Question

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

Answer 1

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


`\mathcal I=\displaystyle\int_0^\infty e^(-st)\cos2t\,\mathrm dt`

and setting
`u=e^(-st)` and
`\mathrm dv=\cos2t\,\mathrm dt`, it follows that


`\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`

`\mathcal I=\frac1{4s}-\frac1{4s^2}\mathcal I`

`(4s^2+1)/(4s^2)\mathcal I=\frac1{4s}`

`\mathcal I=\frac s{s^2+4}`