TITLE: The horizontal pivot line of $\sin^2 x$ here is exactly $\frac{3}{8}$. Why?
QUESTION [6 upvotes]: 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?
$$\frac{1}{2}\frac{\int_0^{\pi} (\sin^4 x) dx}{\int_0^{\pi} (\sin^2x) dx} = \frac{3}{8}.$$

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