Finding power series representation of $ \int_0^{\frac{\pi }{2}} \frac{1}{\sqrt {1 - k^2\sin^2{x}}}\;{dx}$ 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. 

In the very first step, $(2n$ choose $n)$ should be $(-1/2$ choose $n)$ because the exponent is $1/2$ and you're apparently using a standard Taylor expansion for $(1+X)^M$ for $M=-1/2$ and $X=-k^2\sin^2 x$. You should also add $(-1)^n$. 
The combinatorial factor $(-1/2$ choose $n)$ instantly produces the product of odd numbers over the product of even numbers. It is equal to $(-1/2)(-3/2)\dots (-n+1/2) / n!$. Absorbing $(-1)^n$, you get $(1/2)(3/2)\dots (n-1/2)/n!$.
Oh, your description is actually equivalent.
Indeed, you will get another copy of $(1/2)(3/2)\dots (n-1/2)/n!$, together with the $\pi/2$ factor, from the integral $\int_0^{\pi/2} dx\,\sin^2 x$, so your result is correct, and the original formula is only missing the second power of the big ratio and needs to be corrected.
