$\int_0^\pi g(x)\sin(nx) dx$ coverges to $0$ for any function $g$ is bounded variation I have a homework as follows : Prove that $\int_0^\pi g(x)\sin(nx) dx$ coverges to $0$ for any function $g$ is bounded variation on $[0,1]$.
my attempt: for any bounded variation function $g$,
$$\int_0^\pi g(x)\sin(nx) dx =  -n \int_0^\pi g(x)d(\cos(nx))\\
= -\int_0^\pi \sin(nx)dg(x).$$
How to show $\int_0^\pi \sin(nx)dg(x)\to 0$?

$$\int_0^\pi g(x)\sin(nx) dx =  -n \int_0^\pi g(x)d(\cos(nx))\\
= -\int_0^\pi \sin(nx)dg(x)\\= ||\frac{1}{n}\cos(nx)||_{\infty}V(g:[0,\pi]\\ =\frac{1}{n}||\cos(nx)||_{\infty}V(g:[0,\pi].$$
So, $\int_0^\pi g(x)\sin(nx) dx$ as $n\to\infty$. Is it correct?
