Laplace transform convolution

Question

$x(t) = cos(3πt)$
h(t) = $\exp(-2t)u(t)$

Find y(t) = x(t) * h(t) (ie convolution)

Y(s) = X(s)H(s) and then take inverse laplace tranform of Y(s)

$ L(x(t)) = \frac{s}{s^2+9π^2} $
$ L(h(t)) = \frac{1}{s+2} $

I then try to find the partial fractions but it looks more complicated than it should be..

Answer 1

If you grind it out: $$X(s)H(s) = \frac{2}{4 + 9\pi^2}\frac{s}{s^2 + 9\pi^2} + \frac{9}{4 + 9\pi^2}\frac{1}{s^2 + 9\pi^2} + \frac{-2}{(4 + 9\pi^2)}\frac{1}{s+2}$$

Worth double checking!