Hopf Invariant of $f$ To set up notation: Let $f:S^3\to S^2$. For a volume form $\omega$ on $S^2$, $f^*\omega$ is a closed two form on $S^3$, which can be written as $d\alpha$ for some 1 form $\alpha$. The number $$\int_{S^3}\alpha\wedge d\alpha$$
is called the Hopf invariant of $f$. 
My goal is to show that the result is independent of our choice of $\alpha$. Here are my thoughts:
If $\beta$ satisfies $d\beta=f^*\omega$, then we should have $\alpha=\beta+\xi$, where $\xi$ is a closed form on $S^3$. This means that $\alpha\wedge d\alpha=(\beta+\xi)\wedge d\beta=\beta\wedge d\beta +\xi\wedge d\beta$.
So it seems like we must show 
$$
\int_{S^3}\xi\wedge d\beta=0.
$$ 
This is where I am stuck. I feel like I am either way off base, or need to apply Stokes Theorem somehow, but I am not sure how I should proceed.
 A: This has been resolved in the comments: $d(-\xi \wedge \beta) = \xi \wedge d\beta$, so your form is exact and has integral zero.
I want to point out that this is a reasonably common trick. Here you're using it to get homotopy invariants of maps $S^3 \to S^2$; one other place it's important is in foliations. Given a codimension 1 (co-orientable) foliation $\mathcal F$ on a closed 3-manifold, I can find a 1-form $\omega$ such that $\mathcal F = \text{ker}(\omega)$. $\omega$ is usually not closed; if I remember right, the existence of a closed such $\omega$ is equivalent to the foliation being taut. 
Anyway, one then considers $\int \omega \wedge d\omega$. Then this was independent of $\omega$ for basically the same reason as before. This gives a cobordism invariant of codimension 1 foliations on a 3-manifold (whatever that means). One thing that makes this especially interesting is that there are examples of folations for which this invariant (the Gobdillon-Vey invariant) is any specified real number, so foliations are very complicated indeed.
