We think of this as

`G(s)=1/s((omega_0)/(s^2+(omega_0)^2))`

We use rule (3) from above:

If `Lap^{:-1:}{G(s)}=g(t)`, then `Lap^{:-1:}{(G(s))/s}=int_0^tg(t)dt`

Now, `Lap^{:-1:}{(omega_0)/(s^2+(omega_0)^2)}=sin omega_0t`

So

`Lap^{:-1:}{(omega_0)/(s(s^2+omega_0^2))}`

`=int_0^tsin omega_0t\ dt`

`=[-1/w_0\ cos omega_0t]_0^t`

`=-1/omega_0[cos omega_0t-1]`

`=1/omega_0(1-cos omega_0t)`