Elliptic integral $\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk$ Question:
Prove that 

$$\int^1_0 \frac{K(k)}{\sqrt{1-k^2}}\,dk=\frac{1}{16\pi}\Gamma^4\left(
 \frac{1}{4}\right)$$

My attempt 
Start by the transformation 
$$k \to \frac{2\sqrt{k}}{1+k}$$
Hence we have
$$\int^{1}_0 K\left(\frac{2\sqrt{k}}{1+k}\right)\,\frac{1}{\sqrt{k}(1+k)}dk$$
Now we use that
$$K(k)=\frac{1}{k+1}K\left( \frac{2\sqrt{k}}{1+k} \right)$$
So we have
$$\int^1_0 \frac{K(k)}{\sqrt{k}}\,dk=2\int^1_0 K(k^2)\,dk$$
[1] I have no idea how to solve the last integral? 
[2] Should I use another approach to solve the integral ?

By definition we have 
$$K(k) = \int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2\,x^2}}\,$$
 A: A simple method for the last integral:
\begin{align}
\int \limits_0^1 \frac{\operatorname{K}(k)}{\sqrt{k}} \, \mathrm{d} k &= \int \limits_0^1 \int \limits_0^{\pi/2} \frac{\mathrm{d} \phi}{\sqrt{k(1-k^2 \sin^2(\phi))}} \, \mathrm{d} k \stackrel{\text{Tonelli}}{=} \int \limits_0^{\pi/2} \int \limits_0^1 \frac{\mathrm{d} k}{\sqrt{k(1-k^2 \sin^2(\phi))}} \, \mathrm{d} \phi \\
&\!\!\!\!\!\!\stackrel{k = \frac{\sin(\theta)}{\sin(\phi)}}{=} \int \limits_0^{\pi/2} \int \limits_0^\phi \frac{\mathrm{d} \theta \, \mathrm{d} \phi}{\sqrt{\sin(\theta)\sin(\phi)}} = \frac{1}{2} \int \limits_0^{\pi/2} \int \limits_0^{\pi/2} \frac{\mathrm{d} \theta \, \mathrm{d} \phi}{\sqrt{\sin(\theta)\sin(\phi)}} = \frac{1}{2} \left[\int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sqrt{\sin(t)}}\right]^2 \\
&= \frac{1}{8} \operatorname{B}^2\left(\frac{1}{4},\frac{1}{2}\right) = \frac{1}{8} \left(\frac{\operatorname{\Gamma}\left(\frac{1}{4}\right) \operatorname{\Gamma}\left(\frac{1}{2}\right)}{\operatorname{\Gamma}\left(\frac{3}{4}\right)}\right)^2 \stackrel{\Gamma\text{-reflection}}{=} \frac{1}{8} \left(\frac{\operatorname{\Gamma}^2\left(\frac{1}{4}\right) \operatorname{\Gamma}\left(\frac{1}{2}\right) \sin\left(\frac{\pi}{4}\right)}{\pi}\right)^2 \\
&= \frac{\operatorname{\Gamma}^4\left(\frac{1}{4}\right)}{16 \pi} \, .
\end{align}
A: Consider the integral
\begin{align}
I = \int_{0}^{1} \frac{K(x)}{\sqrt{1-x^2}} \ dx
\end{align}
where $K(x)$ is the complete elliptic integral of the first kind. It can be shown that the hypergeometric form is
\begin{align}
K(x) = \frac{\pi}{2} \ {}_{2}F_{1}(\frac{1}{2}, \frac{1}{2}; 1; x^{2}).
\end{align}
By using the series form the integral becomes
\begin{align}
I &= \frac{\pi}{2} \ \sum_{r=0}^{\infty} \frac{(1/2)_{r} (1/2)_{r}}{r! (1)_{r}} \ \int_{0}^{1} x^{2r} (1-x^{2})^{-1/2} \ dx.
\end{align}
By making the substitution $x = \sqrt{t}$ the integral can be cast into Beta function form and yields
\begin{align}
I &= \frac{\pi}{4} \ \sum_{r=0}^{\infty} \frac{(1/2)_{r} (1/2)_{r}}{r! (1)_{r}} \ B(r+1/2, 1/2) \\
&= \left( \frac{\pi}{2} \right)^{2} \ {}_{3}F_{2}(a,a,a; 1,1; 1)
\end{align}
where $a = 1/2$. Now, by using the identity 
\begin{align}
{}_{3}F_{2}(a,a,a; 1,1; 1) = \frac{\Gamma\left(1 - \frac{3a}{2}\right) \ \cos\left( \frac{a \pi}{2} \right)}{\Gamma^{3}\left( 1 - \frac{a}{2}\right)}
\end{align}
which is valid for $Re(a) < 2/3$, the integral value is seen to be
\begin{align}
I = \left( \frac{\pi}{2} \right)^{2} \ \frac{\Gamma\left(\frac{1}{4}\right) \ \cos\left( \frac{\pi}{4} \right)}{\Gamma^{3}\left(\frac{3}{4}\right)}.
\end{align}
Now using the relation
\begin{align}
\Gamma\left(\frac{3}{4}\right) = \frac{\sqrt{2} \pi}{\Gamma\left(\frac{1}{4}\right)} 
\end{align}
the final value is obtained, namely,
\begin{align}
I = \frac{1}{16 \pi} \Gamma^{4}\left(\frac{1}{4}\right).
\end{align}
Hence 
\begin{align}
\int_{0}^{1} \frac{K(x)}{\sqrt{1-x^2}} \ dx = \frac{1}{16 \pi} \Gamma^{4}\left(\frac{1}{4}\right).
\end{align}
