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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?

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I guess since $\int \cos 2 \theta \ d \theta$ is easy compute...it would be a reasonable assumption that $\int \sqrt{\cos 2 \theta} \ d \theta$ is easy to compute. –  PEV Feb 1 '11 at 0:04
    
Did you try the Wikipedia article? This integral might appear if you were interested in computing the circumference of a certain ellipse. –  Qiaochu Yuan Feb 1 '11 at 0:04
    
@Qiaochu Yuan: I checked out the article and if one were to integrate from 0 to $\frac{\pi}{2}$ one would get the circumference of the ellipse of eccentricity $\sqrt{2}$. This is not one of the special cases given in the article. –  Joe Johnson 126 Feb 1 '11 at 0:18
    
@Joe: the indefinite integral (the incomplete as opposed to the complete elliptic integral) measures part, rather than all, of the circumference. –  Qiaochu Yuan Feb 1 '11 at 0:22
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"A student randomly asked me to compute" , what was happening at the time? randomly asking someone to compute an integral that turns out to be elliptic is quite remarkable! What other questions does this student ask? –  Arjang Apr 30 '11 at 4:36

1 Answer 1

up vote 7 down vote accepted

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.

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