The horizontal pivot line of $\sin^2 x$ here is exactly $\frac{3}{8}$. Why?

Question

I noticed that the horizontal pivot line (or $y$-coordinate of the centroid) under the curve $y=\sin^2 x$ between $0$ and $\pi$ is exactly $\frac{3}{8}$. There may be no reason for me to find this strange, but it's just so neat. Does anyone know why this is?

$$\frac{1}{2}\frac{\int_0^{\pi} (\sin^4 x) dx}{\int_0^{\pi} (\sin^2x) dx} = \frac{3}{8}.$$

Answer 1

I had a mistaken argument before. It occurs to me to wonder how you know the value $3/8.$ The simplest way to find the two integrals is by using the identities $$ \sin^2 x = -\frac{1}{2} \cos {2 x} \; + \; \frac{1}{2} $$ and $$ \sin^4 x = \frac{1}{8} \cos {4 x} \; - \; \frac{1}{2} \cos {2 x} \; + \; \frac{3}{8}$$

Answer 2

I suppose a general explanation for this phenomenon might be that the average value of $\sin^n\,x$ or $\cos^n\,x$ over $[0,2\pi]$ is pretty simple; it's just $\frac1{2^n}\binom{n}{n/2}$ if $n$ is even, and $0$ if $n$ is odd. This is clear if you write $\cos^n\,x = \left(\frac12(e^{ix}+e^{-ix})\right)^n$ and apply the binomial theorem; after integration over a period, only the constant term will remain.

A corollary is that if you take any polynomial over $\cos x$ and $\sin x$ whose coefficients are rational numbers, its average value over $[0, 2\pi]$ will be a rational number.

Answer 3

To determine the value of this quotient you don't even need to compute the integrals: $$\int_0^\pi\sin^4 x\ dx= - \sin^3 x\ \cos x\Bigr|_0^\pi + \int_0^\pi 3\sin^2 x\ \cos^2 x\ dx = 3\int_0^\pi \sin^2 x\ dx- 3\int_0^\pi \sin^4 x\ dx\ . $$