Given $f$ such that $f''$ is continious and differentiable on $[0,2\pi]$, and that $f(0)=f(2\pi)$ and $ |f''(x)|\leq1$ for all $x\in[0,2\pi]$ I need to show
$$\left|\int_{0}^{2\pi}f(t)\sin nt\text{d}t\right|\leq\frac{4}{n^2},\ \forall n$$
I started off with integration by parts twice and got that this integral is
$-\underbrace{\frac{1}{n}\left(f(2\pi)\cos2\pi n-f(0)\cos 0\right)}_{\text{=0}}+\frac{1}{n^{2}}\left(f'(2\pi)\underbrace{\sin2\pi n}_{\text{=0}}-f'(0)\underbrace{\sin0}_{\text{=0}}\right)-\intop_{0}^{2\pi}f''(t)\cdot\frac{\sin nt}{n^{2}}\text{d}t$
I checked I have no mistakes for like an hour or so by now. However, this doesn't seem to work out now because
$$\left|\intop_{0}^{2\pi}f''(t)\cdot\frac{\sin nt}{n^{2}}\text{d}t\right|\leq\frac{1}{n^{2}}\left|\intop_{0}^{2\pi}1\cdot\sin nt\text{d}t\right|=0$$ which is problematic.