# Find the laplace transform of this function. f(t)=e^{-2t}\sin 3t \sin t

Find the laplace transform of this function
$f\left(t\right)={e}^{-2t}\mathrm{sin}3t\mathrm{sin}t$
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krolaniaN

$f\left(t\right)={e}^{-2t}\mathrm{sin}3t\mathrm{sin}t$
$f\left(t\right)={e}^{-2t}\mathrm{sin}3t\mathrm{sin}t$
$f\left(t\right)={e}^{-2t}\mathrm{sin}3t\mathrm{sin}t$
$=\frac{{e}^{-2t}}{2}\left[\mathrm{cos}\left(3t-t\right)-\mathrm{cos}\left(3t+t\right)\right]$
$=\frac{{e}^{-2t}}{2}\left[\mathrm{cos}2t-\mathrm{cos}4t\right]$
$\left[\mathrm{sin}\alpha \mathrm{sin}\beta =\frac{1}{2}\left[\mathrm{cos}\left(\alpha -\beta \right)-\mathrm{cos}\left(\alpha +\beta \right)\right]\right]$
$L\left\{f\left(t\right):s\right\}=L\left\{\frac{{e}^{-2t}}{2}\left[\mathrm{cos}2t-\mathrm{cos}4t\right]:s\right\}$
$=\frac{1}{2}L\left\{\left(\mathrm{cos}2t-\mathrm{cos}4t\right):s+2\right\}$
$\left[L\left\{{e}^{-at}f\left(t\right):s\right\}=L\left\{f\left(t\right):s+a\right\}\right]$
$=\frac{1}{2}\left[L\left\{\mathrm{cos}2t:s+2\right\}-L\left\{\mathrm{cos}4t:s+2\right\}\right]$
$L\left[{e}^{ax}:s\right]={\int }_{0}^{\mathrm{\infty }}{e}^{ax}{e}^{-sx}dx$
$={\int }_{0}^{\mathrm{\infty }}{e}^{-\left(s-a\right)x}dx$
$={\left[\frac{{e}^{-\left(s-a\right)x}}{-\left(s-a\right)}\right]}_{0}^{\mathrm{\infty }}$

$\left[\because \underset{x\to \mathrm{\infty }}{lim}{e}^{-\left(s-a\right)x}=0\right]$
so $L\left\{{e}^{i\omega x}:s\right\}=\frac{1}{s-i\omega }$
$=\frac{s+i\omega }{\left(s-i\omega \right)\left(s+i\omega \right)}$
$=\frac{s+i\omega }{{s}^{2}+{\omega }^{2}}$