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