# Laplace transform of $$\displaystyle{f{{\left({t}\right)}}}={t}{e}^{{-{t}}}{\sin{{\left({2}{t}\right)}}}$$

Laplace transform of $f\left(t\right)=t{e}^{-t}\mathrm{sin}\left(2t\right)$
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maggionmoo
Here is an approach.
$\mathcal{L}\left(\mathrm{sin}2t\right)=\frac{2}{{s}^{2}+{2}^{2}}=\frac{2}{{s}^{2}+4}$ , using the table.
$\mathcal{L}\left({e}^{-t}\mathrm{sin}2t\right)=\frac{2}{{\left(s+1\right)}^{2}+4}$ , using frequency shifting.
$\mathcal{L}\left(t{e}^{-t}\mathrm{sin}2t\right)=-\frac{d}{ds}\left(\frac{2}{{\left(s+1\right)}^{2}+4}\right)=\frac{4\left(s+1\right)}{{\left({\left(s+1\right)}^{2}+4\right)}^{2}}$ , using frequency differentiation.
###### Not exactly what you’re looking for?
veselrompoikm
Let $f\left(t\right)=t{e}^{-t}\mathrm{sin}\left(2t\right)=tg\left(t\right)$ and $g\left(t\right)={e}^{-t}\mathrm{sin}\left(2t\right)={e}^{-t}h\left(t\right)$ with $h\left(t\right)=\mathrm{sin}\left(2t\right)$. So $F\left(s\right)=-{G}^{\prime }\left(s\right)$ and $G\left(s\right)=H\left(s+1\right)$ with $H\left(s\right)=\frac{2}{{s}^{2}+4}$