Intergral minimization

Using integration by parts twice, $\begin{aligned} \int_0^{2\pi}f(x)\cos xdx&=\left[f(x)\sin x\right]_0^{2\pi}-\int_0^{2\pi}f'(x)\sin xdx\\ &=0+\left[f'(x)\cos x\right]_0^{2\pi}-\int_0^{2\pi}f''(x)\cos xdx\\ &=f'(2\pi)-f'(0)-\int_0^{2\pi}f''(x)\cos xdx \end{aligned}$ Now, notice that we can write $f'(2\pi)-f'(0)$ as an integral of $f''(x)$ , so the expression becomes $\int_0^{2\pi}f''(x)dx-\int_0^{2\pi}f''(x)\cos xdx=\int_0^{2\pi}f''(x)(1-\cos x)dx$ $(1-\cos x)\geq0$ and $f''(x)\geq0$ for all $x\in[0,2\pi]$ , so the minimum value of the integral is $0$ , which is attained when $f''(x)=0$ .