I think my calc class didn't learn that yet or wont learn it. Is this part of complex numbers unit? Cause my calc department took that away for our final exam since we had snow days on the east coast.
I was able to find the same answer by just distributing it and removing the high degrees
Since you just want the first four nonzero terms, I would do it with a chart and multiply one term of the sine expansion by every term of $e^{-x}$, then line them up by powers of $x$:
$\begin{matrix}\text{multiplication} & \vert & 1 & x & x^2 & x^3 & x^4 & x^5 & x^6 \\ \hline xe^{-x} & \vert & & x & -x^2 & +\dfrac{x^3}{2!} & -\dfrac{x^4}{3!} & +\dfrac{x^5}{4!} & -\dfrac{x^6}{5!} \\ -\dfrac{x^3}{3!}e^{-x} & \vert & & & & -\dfrac{x^3}{3!} & +\dfrac{x^4}{3!} & -\dfrac{x^5}{3!2!} & +\dfrac{x^6}{3!3!} \\ \dfrac{x^5}{5!}e^{-x} & \vert & & & & & & +\dfrac{x^5}{5!} & -\dfrac{x^6}{5!} \\ \hline \text{Total} & \vert & & x & -x^2 & +\dfrac{x^3}{3} & & -\dfrac{x^5}{30} & +\dfrac{x^6}{90}\end{matrix}$
Then you just take the first four terms. There was no need to do find the multiplication of $-\dfrac{x^7}{7!}$ unless I was going out to at least the $x^7$ term.