How about integration by parts...
\int_0^{2\pi} f(x) \cos x\,{\rm d}x = \left[ f(x)\sin x \right]_{x=0}^{x=2\pi}-\int_0^{2\pi} \sin x\,f'(x)\,{\rm d}x
$ \left[ f(x)\sin x \right]_{x=0}^{x=2\pi} = 0$ f'(x) \leq 0
So now you have to prove that
-\int_0^{2\pi} \sin x\,f'(x)\,{\rm d}x \ge 0 or expanded as
\int_0^{2\pi} \sin x\,f'(x)\,{\rm d}x = \int_0^{\pi} \sin x\,f'(x)\,{\rm d}x +\int_\pi^{2\pi} \sin x\,f'(x)\,{\rm d}x \leq 0
Since $\sin x$ is positive between $x=0\ldots\pi$ and negative otherwise the above can be written as
\int_0^{\pi} \sin x\,f'(x)\,{\rm d}x -\int_0^{\pi} \sin x\,f'(x+\pi)\,{\rm d}x \leq 0
or
\int_0^{\pi} \sin x\,\left(f'(x)-f'(x+\pi)\right)\,{\rm d}x \leq 0
Which is true only if f'(x)-f'(x+\pi)\leq0 for $x=0\ldots\pi$. If $f(x)$ is decreasing with negative slope then |f'(\pi)|\leq|f'(0)| and thus the above is true.