We will use: `ccL{t*g(t)}=-G^'(s)=-d/(ds)G(s)`

Let `g(t) = e^(-t)\ cos\ 4t`

Then

`{:(G(s),=ccL{e^(-t)cos\ 4t}),(,=(s+1)/((s+1)^2+16)),(,=(s+1)/(s^2+2s+17)):}`

Now

`d/(ds)(s+1)/((s+1)^2+16)=-(s^2+2s-15)/((s^2+2s+17)^2)`

So `ccL{t*e^(-t)*cos\ 4t}=(s^2+2s-15)/((s^2+2s+17)^2)`