Consider $X,Y$ two smooth vector fields of $\mathbb{R}^2$ and the set $C=\{z\in \mathbb{R}^2;\, det(X,Y)(z)\neq 0\}$.
On the set $C$, we define the form $\alpha$ by :
$ \alpha(X)=1,\quad \alpha(Y)=0. $
Given a trajectory $z(\cdot)$ solution on $[0,T]$ of the system $$ \dot z(t) = X(z(t)) + u(t)\,Y(z(t)) $$ where $u$ is a given function and such that $z(.)\in C$, we define: $$ \int_{z(\cdot)} \alpha = T. $$
On which set do we have $d\alpha=0$ ?
By the intrinsic expression for the exterior derivative, we have $$d\alpha(X,Y)=X(\alpha(Y))-Y(\alpha(X))-\alpha([X,Y])=-\alpha([X,Y]).$$ So, $\alpha$ is closed at every point where $\alpha([X,Y])=0$. But, by construction, the kernel of $\alpha$ at a point $p$ is the line spanned by $Y(p)$. Hence, $d\alpha$ vanishes at $p$ if and only if $Y$ and $[X,Y]$ are colinear at $p$.