Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.1 Definition of the Laplace Transform - Problems - Page 675: 16

$$\dfrac{s}{(s^2-b^2) }$$

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