= Solution
Use the standard convention that these compact <smooth manifolds> have no boundary. The <interior product> identity for a one-form gives $df\wedge\iota_X\omega_g=df(X)\omega_g=X(f)\omega_g$. For example, in the <coframe> from part (b), only the matching index survives in this wedge product, with the same pair of cancelling signs as in part (c). Thus the graded <Leibniz rule> and part (c) give
$$
d(f\nu)=df\wedge\nu+f\,d\nu
=\bigl(X(f)+f\operatorname{div}X\bigr)\omega_g.
$$
Integrate and apply the <Generalized Stokes theorem>. Since $\partial M=\varnothing$, the integral of $d(f\nu)$ is zero. Rearranging proves <integration by parts for Riemannian divergence>:
$$
\boxed{\int_M X(f)\omega_g=-\int_M f\operatorname{div}X\,\omega_g.}
$$
If a boundary were allowed, the exact general formula would instead have the additional term $\int_{\partial M}f\nu$ on the right; <compactness> alone does not remove that boundary contribution.
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