 Solution

A coorientation gives a globally nonvanishing $C^2$ one-form $\zeta$ with $E=\ker \zeta$. For tangent fields $A,B$ in the integrable distribution,since integrability makes their bracket tangent to $E$. Thus $\zeta \wedge d\zeta =0$. Choose a transverse field $Y$ with $\zeta (Y)=1$. The $C^1$ one-form $\eta =-{\iota }_{Y}d\zeta$ then satisfies $d\zeta =\eta \wedge \zeta$, as is checked on $E$ and $Y$.

The Godbillon-Vey invariant in the unnormalized convention isThe three-form is continuous at the specified regularity, so its integral is defined. We prove independence of both choices by producing exact differences with $C^1$ primitives, allowing ordinary Stokes theorem.

First, for a fixed $\zeta$, another admissible auxiliary form has the form ${\eta }^{\prime }=\eta +a\zeta$. From $d^2\zeta =0$ and $d\zeta =\eta \wedge \zeta$, one gets $d\eta \wedge \zeta =0$. Expanding the two three-forms, using this relation and $\zeta \wedge d\zeta =0$, givesIts integral on the closed manifold vanishes.

Next any other defining form is ${\zeta }^{\prime }=h\zeta$ with $h$ nowhere zero. Let $g=\log |h|$. Thenso one may use ${\eta }_0^{\prime }=\eta +dg$. Its three-form differs byAgain the difference integrates to zero; any subsequent auxiliary choice is handled by the first calculation. This includes reversing the chosen coorientation, since $h$ can be negative. Thus the invariant, and the corresponding de Rham cohomology class, depends only on the oriented manifold and the foliation defined by $E$. Reversing the orientation of $N$, unlike reversing the coorientation of $E$, changes the sign of its numerical value.