It's actually still the same answer. You can make the same symmetry argument.
Here's a hint:
Obviously, if [tex]f[/tex] is [tex]2\pi[/tex]-periodic, then [tex] \int_0^{2\pi} f(x)\sin{x} \ dx = \int_{-\pi}^{\pi} f(x)\sin{x} \ dx[/tex]. Thus if [tex]f[/tex] is an even function, since [tex]\sin[/tex] is odd, then [tex]f(x)\sin{x}[/tex] is odd and