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

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  • $\begingroup$ What can be said about $w$, $\theta$ and $T$? $\endgroup$ Commented Oct 31, 2012 at 14:42

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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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  • $\begingroup$ @PeterPhipps, thanks! $\endgroup$ Commented Oct 31, 2012 at 15:11
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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 $\,w,T,\theta\,$

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