Laplace transform of convolution with no function of t

Question

Instructions: Evaluate the given Laplace transform. Do not evaluate the integral before transforming.
Problem Given: $\mathscr{L}\{\int_0^t e^{-\tau} cos\tau d\tau \}$
My Problem: To treat this as a convolution, I think I need to rewrite one of the two functions of $\tau$ as a function of $t-\tau$. My intuition is to use the following trig identity, but this could make things nasty come test time (i.e. tomorrow):

$cos{t-\tau} = {costcos\tau - sintsin\tau}$

$cos\tau = \frac{cos{(t-\tau)} - sintsin\tau}{cost} $

Do I just need to tough it out going this route, or am I approaching this poorly?

Answer 1

Use the fact that $$\mathscr{L} \left( \int_{0}^{t} e^{-\tau} \cos \tau d \tau \right) = \frac{ \mathscr{L} (e^{-\tau} \cos \tau ) }{s} = \frac{\frac{s+1}{(s+1)^2+1}}{s} = \frac{1}{s}\left (\frac{s+1}{(s+1)^2+1} \right).$$

In general, we have $$\mathscr{L} \left( \int_{0}^{t} f(\tau) d\tau \right) = \frac{\mathscr{L} ( f(\tau))}{s} .$$