Let $M$ be a closed $3$-manifold, and let $\xi$ be a $2$-dimensional subbundle of $TM$. I know the following.

- There is a nowhere zero $1$-form $\alpha$ on $M$ with $\alpha(X) = 0$ for any vector field $X$ which is a section of $\xi$.
- Any two $1$-forms $\alpha$, $\alpha'$ with this property satisfy $\alpha = f\alpha'$ for some smooth nowhere zero function $f$.
- $\xi$ is integrable if and only if $\alpha \wedge d\alpha = 0$ for any $\alpha$ as above.
- For integrable $\xi$, we can write $d\alpha = \omega \wedge \alpha$ for some $1$-form $\omega$.
- Any two choices $\omega$, $\omega'$ satisfying this equation have $\omega' - \omega = g\alpha$ for some smooth function $g$.

Assuming $\xi$ is integrable, let $\alpha$, $\alpha'$ be two $1$-forms as above, and let $d\alpha = \omega \wedge \alpha$ and $d\alpha' = \omega' \wedge \alpha'$ for some $1$-forms $\omega$, $\omega'$. I have two questions.

- Is $\omega \wedge d\omega - \omega' \wedge d\omega'$ an exact form or not?
- What is the geometric meaning behind $\omega \wedge d\omega - \omega' \wedge d\omega'$ being an exact or nonexact form?

**Edit.** Now that Tsemo has mentioned that this form is indeed exact, could anybody supply a direct proof of its exactness?