TITLE: How to integrate $\int_0^\pi \frac{1}{\sqrt{1+k^2\sin^2\phi}} d \phi$?
QUESTION [5 upvotes]: I am currently dealing with the integral
$$\int_{0}^{\large\pi}\frac{{\rm d}\phi}
{\,\sqrt{\vphantom{\Large A}\,1 + k^{2}\sin^{2} \phi \,}\,}
$$
I know that if I had a minus sign in the denominator, then this would be similar to an elliptic integral, but in this case, I don't really know what this is.

REPLY [12 votes]: Notice $\sin^2\phi = 1 - \cos^2\phi$ and $\cos\phi = \sin\left(\frac{\pi}{2} - \phi\right)$, you have
$$\int_0^\pi\frac{d\phi}{\sqrt{1+k^2\sin^2\phi}}
= 2\int_0^{\pi/2}\frac{d\phi}{\sqrt{1+k^2\sin^2\phi}}
= \frac{2}{\sqrt{1+k^2}}\int_0^{\pi/2}\frac{d\phi}{\sqrt{1-\frac{k^2}{1+k^2}\cos^2\phi}}
= \frac{2}{\sqrt{1+k^2}}\int_0^{\pi/2}\frac{d\phi}{\sqrt{1-\frac{k^2}{1+k^2}\sin^2\phi}}
= \frac{2}{\sqrt{1+k^2}}K\left(\frac{k}{\sqrt{1+k^2}}\right)
$$