My goal is to find the integral below using Laplace transform, which is supposed to simplify it. However, I am stuck and don't know how to write the full solution. $$I=\int_{0}^{x} (x - y)^2 \cos(3y) dy$$ Here's my attempt at some ideas:
I've been suggested to rewrite $(x-y)^2$ as $y^2$ (due to convulation?) and proceed to do Laplace transform. According to the formula table I get $\mathcal{L}\{y^2\}=\frac{2}{s^{3}}$. Similarly, $\mathcal{L}\{\cos(3y)\}=\frac{s}{s^{2}+9}$. Could someone please help me arrive at the full solution?