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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.

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  • $\begingroup$ I am not much aware of elliptic integrals but can't you write $k^2=-(-k^2)$? You say you need a minus in the denominator so I guess this is what you are looking for. $\endgroup$ – Pranav Arora May 27 '14 at 16:51
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    $\begingroup$ It is far more interesting to write $ k^2 = - (\mathrm{ i} k)^2$. $\endgroup$ – Dmoreno May 27 '14 at 16:57
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Notice $\sin^2\phi = 1 - \cos^2\phi$ and $\cos\phi = \sin\left(\frac{\pi}{2} - \phi\right)$, you have $$\int_0^\pi\frac{d\phi}{\sqrt{1+k^2\sin^2\phi}} = 2\int_0^{\pi/2}\frac{d\phi}{\sqrt{1+k^2\sin^2\phi}} = \frac{2}{\sqrt{1+k^2}}\int_0^{\pi/2}\frac{d\phi}{\sqrt{1-\frac{k^2}{1+k^2}\cos^2\phi}} = \frac{2}{\sqrt{1+k^2}}\int_0^{\pi/2}\frac{d\phi}{\sqrt{1-\frac{k^2}{1+k^2}\sin^2\phi}} = \frac{2}{\sqrt{1+k^2}}K\left(\frac{k}{\sqrt{1+k^2}}\right) $$

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It might be of help to consider this integral as an elliptic one with a complex modulus.

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Hint. You can see $\sqrt{1+k^2\sin^2\phi}$ as the pythagorean sum of $1$ and $k\sin\phi$. That is, try drawing a right-angled triangle with angle $\phi$ and sides $1$ and $k\sin\phi$. Then, re-express your square root in terms of something easier.

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