$$\int_0^{2\pi} f(x) \cos x \, dx = \left. f(x) \sin x \right|_0^{2\pi} - \int_0^{2\pi} f'(x) \sin x \, dx$$ $$= -\int_0^{2\pi} f'(x) \sin x \, dx$$ $$= -\left[\int_0^{\pi} f'(x) \sin x \, dx + \int_{\pi}^{2\pi} f'(x) \sin x \, dx\right]$$ $$= -\left[\int_0^{\pi} f'(x) \sin x \, dx + \int_0^{\pi} f'(x+\pi) \sin(x+\pi) \, dx\right]$$ $$= -\left[\int_0^{\pi} f'(x) \sin x \, dx - \int_0^{\pi} f'(x+\pi) \sin x \, dx\right]$$ $$= \int_0^{\pi} \left(f'(x+\pi) - f'(x)\right) \sin x \, dx \geq 0 \quad \text{since } f'(x) \text{ is increasing}$$