# Why does the integral of $\sin x$ over the interval $[0,\,2\pi]$ not go to 0 in Fourier Analysis? [closed]

Square Wave

Formula for ak, bk

As the square wave function is odd, I know that there will be no terms $a_k$.

So what I'm left with finding is $b_k$.

As $x(t)$ will equal $3$ or $-3$, all I really have to integrate is $\sin x$ over the interval of $[0,\, 2\pi]$. My question is why can I take the integral as equal to double the positive area instead of it just equaling $0$?

## closed as unclear what you're asking by Leucippus, Claude Leibovici, Ben, José Carlos Santos, J. M. is a poor mathematicianAug 28 '17 at 5:19

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The product of an odd function, $x(t)$, with another odd function, $\sin(w t)$, is an even function. In particular, if $y(t)=x(t) \sin(w t)$ then $y(-t)=x(-t) \sin(-w t) = (-x(t))(-\sin(w t))= y(t)$. Consequently, $$\int_{-a}^{a} y(t) dt= \int_{-a}^{0} y(t) dt + \int_{0}^{a} y(t) dt = -\int_{a}^{0} y(\xi) d\xi + \int_{0}^{a} y(t) dt = 2\int_0^{a} y(t) dt.$$