Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.1 Definition of the Laplace Transform - Problems - Page 676: 22

$$\dfrac{2}{s}+\dfrac{2s}{s^2+4b^2}$$

The Laplace Transform can be written as: $L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt $ We are given that $f(t)=4 \cos^2 (bt)$ Now, $L[F(t)]= L[4 \cos^2 (bt)] \\= 4 L[\dfrac{1+\cos 2bt}{2}-L[3] \\=2 [\dfrac{0! } {s^{0+1}}]+2[\dfrac{s}{s^2+4b^2}\\=\dfrac{2}{s}+\dfrac{2s}{s^2+4b^2}$