I found the cosine series of $x^2/2$ to be (by first finding Fourier sine series for $x$ on $(0,l)$ and then integrating that term by term)
$${x^2 \over 2}={l^2 \over 6}+{2 l^2 \over \pi^2}[\sum{(-1)^n \over n^2}\cos({n\pi x \over l})]$$
But my question is I have to find the sine series of $x^3$
using the above. I integrated $x^2/2$ to get
$${x^3 \over 6}={l^2 \over 6}x+{2 l^3 \over \pi^3}[-\sin({\pi x \over l})+{1 \over 2^3}\sin({2\pi x \over l})-{1 \over 3^3}\sin({3\pi x \over l})+\cdots+C]$$ where $C$ is the constant of integration.
What happens to the constant of integration. What's the value of it? And how to find it. (In cosine, the constant could be found by finding the coefficient for the $n=0$ situation) Should $C=-{l^2 \over 6}x$? But I think $C=0$ but then how can I put the series in the Fourier Sine series to be of the form $\sum A_n \sin({n\pi x\over l})$.
$C$ is a constant and cannot ever be equal to $-\frac{\ell^2}{6}x$ if $x$ is the variable being integrated by. The point of the constant of integration is that when you differentiate it, it is equal to 0 (this is why it must be introduced). Additionally, you should have the $C$ value on both sides, so that you can cancel them, and in effect $C=0$.
To see this last point, if it were not the case that we should have the same $C$ value on both sides, consider the following: $1=1$, so that by indefinitely integrating both sides we get $$x+C_1=x+C_2$$ Now, we can cancel $x$ on both sides to get $C_1=C_2$.
(Note, we had to already pick a distinguished antiderivative to do the above, but the idea is that if we choose the same distinguished antiderivative for both sides, then $C_1$ better equal $C_2$.)
As far as dealing with the $\frac{-\ell^2}{6}x$ term, you can write its Fourier cosine series, and add them component-wise.