What is the Laplace transform of $e^{3t} \cdot \sin^2 t$

Question

I am trying to compute the Laplace transform of the following functions $e^{3t} \cdot \sin^2 t$

Please help me to find a solution

Answer 1

Converted my comments into an answer:

The procedure I would use is to find the first find the Laplace Transform of $\sin^2(t)$ using the identity $\sin^2(t)\equiv\dfrac{1-\cos(2t)}{2}$ and then apply the first shifting theorem, which states that: $$\mathcal{L}\{e^{at}f(t)\}=F(s-a) \tag{1}$$ 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.

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Letting $f(t)=\sin^2(t)$, we obtain for $F(s)$: $$F(s)=\mathcal{L}\{\sin^2(t)\}=\frac{1}{2}\mathcal{L}\{1\}-\frac{1}{2}\mathcal{L}\{\cos(2t)\}=\frac{1}{2s}-\frac{s}{2(s^2+4)}=\frac{2}{s(s^2+4)}$$ Thus, it follows from the first shifting theorem that: $$\mathcal{L}\{e^{3t}f(t)\}=F(s-3)=\frac{2}{(s-3)((s-3)^2+4)}=\frac{2}{(s-3)(s^2-6s+13)}$$