Let $\mathbb{X}$ and $\mathbb{Y}$ be vector fields on $\mathbb{R}^n$ such that $[\mathbb{X},\mathbb{Y}]=0$, and let $\Phi_t$ and $\Psi_s$ denote their respective flows. Let $c:[0,1]\times [0,1]\to \mathbb{R}^n$ denote the singular $2$-cube given by $$c(s,t)=\Psi_s(\Phi_t(0)),$$ where $0$ denotes the origin of $\mathbb{R}^n$. Let $\omega=\omega_idx^i$ be a $1$-form on $\mathbb{R}^n$ and suppose that $i_{\mathbb{X}}\omega=i_{\mathbb{Y}}\omega=0$. Show that $$\int_c d\omega =0.$$
Firstly, as the Jacobi bracket vanishes, the flows of the vector fields commutes. Then, $$\frac{\partial c}{\partial t}(s,t)=\frac{\partial}{\partial t}(\Psi_s(\Phi_t(0)))=\frac{\partial}{\partial t}(\Phi_t(\Psi_s(0)))=\mathbb{X}(\Phi_t(\Psi_s(0)))=\mathbb{X}(c(s,t))$$ $$\frac{\partial c}{\partial s}(s,t)=\frac{\partial}{\partial s}(\Psi_s(\Phi_t(0)))=\mathbb{Y}(\Psi_s(\Phi_t(0)))=\mathbb{Y}(c(s,t))$$ If we denote as $c_{(j,\alpha)}$ the $(j,\alpha)$-th face of $c$ and $c_{(j,\alpha)}^*$ its pullback, then $$c_{(1,0)}^*\omega(t)=\omega_i(c(0,t))\mathbb{X}^i(c(0,t))dt$$ $$c_{(1,1)}^*\omega(t)=\omega_i(c(1,t))\mathbb{X}^i(c(1,t))dt$$ $$c_{(2,0)}^*\omega(s)=\omega_i(c(s,0))\mathbb{Y}^i(c(s,0))ds$$ $$c_{(2,1)}^*\omega(s)=\omega_i(c(s,1))\mathbb{Y}^i(c(s,1))ds$$ Using the Stokes theorem, $$\int_c d\omega=\int_{\partial c}\omega = \sum_{j=1}^2\sum_{\alpha=0}^1 (-1)^{j+\alpha}\int_0^1 c_{(j,\alpha)}^*\omega$$
However, I don't know how to conclude the question. I think that the hypothesis of $i_{\mathbb{X}}\omega=i_{\mathbb{Y}}\omega=0$, but I'm not sure where to use it. Any help would be appreciate.
By definition, $\iota_{X} \omega = \omega(X) = \omega_i X^i$ and $\iota_{Y} \omega = \omega(Y) = \omega_i Y^i$
So if $\iota_{X}\omega = \iota_Y\omega = 0$, then all four of your $c_{(j, \alpha)}^\star \omega $'s are identically zero.