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How is $\int_0^T\cos(2\omega t+ 2\theta) dt = 0$ Regards

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    What can be said about $w$, $\theta$ and $T$?2012-10-31

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$\frac{d}{dx}\sin(mx + c) = m cos(mx + c)$

so backwards we have

$\int \cos(2wt + 2 \theta) dt = \frac{1}{2w} \sin(2wt + 2 \theta) + c $

and you can evaluate that at the limits.

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    @PeterPhipps, thanks!2012-10-31
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Following Sperners' solution:

$\int_0^T\cos (2wt+2\theta)\,dt=\left.\frac{1}{2w}\sin (2wt+2\theta)\right|_0^T=\frac{1}{2w}\left[\sin\left(2wT+2\theta\right)-\sin \left(2\theta\right)\right]$

and the above is zero iff

$\sin(2wT+2\theta)=\sin 2\theta\Longleftrightarrow 2wT+2\theta=2\theta\,\,\vee\,\,2wT+2\theta=\pi-2\theta$

So the claim is false unless some relations or given values apply to $\,w,T,\theta\,$