I want to show that $\displaystyle \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt {1 - k^2\sin^2{x}}}\;{dx} = \frac{\pi}{2}\sum_{n \ge 0}k^{2n}\left({\frac{{1 \cdot 3 \cdots \left( {2n - 1} \right)}} {{2 \cdot 4 \cdots \cdot 2n}}} \right)$, where $ -1 < k < 1$.
Here is what I did:
$\displaystyle \begin{aligned}\int_0^{\frac{\pi }{2}} \frac{1}{{\sqrt {1 - k^2\sin^2{x}}}}\;{dx} & = \int_{0}^{\pi/2}\sum_{n \ge 0} \frac{k^{2n}}{2^{2n}}\binom{2n}{n}\sin^{2n}{x}\;{dx} \\& = \sum_{n \ge 0}\int_{0}^{\pi/2} \frac{k^{2n}}{2^{2n}}\binom{2n}{n}\sin^{2n}{x}\;{dx} \\& = \frac{\pi}{2} \sum_{n \ge 0} ~ k^{2n} \bigg(\frac{1}{2^{2n}}\binom{2n}{n}\prod_{1 \le r \le n}\frac{2r-1}{2r} \bigg) \\& = \frac{\pi}{2} \sum_{n \ge 0} ~ k^{2n} \bigg(\prod_{1 \le r \le n}\frac{2r-1}{2r} \cdot \prod_{1 \le r \le n}\frac{2r-1}{2r} \bigg) \\& = \frac{\pi}{2}\sum_{n \ge 0}k^{2n}\bigg(\prod_{1 \le r \le n}\frac{2r-1}{2r}\bigg)^2.\end{aligned}$
However, there no power on the coefficients in the given series, so they obviously don't match, and I couldn't whatsoever discern a mistake in my calculations. Thanks in advance.