# Use Laplace transform to evaluate the integral int_0^infty te^{-2t} sin(2t)dt a) frac{1}{B} b) not defined c) frac{4s}{(s^2+4)^2} d) frac{4(s+2)}{(s^2+4s+8)^2}

Use Laplace transform to evaluate the integral
${\int }_{0}^{\mathrm{\infty }}t{e}^{-2t}\mathrm{sin}\left(2t\right)dt$
a) $\frac{1}{B}$
b) not defined
c) $\frac{4s}{\left({s}^{2}+4{\right)}^{2}}$
d) $\frac{4\left(s+2\right)}{\left({s}^{2}+4s+8{\right)}^{2}}$
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Step 1
Here, the objective is to evaluate the given integral using Laplace transformation
${\int }_{0}^{\mathrm{\infty }}t{e}^{-2t}\mathrm{sin}\left(2t\right)dt$
Step 2
According to the definition of Laplace transformation
$F\left(s\right)=L\left(f\left(t\right)\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)ds$
Compare the given integral with (1)
$f\left(t\right)=t\mathrm{sin}2t$
To evaluate the given integral, find the Laplace transformation of f(t)
$F\left(s\right)=L\left(f\left(t\right)\right)$
$F\left(s\right)=L\left(t\mathrm{sin}2t\right)$
Since $L\left(\mathrm{sin}\left(at\right)\right)=\frac{a}{\left({s}^{2}+{a}^{2}\right)}$
Use
Therefore
$F\left(s\right)=L\left(t\mathrm{sin}2t\right)=-\frac{d}{ds}\cdot \left(\frac{2}{\left({s}^{2}+{2}^{2}\right)}\right)$

$F\left(s\right)=\frac{4s}{\left({s}^{2}+4{\right)}^{2}}$
Step 3
Hence
${\int }_{0}^{\mathrm{\infty }}t{e}^{-st}\mathrm{sin}2tdt=\frac{4s}{\left({s}^{2}+4{\right)}^{2}}$