to show integration is $\ge 0$

Question

I am given that $f(x)$ is continous on $[0,2\pi]$ and and $f''(x)\ge 0$ on the prescribed interval, I have to show $$\int_{0}^{2\pi}f(x)\cos x \; dx \ge 0;$$ well $\cos x\ge 0$ on $[0,\pi/2]$ and $\cos x\le 0$ on $[\pi/2,3\pi/2]$ and again $\cos x\ge 0$ on $[3\pi/2,2\pi]$, what I did, multiplied with $f''(x) \cos x$ and took the integration by parts, but nothing fruitful I came across; please give hint.

Answer 1

Integration by parts works here, though an appropriate choice of antiderivative is needed. Indeed, observe that $$\begin{align*} \int_{0}^{2\pi} f(x)\cos x \; dx &= \left[ f(x) \sin x \right]_{0}^{2\pi} - \int_{0}^{2\pi} f'(x)\sin x \; dx \\ &= - \int_{0}^{2\pi} f'(x)\sin x \; dx \\ &= -\left[ f'(x) (1-\cos x) \right]_{0}^{2\pi} + \int_{0}^{2\pi} f''(x)(1-\cos x) \; dx \\ &= \int_{0}^{2\pi} f''(x)(1-\cos x) \; dx. \end{align*}$$ Now we find that the integrand $f''(x)(1-\cos x)$ is always non-negative, hence the desired inequality follows.