2022-03-26

Find Laplace transform of L[te
t
sin4t] is

nick1337

To find the Laplace transform of $ℒ\left\{\mathrm{sin}\left(4t\right)\right\}$, we will use the definition of the Laplace transform:
$ℒ\left\{f\left(t\right)\right\}=F\left(s\right)={\int }_{0}^{\infty }{e}^{-st}f\left(t\right)dt$
Substituting $f\left(t\right)=\mathrm{sin}\left(4t\right)$ into the formula, we get:
$F\left(s\right)={\int }_{0}^{\infty }{e}^{-st}\mathrm{sin}\left(4t\right)dt$
To solve this integral, we will use integration by parts:
$\int udv=uv-\int vdu$
Letting $u=\mathrm{sin}\left(4t\right)$ and $dv={e}^{-st}dt$, we get:
$du=4\mathrm{cos}\left(4t\right)dt\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}v=-\frac{1}{s}{e}^{-st}$
Substituting into the formula, we get:
$F\left(s\right)={\left[-\frac{1}{s}\mathrm{sin}\left(4t\right){e}^{-st}\right]}_{0}^{\infty }+\frac{4}{s}{\int }_{0}^{\infty }\mathrm{cos}\left(4t\right){e}^{-st}dt$
Evaluating the first term using the limits of integration, we get:
${\left[-\frac{1}{s}\mathrm{sin}\left(4t\right){e}^{-st}\right]}_{0}^{\infty }=\frac{1}{s}{lim}_{t\to \infty }\mathrm{sin}\left(4t\right){e}^{-st}-\frac{1}{s}\mathrm{sin}\left(0\right){e}^{-s0}=0-0=0$
Simplifying the second term, we get:
$\frac{4}{s}{\int }_{0}^{\infty }\mathrm{cos}\left(4t\right){e}^{-st}dt=\frac{4}{s}\frac{s}{{s}^{2}+{4}^{2}}=\frac{4}{{s}^{2}+{4}^{2}}$
Therefore, the Laplace transform of $\mathrm{sin}\left(4t\right)$ is:
$ℒ\left\{\mathrm{sin}\left(4t\right)\right\}=F\left(s\right)=\frac{4}{{s}^{2}+{4}^{2}}$

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