One possible start is the computation of the integral...
... that can be done by parts...
Kind regards
Wat the "TheEmptySet" has showed is perfectly right but i like to show the same proof in another way ...We know by theorem Laplace {{e^at} f(t)}=F(s-a)
Proof of the theorem:
We know L{f(t)}=Intergral 0-infinity ({e^-st} f(t) dt)=F(s)
Therefore Intergral {e^at f(t)} = Intergral 0-infinity ({e^-st}{e^at} f(t) dt)
=Intergral 0-infinity ({e^-(s-a)t} f(t) dt)
=F(s-a) {Reason:Since Intergral 0-infinity ({e^-st} f(t) dt)=F(s)}
Hence Laplace {{e^at} f(t)}=F(s-a)..
In the qn here in the place of f(t) we have coskt..
Therefore we know laplace {coskt}=s/(s^2-k^2) = F(s)
Therefore F(s-a)=(s-a)/{(s-a)^2-k^2}
I think ive done my best to make the script understandable..
If not im really sorry about the inconvenience..