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Let $M$ be a closed (smooth) $2$-manifold, let $\omega$ be a (smooth) $1$-form on $M$, and let $V$ be a (smooth) vector field on $M$. I am trying to show that if $i_V d\omega=df$ for some $f\in C^\infty(M)$, where $i_V$ is the interior multiplication by $V$, then $$ \int_M fd\omega=\int_M i_V \omega ~d\omega$$

A consequence of the assumptions is : $L_V \omega=i_V d\omega+di_V \omega=df+di_V \omega=d(f+i_V \omega)$, so the Lie derivative $L_V\omega$ of $\omega$ by $V$ is exact. (The first equality is Cartan's magic formula) In particular, $L_V d\omega$ is zero since $L_V d\omega =dL_V \omega$.

Actually I'm not sure that this is a relevant information. Any hints?

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The magic formula isn't so useful in this context. Stoke's theorem, and the "product rule" for exterior differentiation and interior products is more relevant. Note that \begin{align} f d \omega &= d(f \omega) + \omega \wedge df, \end{align} So, \begin{align} \int_M f \, d \omega &= \int_M d(f \omega) + \int_M \omega \wedge df \\ &= \int_{\partial M} f \omega + \int_M \omega \wedge (i_V d \omega) \\ &= \int_M \omega \wedge (i_V d \omega) \tag{$*$} \end{align} where I used Stoke's theorem in the second line, and in the third, the fact that the boundary is empty. Next, note that \begin{align} i_V(\omega \wedge d \omega) &= (i_V \omega) \wedge d \omega + (-1)^{|\omega|} \omega \wedge i_V(d \omega) \end{align} On the LHS, $\omega \wedge d \omega$ is a $3$-form on the $2$-dimensional manifold $M$, so it is $0$; hence taking the interior product with $V$ still keeps it $0$. On the RHS, note that $i_V \omega$ is a smooth function ($0$-form), so we may write the $\wedge$ more commonly as a $\cdot$, and also use the fact that $|\omega| = 1$; i.e it is a $1$-form. Hence, we find that \begin{align} \omega \wedge i_V(d \omega) &=(i_V \omega)\cdot d \omega \tag{$**$} \end{align} By plugging $(**)$ into $(*)$ we immediately get the desired result \begin{align} \int_M f\, d \omega &= \int_M(i_V \omega)\cdot d \omega. \end{align}

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