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$$ \int_0^{2\pi}{ \sqrt{ 1 - \sin{ \theta } \sin{ 2\theta } + \cos\theta \cos{2\theta} } %sqrt \; d\theta } %int $$

I tried removing the $2\theta$ terms, choosing the identity $ \cos{2\theta} = 1 - 2 \sin^2{\theta} $, but this results in the unsavory dish:

$$ \int_0^{2\pi}{ \sqrt{ 1 - 4\sin^2{ \theta } \cos{ \theta } + \cos{\theta} } %sqrt \; d\theta } %int $$

What is a reasonable next step, then? Is removing $2\theta$ terms a good strategy for this problem?

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The first step could be to make the integral $\int_0^{2\pi} \sqrt{ 1 - \cos 3\theta} d\theta$ using the identity $\cos A \cos B - \sin A \sin B = \cos (A+B)$. – Srivatsan Jan 28 '12 at 17:23
The second step could be to note that $1-\cos(3\theta)=2\sin^2(3\theta/2)$ using the identity $\cos(2A)=1-2\sin^2(A)$, and the third step could be to note that $|\sin(3\theta/2)|$ has period $2\pi/3$. – Did Jan 28 '12 at 17:27

First notice that the trig terms in the integral are of the form $\cos A \cos B - \sin A \sin B$, which by a well-known identity is equal to $\cos (A+B)$; namely, $\cos \theta \cos 2\theta - \sin \theta \sin 2\theta = \cos 3\theta$, and so you can write your integral as

$$ \int_0^{2\pi} \sqrt{1-\cos 3\theta}\, d \theta $$

Next you can use the fact that $\cos 2A = 1 - 2\sin^2 A$, i.e. $1 - \cos 2A = 2\sin^2 A$, to get that $1 - \cos 3\theta = 2 \sin^2 \frac{3\theta}{2}$. Plugging this in gives

$$\int_0^{2\pi} \sqrt{2 \sin^2 \frac{3\theta}{2}}\, d\theta \\ = \int_0^{2\pi} \sqrt{2} \left| \sin \frac{3\theta}{2} \right|\, d\theta$$

Then you can split the integral up into where $\sin \frac{3\theta}{2}$ is positive and where it's negative, and evaluation is elementary.

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