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

$\dfrac{1}{s} (1-2e^{-2s})$

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, $L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt \\= \int_{0}^{2} e^{-st} f(t) \ dt+\int_{2}^{\infty} e^{-st} f(t) \ dt \\=[\dfrac{-1}{se^{st}}]_{0}^{2} -\dfrac{1}{se^{2s}} \\=\dfrac{1}{s} (1-2e^{-2s})$