Let $M$ denote a smooth $n$-dimensional manifold.
(a) Let $\phi$ denote a smooth $n$ form which is nowhere zero. Show that every $x_{0} \in M$ has a neighborhood on which we can find smooth local coordinates $x^{1}, ...x^{n}$ such that : $\phi=dx^{1}\wedge...\wedge dx^{n}$
(b) Let $\psi$ denote a closed smooth $n-1$ form on $M$ which is nowhere zero. Show that each near point, there can be found smooth coordinates $x^{1}, ...x^{n}$ such that : $\psi=dx^{2}\wedge...\wedge dx^{n}$
Now, in any local coordinates I can write $\phi=f(y)dy^{1}\wedge...\wedge dy^{n}$ but I have no idea how to turn this into the required coordinates. Similarly I can write $\psi$ as sum of $n-1$ forms. The only other thing I see is that $\phi$ looks sort of like an orientation form, but that is all I can gleam from this.
Many thanks for any help!
I've done this question before. I can't remember how to do part (b) right now, but I have included an argument for part (a).
Fix $p \in M$ and let $\hat{x}^1, \dots, \hat{x}^n$ be local coordinates in a neighbourhood $U$ of $p$. Then we have
$$\phi|_U = f(\hat{x}^1, \dots, \hat{x}^n)d\hat{x}^1\wedge\dots\wedge d\hat{x}^n$$
for some smooth $f$. Let $g(\hat{x}^1) = \int_a^{\hat{x}^1}f(t, \hat{x}^2, \dots, \hat{x}^n)dt$; note that
$$d(g(\hat{x}^1)) = f(\hat{x}^1, \dots, \hat{x}^n)d\hat{x}^1.$$
As $\phi$ is nowhere zero, $g'(\hat{x}^1(p)) = f(\hat{x}^1(p), \dots, \hat{x}^n(p)) \neq 0$, so by the Inverse Function Theorem, there is some possibly smaller neighbourhood $V$ of $p$ such that $g\circ\hat{x}^1$ is injective. This allows us to introduce new coordinates $x^1 = g\circ\hat{x}^1, x^2 = \hat{x}^2, \dots, x^n = \hat{x}^n$ in which we have
$$\phi|_U = f(\hat{x}^1, \dots, \hat{x}^n)d\hat{x}^1\wedge\dots\wedge d\hat{x}^n = dx^1\wedge\dots\wedge dx^n.$$