Well, I definitely trust you guys as far as math is concerned, maybe not so much in other topics. :P
But yeah, it seemed like nonsense and so I couldn't figure out where they were getting the statement from. When multiple people suggested the same argument, though . . .

And then conversely, if $\alpha\wedge d\alpha = 0$, then we have that if $X,Y$ are again in $\xi$, then $d\alpha(X,Y) = -\alpha([X,Y])$, and thus:
$$0 = (\alpha\wedge d\alpha)(X,Y,Z) = \alpha(X)d\alpha(Y,Z) - \alpha(Y)d\alpha(X,Z) + \alpha(Z)d\alpha(X,Y) = \alpha(Z)d\alpha(X,Y) = -\alpha(Z)d\alpha([X,Y]),$$
where $\alpha(Z)\neq 0$.

And we a $d$-field $\eta$ is integrable if for each point $p\in M$, there exists a $d$-submanifold $I\subseteq M$ such that for each $q\in I$:
$$T_qI = \eta_q.$$