$$f(t)=2\int_0^t f(u)\cos (2(t-u))~du+3\sin(2t),t\geq 0$$
I find it hard to transform the first term, $f(u)\cos(2(t-u))$.
I think I should use this formula,
$$\cos at = \frac{s}{s^2 + a^2}$$
to get this solution as below,
$$F(s)=2F(s)\frac{s}{s^2+4}+3\frac{2}{s^2+4}$$
But I still don't understand how they transformed $f(u)\cos(2(t-u))$ to $F(s)\cdot \dfrac{s}{s^2 + 4}$
How did they manage to transform it? What happens to "$t-u$"?
I am aware of the Dirac Delta function, but this is not $U(t-u)$, its $cos(t-u)$, and it confuses me.
Hint:
Remember that if $f(t)$ and $g(t)$ are piecewise continuous functions on $[0,\infty)$, then the convolution integral of $f(t)$ and $g(t)$, which we denote by $(f\ast g)(t)$ is: $$(f\ast g)(t)=\int_0^t f(t-\tau)g(\tau) d\tau \tag{1}$$ Also, recall the following properties: $$(f\ast g)(t)=(g\ast f)(t) \tag{2}$$ You can prove $(2)$ straight from the definition: Use the change of variable $u=t-\tau$. $$\mathscr{L}\{(f\ast g)(t)\}=F(s)G(s) \tag{3}$$ Where $\mathscr{L}\{f(t)\}=F(s)$ and $\mathscr{L}\{g(t)\}=G(s)$.