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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 $0$?

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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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, $$\int_{-a}^{a} y(t) dt= \int_{-a}^{0} y(t) dt + \int_{0}^{a} y(t) dt = -\int_{a}^{0} y(\xi) d\xi + \int_{0}^{a} y(t) dt = 2\int_0^{a} y(t) dt.$$

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I just realized I should have searched the internet more thoroughly. How does knowing a function as even or odd help in integration ?? answers my question (after referencing @Math Lover's answer).

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