Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.9 The Convolution Integral - Problems - Page 716: 24

See below

Using the Convolution Theorem $L^{-1}[F(s)* G(s)]=L^{-1}[\frac{s+4}{s^2+8s+25}.\frac{se^{-\frac{\pi}{2}}}{s^2+16}]\\ =L^{-1}[\frac{s+4}{s^2+8s+25}].L^{-1}[\frac{se^{-\frac{\pi}{2}}}{s^2+16}]\\ =e^{-4t}\cos 3t * \cos 4tu_{\frac{\pi}{2}} (t)\\ =\int^t_\frac{\pi}{2} e^{-4(t-x)}\cos 3(t-x) \cos 4xdx$