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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?

  • 4
    The first step could be to make the integral $\int_0^{2\pi} \sqrt{ 1 - \cos 3\theta} d\theta$ using the identity $\cos A \cos B - \sin A \sin B = \cos (A+B)$.2012-01-28
  • 4
    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 $2\pi/3$.2012-01-28

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