The question asks us to find a Fourier series for an odd extension of function $y=1/4x^4-1/3x^3$ in (1,-1).
Attempted solution:
Using the integral formulas: $$\int x^3 sin (ax) dx = \frac{3(a^2 x^2-2)sin(ax)-ax(a^2 x^2-6)cos(ax)}{a^4}$$ and $$\int x^4 sin(ax)dx=\frac{4ax(a^2x^2-6)}{a^5}sin(ax)-\frac{(a^4x^4-12a^2x^2+24)}{a^5}cos(ax)$$
Let $$\frac{1}{4}x^4-\frac{1}{3}x^3=\sum_{n=1}^{\infty}A_m sin(m\pi x)$$ so $$A_m=2\int_0^1 \frac{1}{4}x^4 sin(m\pi x) dx - 2\int_0^1 \frac{1}{3}x^3sin(m\pi x)dx$$
$$\begin{align}
\frac{1}{2}\int_0^1 x^4 sin(m\pi x) dx &= \frac{1}{2} \left[ \frac{4m\pi x(m^2\pi^2 x^2 - 6)sin(m\pi x)}{m^5 \pi^5}-\frac{m^4\pi^4 x^4 - 12m^2 \pi^2 x^2+24}{m^5\pi^5}cos(m\pi x) \right]^1_0 \\
&= \begin{cases} \frac{-1}{2} \left[ \frac{-m^4\pi^4 + 12m^2\pi^2-48}{m^5\pi^5} \right] & \text{if $m$ is odd} \\
\frac{-1}{2} \left[ \frac{m^4\pi^4-12m^2\pi^2}{m^5 \pi^5} \right] & \text{if $m$ is even} \end{cases} \end{align}$$
$$\begin{align} \frac{-2}{3} \int_0^1 x^3 sin(m\pi x)dx
&= \frac{-2}{3} \left[ \frac{3(m^2\pi^2x^2-2)sin(m\pi x) - m\pi x (m^2\pi^2x^2-6)cos(m\pi x)}{m^4\pi^4} \right]_0^1
&= \begin{cases} \frac{-2}{3} \left[ \frac{m\pi(m^2\pi^2-6)}{m^4\pi^4} \right] & \text{if $m$ is odd} \\ \frac{2}{3} \left[ \frac{m\pi (m^2 \pi^2-6)}{m^4\pi^4} \right] & \text{if $m$ is even} \end{cases} \end{align}$$
Hence for m=odd,
$$A_m=\frac{-1}{6m\pi}-\frac{2}{m^3\pi^3}+\frac{24}{m^5\pi^5}$$
For m=even,
$$A_m=\frac{1}{6m\pi}+\frac{2}{m^4\pi^3}$$
Adding them up I get
$$\frac{1}{4}x^4 -\frac{1}{3}x^3 = \sum_{n=1}^{\infty} (-1)^n \left( \frac{1}{6n\pi} + \frac{2}{n^3\pi^3} \right) sin(n\pi x) + \sum_{n=1}^{\infty} \frac{24}{(2n-1)^5\pi^5} sin[(2n-1)\pi x] $$
However the question asks the answer to be in the form of $$\frac{1}{4}x^4 - \frac{1}{3}x^3 = \sum_{n=1}^{\infty} A_n sin \left( \frac{(2n+1)\pi x}{2} \right) $$
I honestly don't know how to proceed from here... Any help would be much appreciated!