# What is the Laplace transform of e^(3t) * sin^2 t

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