Answer
$\dfrac{1}{2(s-2)}+\dfrac{s-2}{2[(s-2)^2+4]}$
Work Step by Step
The Laplace transform of function $t^3$ is given as:
$L(e^{2t}/2)=\dfrac{1}{2(s-2)}$ and $L( \dfrac{\cos 2t}{2})=\dfrac{s}{2(s^2+4)}$
The first shifting Theorem for $a= 2$ can be expressed as:
$L[e^{2t} (\cos \dfrac{1+\cos 2t}{2})]=\dfrac{1}{2(s-2)}+\dfrac{s-2}{2[(s-2)^2+4]}$
