So we are letting `g(t)=t\ cos\ t`.

This time we need the 2nd derivative of `G(s)`.

`G(s)=` `Lap{t\ cos\ t}=(s^2-1)/((s^2+1)^2)`

(This is from the Table of Laplace Transforms.)

First derivative:

`d/(ds)(s^2-1)/((s^2+1)^2)=-2s(s^2-3)/((s^2+1)^3)`

Second derivative:

`(d^2)/(ds^2)(s^2-1)/((s^2+1)^2)=6(s^4-6s^2+1)/((s^2+1)^4)`

Now `(-1)^2=1`.

So

`Lap{t^3\ cos\ t}=6(s^4-6s^2+1)/((s^2+1)^4)`

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