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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?

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

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Pardon me, but What is the definition of "horizontal pivot line"? – NoChance Feb 11 '12 at 4:57
The OP is presumably referring to the $y$-coordinate of the centroid. "Why" is a question worth thinking about. The usual argument for the location of the centroid tells us that the $y$-coordinate is given by the integral. There may be a more intuitive argument. – André Nicolas Feb 11 '12 at 5:11
When finding the centre of mass of a given area under a curve, you would do so by finding the horizontal and vertical 'pivot lines'. – Korgan Rivera Feb 11 '12 at 5:11
@AndréNicolas exactly. :) – Korgan Rivera Feb 11 '12 at 5:12
Thanks for the clarification. – NoChance Feb 11 '12 at 5:22

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 $$ \sin^4 x = \frac{1}{8} \cos {4 x} \; - \; \frac{1}{2} \cos {2 x} \; + \; \frac{3}{8}$$

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Very nice use of the symmetry! – André Nicolas Feb 11 '12 at 5:53
@Andre, thanks. – Will Jagy Feb 11 '12 at 6:13
Is it obvious that the area below $\sin^2 x$ with that rectangle removed has its center of mass at $y = \frac12$? – Rahul Feb 11 '12 at 6:46
@RahulNarain, it breaks up into four identical pieces, first with $0 \leq x \leq \pi / 4,$ then rotated 180 degrees to $\pi / 4 \leq x \leq \pi / 2,$ then reflected to $\pi / 2 \leq x \leq 3 \pi / 4,$ then rotated again to $3 \pi / 4 \leq x \leq \pi.$ We do not need to find the center of mass of any of the four pieces, they are identical, and the rotation takes care of forcing the $y$-coordinate of their collective center of mass being $1/2.$ – Will Jagy Feb 11 '12 at 7:00
If you take the piece below $y = \sin^2 x$ over $0 \le x \le \frac\pi4$ and rotate it 180° about $(\frac\pi4,\frac12)$, don't you get the piece above $y = \sin^2 x$ instead? Whereas what you want is the one above $y=\frac12$ and below $y = \sin^2 x$. – Rahul Feb 11 '12 at 7:22

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.

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I find it striking that the average value of $\sin^n x$ or $\cos^n x$ over $[0, 2\pi]$ is the same as the value of $\sum_{k=0}^n\binom{-1/2}{k}\binom{-1/2}{n-k} (-1)^k$. (See this answer.) I wonder if there is a direct connection? – Mike Spivey Feb 11 '12 at 21:31
@Mike: That's very interesting! But I can't get a good mental handle on your expression. Is there a combinatorial or other natural interpretation of it or of its summands? – Rahul Feb 12 '12 at 5:29
I asked for a combinatorial proof of an equivalent sum here. In the question I proposed using lattice paths, but the answer I ended up finding uses colored permutations. I would love to see a connection between either of these and the average value of $\sin^n x$ or $\cos^n x$. I don't immediately see such a connection in either case, though. – Mike Spivey Feb 12 '12 at 14:06

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

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