# use properties of the Laplace transform and the table of Laplace transforms to determine L[f] f(t)=int_0^t (t-w)cos(2w)dw

use properties of the Laplace transform and the table of Laplace transforms to determine L[f] $f\left(t\right)={\int }_{0}^{t}\left(t-w\right)\mathrm{cos}\left(2w\right)dw$
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Step 1
We have given,
$f\left(t\right)={\int }_{0}^{t}\left(t-w\right)\mathrm{cos}\left(2w\right)dw$
This can be written as,
$f\left(t\right)={\int }_{0}^{t}tcos\left(2w\right)dw-{\int }_{0}^{t}w\mathrm{cos}\left(2w\right)dw$
We have to find,
$L\left\{f\left(t\right)\right\}=L\left\{{\int }_{0}^{t}t\mathrm{cos}\left(2w\right)dw\right\}-L\left\{{\int }_{0}^{t}w\mathrm{cos}\left(2w\right)dw\right\}$
Step 2
We know that,
$L\left\{f\left(t\right)\right\}=F\left(s\right)$
$L\left\{{\int }_{0}^{t}f\left(u\right)du\right\}=\frac{1}{s}F\left(s\right)$
So $L\left\{{\int }_{0}^{t}t\mathrm{cos}\left(2w\right)dw\right\}=\frac{t}{s}\left(\frac{s}{{s}^{2}+4}\right)$
$L\left\{{\int }_{0}^{t}t\mathrm{cos}\left(2w\right)dw\right\}=\frac{t}{{s}^{2}+4}$
And
$L\left\{{\int }_{0}^{t}w\mathrm{cos}\left(2w\right)dw\right\}=\frac{1}{s}\left(-\frac{-{s}^{2}+4}{\left({s}^{2}+4{\right)}^{2}}\right)$
On plugging these values in our equation , we get,
$L\left\{f\left(t\right)\right\}=L\left\{{\int }_{0}^{t}t\mathrm{cos}\left(2w\right)dw\right\}-L\left\{{\int }_{0}^{t}w\mathrm{cos}\left(2w\right)dw\right\}$
$L\left\{f\left(t\right)\right\}=\left(\frac{t}{{s}^{2}+4}\right)+\frac{1}{s}\left(\frac{-{s}^{2}+4}{\left({s}^{2}+4{\right)}^{2}}\right)$