TITLE: What's With The Integral $\int\sqrt{\cos(2\theta)}\, \mathrm d\theta$?
QUESTION [5 upvotes]: 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?

REPLY [7 votes]: 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.