integration - TaylorSeries of complete elliptic integral of the first kind

I want compute $K(k)$ as a Taylor Series; $k\in\mathbb{R}$ and $\vert k \vert < 1$. Can someone help me? $$K(k):= \int^{\frac{\pi}{2}}_0 \dfrac{dt}{\sqrt{1-k^2 sin^2t}}$$ Results so far: $$K(k):= \int^{\frac{\pi}{2}}_0 \dfrac{dt}{\sqrt{1-k^2 sin^2t}} = \int^{\frac{\pi}{2}}_0 (1-k^2 sin^2t)^{-\frac{1}{2}}dt$$ With using binomial Series we get $$\int^{\frac{\pi}{2}}_{0} \sum^\infty_{\Phi=0} {-\frac{1}{2} \choose \Phi}(-k^{2\Phi}){\sin^{2\Phi}{t}} \ dt = \sum^\infty_{\Phi=0} {-\frac{1}{2} \choose \Phi}(-k^{2\Phi})\int^{\frac{\pi}{2}}_{0}\sin^{2\Phi}t \ dt$$ For $\Phi$ even: $$\int^{\frac{\pi}{2}}_{0}\sin^{2\Phi}t \ dt = \frac{\pi}{2}\frac{1}{2}\frac{3}{4}\frac{5}{6}...\frac{n-1}{n} = S$$ thus we get: $$1. \sum^\infty_{\Phi=0} {-\frac{1}{2} \choose \Phi}(-k^{2\Phi})\cdot S$$ now i need some help to compute 1. as taylor series, can someone help?

Thanks! Landau.

Answer

The generating function for central binomial coefficient is given by mathworld site.

$$\begin{align*} \int_{0}^{\frac \pi 2} \frac{1}{\sqrt{1 - 4 \left( \frac {k^2} 4 \sin^2 (t) \right )}} dt &= \int_0^{\frac \pi 2 } \sum_{n=0}^{\infty} \binom{2n}{n} \left(\frac {k^2} 4 \sin^2 (t) \right )^n dt \\ &= \sum_{n=0}^{\infty} \binom{2n}{n} \frac{k^{2n}}{4^n} \cdot \frac 1 2 \cdot \beta (n+1/2, 1/2) \\ &= \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2 2^{2n} }\cdot \frac 1 2 \left( \frac{\Gamma(n+1/2)\Gamma(1/2)}{\Gamma(n+1)} \right ) k^{2n}\\ &= \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2 2^{2n} }\cdot \frac 1 2 \left( \frac{(2n)! \sqrt{\pi} \sqrt{\pi}}{(n!)^2} \right ) k^{2n} \\ &= \frac{ \pi }{2 } \sum_{n=1}^\infty \left( \frac{(2n)!}{(n!)^{2} 2^{2n}} \right )^2 k^{2n}\\ &= \frac{\pi}{2} \sum_{n=0}^\infty P_{2n}(0)k^{2n} \end{align*}$$

Where $P_{n}(0)$ is Legendre polynomial. Seems that wolf gives the sum of the right side as EllipticK[k^2]. Also it is given here on wiki.