$\int_0^{\pi /2} \frac{d\phi}{(1+\frac{1}{2}\sin^2 \phi)^{1/2}}$ in terms of gamma function Using $u=\frac{1}{2}\sin^2 \phi$
$du=\sin \phi \cos \phi d\phi $
$du=(2u)^{1/2}\sqrt{1-2u}$
So we are left with the integral
$\int_0^{1/2} \frac{(2u)^{-1/2}(1-2u)^{-1/2}du}{(1+u)^{1/2}}$
Using now $2u=x$
$du=\frac{1}{2}dx$
We are left with
$\frac{1}{\sqrt{2}}\int_0^1\frac{dx}{x^{1/2}(1-x)^{1/2}(2+x)^{1/2}}$
Which is pretty close to be in the form of a beta function, but the factor $(2+x)^{1/2}$ is annoying me. What can be done?
PS: I also tried $\phi = \sin^{-1} x$ but ended up with the same problem.
 A: Using $\sin^{2}\phi=1-\cos^{2}\phi$ we see that the integral is equal to $$\sqrt{\frac{2}{3}}\int_{0}^{\pi/2}\frac{d\phi}{\sqrt{1-(1/3)\cos^{2}\phi}}=\sqrt{\frac{2}{3}}K\left(\frac{1}{\sqrt{3}}\right)$$ The modulus $1/\sqrt{3}$ does not appear to be one of singular moduli and hence it is difficult to say if the corresponding elliptic integral can be expressed in terms of Gamma function. Why do you think it has a closed form in terms of Gamma function?
A: Using
$$
\binom{-1/2}{k}=(-1)^k\frac{\binom{2k}{k}}{4^k}
$$
and
$$
\frac{\Gamma\left(k+\frac12\right)\Gamma\left(\frac12\right)}{\Gamma(k+1)}=\pi\frac{\binom{2k}{k}}{4^k}
$$
we get
$$
\begin{align}
\int_0^{\pi/2}\frac{\mathrm{d}\phi}{\left(1+\frac12\sin^2(\phi)\right)^{1/2}}
&=\int_0^{\pi/2}\frac{\mathrm{d}\sin(\phi)}{\cos(\phi)\left(1+\frac12\sin^2(\phi)\right)^{1/2}}\\
&=\int_0^1\frac{\mathrm{d}t}{\sqrt{1-t^2}\left(1+\frac12t^2\right)^{1/2}}\\
&=\frac12\int_0^1\frac{\mathrm{d}t}{\sqrt{t\,(1-t)\left(1+\frac12t\right)}}\\
&=\frac12\int_0^1\sum_{k=0}^\infty\binom{-1/2}{k}\left(\frac t2\right)^k\frac{\mathrm{d}t}{\sqrt{t\,(1-t)}}\\
&=\frac12\int_0^1\sum_{k=0}^\infty(-1)^k\binom{2k}{k}\left(\frac t8\right)^k\frac{\mathrm{d}t}{\sqrt{t\,(1-t)}}\\
&=\frac12\sum_{k=0}^\infty(-1)^k\binom{2k}{k}\frac{\Gamma\left(k+\frac12\right)\Gamma\left(\frac12\right)}{8^k\Gamma(k+1)}\\
&=\frac\pi2\sum_{k=0}^\infty\left(-\frac1{32}\right)^k\binom{2k}{k}^2\\
\end{align}
$$
Adding $100$ terms gives over $30$ places:
$$
\frac\pi2\sum_{k=0}^\infty\left(-\frac1{32}\right)^k\binom{2k}{k}^2
=1.41573720842595619889216596542
$$
Mathematica says this is $\operatorname{EllipticK}\left(-\frac12\right)$.
