I'm looking for a proof to this theorem stated in Abraham-Marsden book, foundation of mechanics.
Let $\omega$ be a closed 2-form such that $\hat{\omega}:TM\to T^{*}M$, given by $\hat{\omega}(u)(v)=\omega(u,v)$, has constant rank. Then, there exist for each $p\in M$ a smooth chart $(U,(x^1,\cdots ,x^n,y^1,\cdots ,y^n,w^{1},\cdots ,w^k))$ such that: $$\left.\omega\right\vert_{U} = \sum_{i=1}^{n}dx^{i}\wedge dy^i$$
I've tried the following.
We can assume the existence of a smooth chart $(V,x^1,\cdots,x^{n},y^1,\cdots,y^n,w^1,\cdots,w^k)$ about $p$, s.t. $\omega_p=\sum_{i=1}^{n}dx^i_{p}\wedge dy^i_p$, this follows by choosing any smooth chart about $p$, thus we know of the existence of a basis of $T_pM$ s.t. $\omega_p$ has the desire expression, that allows us to define a transformation $T:TpM\to T_pM$, so we use $T$ to generate the smooth chart we want.
After that, we have two constant rank, closed 2-form on $V$, $\omega_0=\omega$ and $\omega_1 = \sum_{i=1}^n dx^i\wedge dy^i$, let $\eta=\omega_1-\omega_0$ and $\omega_t=\omega_0+t\eta$.
We notice, $\hat{\omega}_t(u)=t\hat{\omega}_1(u)+(1-t)\hat{\omega}_0(u)$ so we have $\text{ker}(\hat{\omega}_t)_p=\text{ker}(\hat{\omega}_0)_p$, moreover $d\omega_t=0=d\eta$.
What I'm trying to do is to use Moser trick by defining some t-dependent vector field, so I'm looking for some $X_t$ s.t. $d(i_{X_t}\omega_t)+\eta=0$. But I don't know how to use constant rank condition of $\omega$.
In the book you can find a proof but I cannot understand it ...