I would like to prove the following (**without the Frobenius Theorem**):

On $\mathbb{R}^n$, if $\omega$ is nowhere vanishing $1$-form such that $d\omega\wedge\omega=0$ then there exists (at least locally) $f$ a positive function such that $d(f\omega)=0$.

In fact, I am looking for an easier proof of the Frobenius theorem for distribution of hyperplanes. Indeed the proof I know are inductive, we start with distribution of lines, then inductively we arrive to distribution of hyperplanes.

My statement implies the Frobenius theorem for hyperplanes, so I wonder if we can prove it with as less as technology as possible?

Thanks for your help

Added: It is easy to prove that there exists a 1-form $\theta$ such that $d\omega =\theta \wedge \omega$, then $d(f\omega)=(df+f\theta)\wedge \omega$, so the problem can be rephrased as, does there exisst $f$ and $g$ such that $df+f\theta =g \omega$? I stuck here.