Fourier-series odd vs even square wave?
For a trigonometric Fourier series:

$$
\tilde{x}(t) = a_0 + \sum\limits_{n=1}^{\infty} a_n \cos(n \omega_0 t) + \sum\limits_{n=1}^{\infty} b_n \sin(n \omega_0 t)
$$
with
$$
a_0 = \frac{1}{T} \int_{-T/2}^{T/2}  x(t) \ dt
$$

$$
a_n = \frac{2}{T} \int_{-T/2}^{T/2}  x(t) \cos (n \omega_0 t) \ dt, \quad n = 1, 2, \ldots
$$

$$
b_n = \frac{2}{T} \int_{-T/2}^{T/2}  x(t) \sin (n \omega_0 t) \ dt, \quad n = 1, 2, \ldots
$$
and $ \omega_0 = \frac{2\pi}{T} $ .

the Fourier series of an odd square wave will be a sum of only sine terms, and the Fourier series of an even square wave will be a sum of only cosine terms. 

This is because the product of an odd and an even function is an odd function, which when integrated over a symmetric interval (i.e. from $-T/2$ to $T/2$), will return zero. For an odd square wave, this means that all the $a_n$ will be zero, and for an even square wave, all the $b_n$ will be zero. 

You can save yourself time and potential mistakes by exploiting this fact and only computing the coefficients that are nonzero.