# Find the laplace transform of L\left\{f(x)\right\} being f(x)=\int_0^3 \cos (3t)dt

Find the laplace transform of $L\left\{f\left(x\right)\right\}$ being $f\left(x\right)={\int }_{0}^{3}\mathrm{cos}\left(3t\right)dt$

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Neelam Wainwright

Find $L\left\{f\left(x\right)\right\}$ where $f\left(x\right)={\int }_{0}^{3}\mathrm{cos}\left(3t\right)dt$
Given that $f\left(x\right)={\int }_{0}^{3}\mathrm{cos}\left(3t\right)dt$ , so we first calculate this integral.
$f\left(x\right)={\int }_{0}^{3}\mathrm{cos}\left(3t\right)dt={\left[\frac{\mathrm{sin}\left(3t\right)}{3}\right]}_{0}^{3}=\frac{1}{3}\left[\mathrm{sin}9-\mathrm{sin}0\right]$
$⇒f\left(x\right)=\frac{\mathrm{sin}9}{3}$
$⇒f\left(x\right)$ is a constant function
Now $F\left(s\right)=L\left\{f\left(x\right)\right\}={\int }_{0}^{\mathrm{\infty }}f\left(x\right){e}^{-sx}dx$
$={\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{sin}9}{3}{e}^{-sx}dx=\frac{\mathrm{sin}9}{3}{\int }_{0}^{\mathrm{\infty }}{e}^{-sx}dx$
$=\frac{\mathrm{sin}9}{3}{\left[\frac{{e}^{-sx}}{-s}\right]}_{0}^{\mathrm{\infty }}=\frac{-\mathrm{sin}9}{3s}\left[{e}^{-\mathrm{\infty }}-{e}^{0}\right]$
$⇒F\left(s\right)=L\left\{f\left(x\right)\right\}=\frac{1}{3s}\mathrm{sin}9$