# $\int_M fd\omega=\int_M i_V \omega ~d\omega$ where $M$ is a closed $2$-manifold, $\omega$ is $1$-form, $V$ is vector field, with $i_V d\omega=df$

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?

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}