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Use the definition of the Laplace transform for cos(2t) => [tex] \int\limits^a_b { e^{-st}cos(2t) } \, dx[/tex] where a is ∞ and b is 0

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LammettHash
This is really just an exercise in integrating by parts. If

[tex]\mathcal I=\displaystyle\int_0^\infty e^{-st}\cos2t\,\mathrm dt[/tex]

and setting [tex]u=e^{-st}[/tex] and [tex]\mathrm dv=\cos2t\,\mathrm dt[/tex], it follows that

[tex]\mathcal I=\displaystyle\left(\frac12e^{-st}\sin2t\right)\bigg|_{t=0}^{t\to\infty}-\frac1{4s}\left(e^{-st}\cos2t\right)\bigg|_{t=0}^{t\to\infty}-\frac1{4s^2}\mathcal I[/tex]
[tex]\mathcal I=\dfrac1{4s}-\dfrac1{4s^2}\mathcal I[/tex]
[tex]\dfrac{4s^2+1}{4s^2}\mathcal I=\dfrac1{4s}[/tex]
[tex]\mathcal I=\dfrac s{s^2+4}[/tex]

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