Use integration to find which function has Fourier series $\frac{4}{\pi} \sum\limits^{\infty}_{n = 1, n\text{ odd}} \frac{\sin(nt)}{n^3}$ Problem
It can be shown that the function 
$$
f(t) = \begin{cases}
\dfrac{\pi}{2} + t & , -\pi < t < 0,\\
\dfrac{\pi}{2} - t & , 0 < t < \pi,
\end{cases}
$$
has Fourier series $FS_f(t) = \dfrac{4}{\pi} \sum^{\infty}_{n = 1, odd} \dfrac{\cos(nt)}{n^2}$.
Use integration to find which function has Fourier series $\dfrac{4}{\pi} \sum^{\infty}_{n = 1, odd} \dfrac{\sin(nt)}{n^3}$.

I understand that term-by-term integration of a $2\pi$-periodic Fourier series is the formula $\int^t_{-\pi} f(\alpha) \ d\alpha = \sum_{n = 1}^{\infty} \dfrac{a_n}{n}\sin(nt) - \sum_{n = 1}^{\infty} \dfrac{b_n}{n}(\cos(nt) - \cos(n\pi))$. However, I do not understand how to solve this problem.
I would greatly appreciate it if people could please take the time to explain the reasoning involved in solving this problem.

Using the property that the function evaluated at $t=0$ is the average of the left and right piece-wise function evaluated at zero and 
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 \, n + 1)^3} = \frac{\pi^3}{32}$$
then the following is obtained:
Integrating
$$
f_{c}(t) = \begin{cases}
\frac{\pi}{2} + t & , -\pi < t < 0,\\
\frac{\pi}{2} - t & , 0 < t < \pi,
\end{cases}
$$
becomes
$$
f_{s}(t) = \begin{cases}
c_{0} + \frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\
c_{1} + \frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi.
\end{cases}
$$
Now, using $f_{s}(0) = (f_{s}(0-) + f_{s}(0+))/2$, then $c_{1} = - c_{0}$ and
$$
f_{s}(t) = \begin{cases}
c_{0} + \frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\
-c_{0} + \frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi.
\end{cases}
$$
In order to determine $c_{0}$ evaluate the series at the point $t = \pi/2$ leading to
$$f_{s}\left(\frac{\pi}{2}\right) = - c_{0} + \frac{\pi^2}{4} - \frac{\pi^2}{8} =  - c_{0} + \frac{\pi^2}{8} $$
From the series it is determined that
$$f_{s}(t) = \frac{4}{\pi} \, \sum_{n=0}^{\infty} \frac{\sin(2\,n +1)t}{(2\, n +1)^3}$$
and
$$f_{s}\left(\frac{\pi}{2}\right) = \frac{\pi^2}{8}.$$
This yields $c_{0}=0$ and 
$$
f_{s}(t) = \begin{cases}
\frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\
\frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi.
\end{cases}
$$
One can verify this by using
\begin{align}
f_{s}(t) &= \sum_{n=0}^{\infty} B_{n} \, \sin((2\, n +1) \, t) \\
B_{n} &= \frac{1}{\pi} \, \int_{- \pi}^{\pi} g(t) \, \sin((2 \, n +1) t) \, dt \\
g(t) &= \begin{cases}
\frac{\pi \, t}{2} + \frac{t^2}{2} & , -\pi < t < 0,\\
\frac{\pi \, t}{2} - \frac{t^2}{2} & , 0 < t < \pi.
\end{cases}
\end{align}
