closed differential $1$-forms and integral distribution

Question

If for a distribution $D$, for all $p\in U$, $F_p=ker\omega_1(p)\cap...\cap ker\omega_g(p)$, s.t. each $\omega_i$ is a closed $1$-form, then why is $D$ integrable on $U$?

Answer 1

This is a consequence of the Frobenius theorem. Frobenius theorem says that the distribution is integrable if and only if for every $x$, there exists a neighborhood $U$ of $x$, $1$-form $\alpha_{jk}$ defined on $U$ such that $d\omega_i=\sum\omega_k\alpha_{jk}$, take $\alpha_{jk}=0$.