Converted my comments into an answer: The procedure I would use is to find the first find the Laplace Transform of ${\mathrm{sin}}^{2}(t)$ using the identity ${\mathrm{sin}}^{2}(t)\equiv {\displaystyle \frac{1-\mathrm{cos}(2t)}{2}}$ and then apply the first shifting theorem, which states that: $$\begin{array}{}\text{(1)}& \mathcal{L}\{{e}^{at}f(t)\}=F(s-a)\end{array}$$ Where $F(s)=\mathcal{L}\{f(t)\}$. It is easy to prove the above using the definition of the Laplace Transform, which I leave as an exercise. Letting $f(t)={\mathrm{sin}}^{2}(t)$, we obtain for F(s): $$F(s)=\mathcal{L}\{{\mathrm{sin}}^{2}(t)\}=\frac{1}{2}\mathcal{L}\{1\}-\frac{1}{2}\mathcal{L}\{\mathrm{cos}(2t)\}=\frac{1}{2s}-\frac{s}{2({s}^{2}+4)}=\frac{2}{s({s}^{2}+4)}$$ Thus, it follows from the first shifting theorem that: $$\mathcal{L}\{{e}^{3t}f(t)\}=F(s-3)=\frac{2}{(s-3)((s-3{)}^{2}+4)}=\frac{2}{(s-3)({s}^{2}-6s+13)}$$