# Why is this integral zero for every $n \in \mathbb{Z}$? $\int_0^\pi x(\pi -x) \sin 2nx$.

The integral $$\int_0^\pi x(\pi -x) \sin 2nx$$ evaluates to zero, but the function can't be said to be even or odd. What argument, other than pure calculation (as I did) would give the value $0$ immediately? The thing multiplying the sine in the integrand is symmetric (w.r.t to reflection against $x=\pi/2$) on the interval, but that's about it I can really say. Can something more general be extrapolated from this?

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Out of curiosity...are you computing the Fourier series of that function? :-) –  Avitus Jun 24 '13 at 15:51
Yeah, it popped up in my earlier question: math.stackexchange.com/questions/428341/… –  DepeHb Jun 24 '13 at 15:58

You're on the right track. You need to think about the symmetries and antisymmetries about $x= \pi/2$. The quadratic part you've already noted, now think about the sine function. You may want to draw a few of the sine functions out if you're stuck. After this it should be clear why the integral is zero.

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Define $f(x)=x(\pi-x)\sin 2nx$.

Show that $f(\pi-x)=-f(x)$.

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Put $x=y+\frac{\pi}{2}$; the integral is equal to

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(-y^2+\frac{\pi^2}{4})\sin(2\pi y+n\pi)dy;$$

as $\sin(2\pi y+n\pi)=\sin(2\pi y)\cos(n\pi)=(-1)^{n}\sin(2\pi y)$ and $(-y^2+\frac{\pi^2}{4})(-1)^{n}\sin(2\pi y)$ is odd, then the result follows.

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