Integral problem

**Cursed** · February 12th 2010, 01:31 PM

Given: $\int_0^{2\pi} \cos^2{6\theta} \, d\theta$
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Attempt: $\int_0^{2\pi} \cos^2{6\theta} \, d\theta = \int_0^{2\pi}\$ $\frac{1-sin^2{12\theta}}{2}\, d\theta = \int_0^{2\pi}\ \frac{1}{2}\, d\theta$ $-\frac{1}{2}\int_0^{2\pi}$ $2sin{6\theta}\, cos{6\theta}\, d\theta$ $= \frac{1}{2}( 2\pi - 2) = \pi - 1$

The answer is supposed to be just $\pi$ . Can I get any help here? Thanks.

**pickslides** · February 12th 2010, 01:46 PM

Try instead

$<br /> \int \cos^2(6\theta) ~d\theta = \int \frac{\cos (12\theta)-1}{2} ~d\theta<br />$

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