## Area Under a Sine Arch

2017 Mar 9

I noticed something funny the other day.

Consider a sine graph with equation $$y=c*\sin(cx)$$, where $$c$$ is a constant. The period of such a sine wave is $$\frac{2\pi}{c}$$, so the area under a single one of its arches is
$A=\int_0^{\frac{\pi}{c}} c\sin cx \,dx$

Which is
$A=-\cos \pi + \cos 0$
$A=2$

That struck me as a little odd. Regardless of the value of $$c$$, the area underneath a single arch was always $$2$$? But I suppose it makes sense, because as $$c$$ increases, each arch is stretched vertically but compressed horizontally. However, it gets weirder. If $$y=f'(x)\sin f(x)$$, the area is still $$2$$, because the x-coordinates of the start and endpoints of the nth arch are $$f^{-1}(\pi n-\pi)$$ and $$f^{-1}(\pi n)$$ for integral $$n$$, the area under the nth arch is
$A=\int_{f^{-1}(\pi n-\pi)}^{f^{-1}(\pi n)} c\sin cx \,dx$

Which simplifies to
$-\cos f(f^{-1}(\pi n))+\cos f(f^{-1}(\pi n-\pi))$
$-\cos \pi n+\cos (\pi n-\pi)$

We are interested in only positive areas, and the positive difference between $$-\cos \pi n$$ and $$\cos (\pi n-\pi)$$ is the difference between two consecutive peaks of the cosine arch, which is also always $$2$$.

That means that all of the areas underneath all of the arches in the following pictures is equal to $$2$$: