Let $\omega$ be a smooth 1-form on a smooth manifold $M$. A smooth positive function $\mu$ on some open subset $U\subset M$ is called an integrating factor for $\omega$ if $\mu\omega$ is exact on U.
In Lee's ISM, there is a necessary and sufficient condition for a nonvanishing 1-form to admit integrating factor.
If $\omega$ is nowhere-vanishing, then $\omega$ admits an integrating factor in a neighborhood of each point if and only if $d\omega \wedge \omega \equiv 0$.
I have no idea how to prove the "if" part. Assume $d\omega \wedge \omega \equiv 0$ and that $\omega$ is nowhere-vanishing. What does these conditions tell us about $\omega$ and how can we construct a smooth positive function $\mu$ on some open neighborhood $U$ of any point $p$?
Partial solution:
Ted indicates that I need to use the Frobenius theorem for the proof. I assume the version of Frobenius theorem is the following one in Warner's book:
Let $\mathcal I \subset E^*(M)$ be a differential ideal locally generated by $d-p$ independent 1-forms. Let $m\in M$. Then there exists a unique maximal, connected, integral manifold of $\mathcal I$ through $m$, and this integral manifold has dimension $p$.
For any $m\in M$, we want to prove that there exists a smooth positive function $\mu$ on some open subset $m\in U\subset M$ such that $\mu\omega$ is exact on U. To use the Frobenius theorem, we need a involutive distribution $\mathcal D$ and let $\mathcal I:=\mathcal {I(D)}$(annihilator of $\mathcal D$). But what is our $\mathbb D$? What good can the existence of a unique maximal, connected, integral manifold of $\mathcal I$ through $m$ do to us in order to find $\mu$?
Thanks in advance.
Once you know the Frobenius Theorem, you know that the integral manifolds of $\omega = 0$ are (locally) given by level surfaces of some function $f$. This means that $df = \mu\omega$ for some nonzero function $\mu$.