Integrability of a distribution on $S^{2n-1}$ Let $\theta$ be the restriction of 
$$ \eta = x^2 dx^1 - x^1 dx^2 + \cdots + x^{2n} dx^{2n-1} - x^{2n-1} dx^{2n} $$
to the unit sphere $S^{2n-1} \subset \mathbb{R}^{2n}$. Since $\theta$ is a nowhere vanishing $1$-form, $\ker \theta$ defines a distribution on $S^{2n-1}$. Is it integrable? As I understand it an equivalent condition for integrability here is that $\theta \wedge d \theta = 0$, which holds trivially for $n = 1$. If I computed correctly, I've gotten that $\eta \wedge d \eta \neq 0$ on $\mathbb{R}^{2n}$ except when $n = 1$ (so $\ker \eta$ is not integrable on $\mathbb{R}^{2n}$ for $n \geq 2$), but this does not necessarily mean that the pullback $i^*(\eta \wedge d \eta) = \theta \wedge d \theta$ along the inclusion map $S^{2n-1} \to \mathbb{R}^{2n}$ is nonzero as well, right? In which case I'm not sure how to go about this, since checking whether $\theta \wedge d \theta = 0$ directly in local coordinates on $S^{2n-1}$ seems messier than an intended solution.. (this is an old exam problem). Any help would be appreciated!
 A: Here's a sketch for the case $n=2$, but it will generalize without difficulty. Note first of all that on $S^3$ we have the equation
$$\sum_{i=1}^4 x^i\,dx^i = 0.$$ In particular, at the point $P=(0,0,0,1)$, we have $dx^4=0$ (as a form on $S^3$), and so (again, as forms on $S^3$)
$$\theta(P) = dx^3 \quad\text{and}\quad d\theta(P) = -2dx^1\wedge dx^2,$$
so clearly $(\theta\wedge d\theta)(P) \ne 0$. Now you can argue that this $3$-form is $SO(4)$-invariant and hence is everywhere nonzero on $S^3$.
Indeed, one way to see that gives perhaps a better direct argument. By direct computation on $\Bbb R^4$, $\star(\eta\wedge d\eta) = 2\sum x^i\,dx^i = d(\|x\|^2)$, which is clearly $SO(4)$-invariant. Moreover, letting $\phi = \eta\wedge d\eta$, we see that $\star(\phi\wedge\star\phi) = \|\phi\|^2 = \|{\star}\phi\|^2 \ne 0$ on $\Bbb R^4-\{0\}$. Since $\star\phi$ annihilates the tangent space of $S^3$ it follows that $\phi$ cannot vanish on $S^3$.
A: In fact, we can show more: That $$\theta \wedge d\theta \wedge \cdots \wedge d\theta$$ vanishes nowhere. As you've pointed out, this fact will imply that $\ker \theta$ is nonintegrable and moreover by definition that $\theta$ is a contact form---so, maximally nonintegrable---which is a strictly stronger condition than nonintegrability for $n \geq 3$.
First, observe that the volume form $\Omega = dx^1 \wedge \cdots \wedge dx^n$ associated to the standard metric $\bar g$ on $\Bbb R^{2n}$ and the volume form $\omega^{\wedge n}$ on $\Bbb R^n$ induced by the standard symplectic form $$\omega := dx^1 \wedge dx^2 + \cdots + dx^{2 n - 1} \wedge dx^{2n}$$ are related by
$$\omega^{\wedge n} = n! \,dx^1 \wedge \cdots \wedge dx^{2n} = n! \Omega .$$
Also, with respect to $\bar g$ the (outward-pointing) unit normal vector field on the sphere is the restriction of the Euler vector field $$X := (x^1 \partial_{x^1} + \cdots + x^{2n} \partial_{x^{2n}})$$ to $\Bbb S^{2 n - 1}$, so the volume form $\Omega$ induces thereon is $i^*(\iota_X \Omega)$. Now, compute $\iota_X \omega$ and $d \eta$, and use the results to write the volume form $i^*(\iota_X \Omega)$ in terms of $\theta$.

Computing gives $\iota_X \omega = \eta$ and $d \eta= -2 \omega$, and thus \begin{multline}\iota_X \Omega = \iota_X \left(\frac{1}{n!} \omega^{\wedge n}\right) = \frac{1}{n!} \cdot n (\iota_X \omega) \wedge \omega^{\wedge (n - 1)} \\= \frac{1}{(n - 1)!} (\eta) \wedge \left(-\frac{1}{2} d\eta\right)^{n - 1} = \frac{1}{(n - 1)!}\left(-\frac{1}{2}\right)^{n - 1} \eta \wedge (d\eta)^{n - 1} .\end{multline} Pulling back by $i$ gives that $$i^*(\iota_X \Omega) =-\frac{1}{(n - 1)!}\left(-\frac{1}{2}\right)^{n - 1} \theta \wedge (d\theta)^{\wedge (n - 1)} = \frac{1}{(n - 1)!}\left(-\frac{1}{2}\right)^{n - 1} \theta \wedge d\theta \wedge (d\theta)^{\wedge (n - 2)},$$ where the last equality requires $n \geq 2$. In particular, since the l.h.s. is nonvanishing (it is a volume form on $S^{2 n - 1}$, so is $\theta \wedge d\theta$.

Remark In fact, $(S^{2 n - 1}, \ker \theta)$ is one of the standard examples of a contact structure. Moreover, the bundle endomorphism $J : TM \to TM$ characterized by $g(JX, \cdot) = \omega(X, \cdot)$ is a complex structure on $\Bbb R^{2 n}$, and by construction, the restriction $J'$ of $J$ to the hyperplane distribution $\ker \theta \subset T^* S^{2 n - 1}$ defines the prototypical example $(S^{2 n - 1}, \ker \theta, J')$ of CR geometry, a rich and well-studied type of geometry.
