Suppose we have $f(t) = 1$ and $g(t) = 100\cos(20t)$.
Find $p(t) = f(t)*g(t)$.
Solution
$$(f*g)(t) = \int f(t-x)g(x)\,dx$$
$$f(t-x)=1$, $g(x)=100\cos(20x)$$
$$=100 \int \cos(20x)\,dx$$
$$p(t) = 5\sin(20t)$$
Now, when trying to do the other way around which is $p(t)=g(t)*f(t)$ I get stuck!
$$g(t-x)=100\cos[20(t-x)]=100\cos(20t-20x)$, $f(x)=1$$
$$(g*f)(t)= \int g(t-x)f(x)\,dx$$
$$=100 \int \cos(20t-20x)\,dx$$
Now what?