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A student randomly asked me to compute

$\int\sqrt{\cos(2\theta)}\, \mathrm d\theta.$

I was unable to do so, as were several other instructors. I typed the integral into Wolfram and it says that it is an elliptic integral of type 2. I am not familiar with elliptic integrals and I don't have any idea why this student asked me to do this since we are not even doing derivatives yet. Is this integral important? Does it have a closed form answer?

  • 0
    Stupid question: If $\cos{2\theta}$ can be negative, how is the square root (thus the indefinite integral) evaluated?2011-04-30

1 Answers 1

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The elliptic integral of the second kind crops up since

$\int\sqrt{\cos\;2\theta}\;\mathrm d\theta=\int\sqrt{1-2\sin^2\theta}\;\mathrm d\theta=E(\theta|2)+C$

where $E(\phi|m)=\int_0^\phi \sqrt{1-m\sin^2\theta}\;\mathrm d\theta$ is the incomplete elliptic integral of the second kind.

All well and good, but traditionally one transforms elliptic integrals such that the parameter $m$ is within the interval $[0,1]$ (corresponding to the fact that ellipses have their eccentricity in the interval $(0,1)$). Using certain transformations, one now has

$E(\theta|2)=\sqrt{2}E\left(\arcsin(\sqrt{2}\sin\;\theta)|\frac12\right)-\frac1{\sqrt{2}}F\left(\arcsin(\sqrt{2}\sin\;\theta)|\frac12\right)$

where $F(\phi|m)=\int_0^\phi\frac{\mathrm d\theta}{\sqrt{1-m\sin^2\theta}}$ is the incomplete elliptic integral of the first kind, which is probably what you want.