ with elementary functions.
I have tried the some linear and quadratic transformations for $F(a,b;c;z)$ from hereand and here, respectively; but nothing seems to work. I guess that this integral could be somewhat related to this result. or the the modular lambda function
