wolfram alpha - Evaluation of complete elliptic integral of second kind

I am currently working to try and solve the integral:

$$\int_{0}^{\frac{\pi}{2}} \sqrt{1-a^2\cos(x)^2} = \sqrt{1-a^2} \; E\left ( 1+\frac{1}{-1+a^2} \right )$$

where $E(m)$ is the complete elliptic integral of second kind with the parameter $m=k^2$ and $0.

This is the soultion I found using WolframAlpha

In this case $m$ is always smaller then $0$, yet $m$ is supposed to be the square of some $k$.

The implementation I found in NumericalRecipies seems to only accept $k$ as argument, evident through a test, plotting values of the positive argument-branch and comparing them with WolframAlphas elliptic function.

However, the NR implementation doesn't use any sort of complex variables (therefor, won't accept them as input).

Do I need another implementation, am I completely off with this.

I am very confused and uncertain about this.

Help would be greatly appreciated, Thanks.

Answer

I found the solution to my problem here:

http://analyticphysics.com/Mathematical%20Methods/A%20Miscellany%20of%20Elliptic%20Integrals.htm

To quote:

$$E(-m)=\sqrt{m+1} E \left (\frac{m+1}{m} \right )$$

Simply building an if clause into my driver to choose $-k$ whenever $k<0$ and then use the formula indicated above gave me the result WolfrAmalpha shows in wolframalpha.com/input/?i=plot+EllipticE%5Bx%5D.

The discussion here was what got me to ask the right question. Thanks Jack D'Aurizio.