# Find Laplace of given fn: f(t)=int_0^t sin(t-tau)costau d tau

Find Laplace of given fn:
$f\left(t\right)={\int }_{0}^{t}\mathrm{sin}\left(t-\tau \right)\mathrm{cos}\tau d\tau$
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Step 1
Given that $f\left(t\right)={\int }_{0}^{t}\mathrm{sin}\left(t-\tau \right)\mathrm{cos}\tau d\tau$
The objective is to find the Laplace transform of f(t).
$\mathrm{sin}\left(t-\tau \right)=\mathrm{sin}t\mathrm{cos}\tau -\mathrm{cos}t\mathrm{sin}t$
So,
$\mathrm{sin}\left(t-\tau \right)\mathrm{cos}\tau =\mathrm{sin}t{\mathrm{cos}}^{2}\tau -\mathrm{cos}t\mathrm{sin}\tau \mathrm{cos}\tau$
$=\mathrm{sin}t\left(\frac{1+\mathrm{cos}2\tau }{2}\right)-\mathrm{cos}t\left(\frac{\mathrm{sin}2\tau }{2}\right)$

$f\left(t\right)=\frac{\mathrm{sin}t}{2}{\int }_{0}^{t}\left[1+\mathrm{cos}2\tau \right]d\tau -\frac{\mathrm{cos}t}{2}{\int }_{0}^{t}\mathrm{sin}2\tau d\tau$
$=\frac{\mathrm{sin}t}{2}\left[t+\frac{\mathrm{sin}2t}{2}\right]-\frac{\mathrm{cos}t}{2}{\left[-\frac{\mathrm{cos}2t}{2}\right]}_{0}^{t}$
$=\frac{t\mathrm{sin}t}{2}+\frac{\mathrm{sin}2t\mathrm{sin}t}{4}-\frac{\mathrm{cos}t}{2}+\frac{\mathrm{cos}t\mathrm{cos}2t}{4}$
$=\frac{t\mathrm{sin}t}{2}-\frac{\mathrm{cos}t}{2}+\frac{1}{4}\mathrm{cos}\left(t\right)$
$=\frac{t\mathrm{sin}t}{2}-\frac{\mathrm{cos}t}{4}$
Step 2
Now $f\left(t\right)=\frac{t\mathrm{sin}t}{2}-\frac{\mathrm{cos}t}{4}$
So
$L\left(f\left(t\right)\right)=L\left(\frac{t\mathrm{sin}t}{2}-\frac{\mathrm{cos}t}{4}\right)$
$=L\left(\frac{t\mathrm{sin}t}{2}\right)+L\left(-\frac{\mathrm{cos}t}{4}\right)$
$=\frac{1}{2}L\left(t\mathrm{sin}t\right)-\frac{1}{4}L\left(\mathrm{cos}t\right)$
It is known that laplace transform of $\mathrm{cos}t$ is $\frac{s}{\left({s}^{2}+1\right)}$
Now let's find the $L\left(t\mathrm{sin}t\right),$
$L\left[t\mathrm{sin}t\right]=\frac{d}{\left(ds\right)}\left[L\left(\mathrm{sin}t\right)\right]$
$=\frac{d}{\left(ds\right)}\left[\frac{1}{{s}^{2}+1}\right]$
$=\frac{-2s}{\left({s}^{2}+1{\right)}^{2}}$
Hence the laplace transform for $f\left(t\right)={\int }_{0}^{t}\mathrm{sin}\left(t-\tau \right)\mathrm{cos}\tau d\tau$ is $L\left(f\left(t\right)\right)=\frac{-s}{\left({s}^{2}+1{\right)}^{2}}-\frac{s}{4\left({s}^{2}+1\right)}$