Let $\omega$ be a closed $k$-form on $\mathbb{R}^n$ and $c:I^k \rightarrow \mathbb{R}^n$ a $k$-cube on $\mathbb{R}^n$. Let $\mathbb{X}$ be a vector field on $\mathbb{R}^n$ with flow $\Phi_t$. Show that
$$\frac{d}{dt}\int \Phi_t^*\omega=\int_{\Phi_t \circ \partial c} i_{\mathbb{X}}\omega $$
where for a $(k-1)$-chain $\mathcal{C}=\sum_r a_rc_r$, the $(k-1)$-chain $\Phi_t \circ \mathcal{C}$ is defined to be $\sum_r a_r(\Phi_t \circ c_r)$.
I get that $$\begin{align} \frac{d}{dt} \int_\mathcal{C} \Phi_t^*\omega &= \int_\mathcal{C} \frac{\partial}{\partial t} \Phi_t^*\omega \\ &= \int_\mathcal{C} \Phi_t^*L_{\mathbb{X}_t}\omega \\ &= \int_\mathcal{C} \Phi_t^* [di_{\mathbb{X}_t}\omega+i_{\mathbb{X}_t}d\omega]\\ &= \int_\mathcal{C} \Phi_t^* di_{\mathbb{X}_t}\omega \\ &= \sum_r a_r \int_{\mathcal{C}_r}\Phi_t^*di_{\mathbb{X}_t}\omega \\ &= \sum_r a_r \int_{\mathcal{C}_r}d\Phi_t^*i_{\mathbb{X}_t}\omega \\ &= \sum_r a_r \ d (\int_{\mathcal{C}_r}\Phi_t^*i_{\mathbb{X}_t}\omega) \\ &= ???? \end{align} $$
I really cant see how to continue with this question.