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]
