<s>[INST] Fourier sine series for $x^3$ It is asked to find the Fourier Sine Series for $x^3$ given that
$$\frac{x^2}{2} = \frac{l^2}{6} + \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} \cos\left(\frac{n \pi x}{l} \right)$$
integrating term by term. (This result was found in another exercise). As suggested, I integrated:
$$\int \frac{x^2}{2} dx  = \frac{x^3}{6} = \frac{l^2x}{6} + \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n \frac{1}{n^2}  \int  \cos\left(\frac{n \pi x}{l} \right)dx + C$$
$$  \Rightarrow \frac{x^3}{6} = \frac{l^2x}{6}+ \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n  \frac{1}{n^2} \frac{l}{n \pi} \sin\left(\frac{n \pi x}{l} \right) + C $$
$$\Rightarrow x^3 = l^2x + \frac{12l^3}{\pi^3}\sum_{n=1}^\infty (-1)^n \frac{1}{n^3} \sin\left(\frac{n \pi x}{l} \right) + C$$
It looks like that wolfram gives a different answer. I don't know if the problem is the constant $C$ or if I made something wrong. Please, follow the book approach, don't try to calculate the coefficient of Fourier sine series.
Thanks!
 [/INST] 1) You can simply conclude that $C=0$ as the equation must hold when $x=0$.
2) The answer of wolfram is different from you in appearance. The reason is that in the answer by wolfram, $x$ is also written in sin Fourier series and then combined with other terms in your equation.
</s>