Integral of a $1$-form on $S^1$ with respect to a parametrization (Guillemin and Pollack, 4.4.6)

Question

I have a question on problem 4.4.6 in Guillemin and Pollack's Differential Topology. Alright, so we've been given a parameterization of $S^1$ $$h: \mathbb{R} \to S^1,\ t\mapsto (\cos t,\sin t)$$ and we want to show that for any $1$-form $\omega$, we have $$\int_{S^1}\omega = \int_0^{2\pi}h^*\omega$$ Discussing with my peers, we have the following two strategies:

1. Let $U_N$ and $U_S$ be the charts on $S^1$ obtained by excluding the north and south pole, let $h$ induces a diffeomorphism of $(0,2\pi)$ on $U_N$ and $(-\pi,\pi)$ on $U_S$. So $$\int_{U_N}\omega = \int_0^{2\pi}h^*\omega,\qquad \int_{U_S}\omega = \int_{-\pi}^{\pi}h^*\omega = \int_0^{2\pi}h^*\omega$$ the latter inequality coming from the relevant substitution of the integral over $\mathbb{R}$. Is this enough to deduce the answer, if it is, I don't see why.
2. Since $h$ is a subjective quotient map and periodic, so it factors through a diffeomorphism of $S^1$, say $\tilde{h}$ i.e. $h = \tilde{h}\circ \pi$, where $\pi:\mathbb{R} \to \mathbb{R}/2\pi\mathbb{Z}$ is the quotient map. So we have $$\int_{S^1}\omega = \int_{q(0)}^{q(2\pi)}\tilde{h}^*\omega$$ Can we say $$\int_{q(0)}^{q(2\pi)}\tilde{h}^*\omega = \int_{0}^{2\pi}q^*\tilde{h}^*\omega$$

I suppose a third strategy would be verify in local coordinates but we're a bit confused on that as well. Any remarks will be appreciated!

Answer 1

I think the following Proposition (16.8) from John Lee's Introduction to Smooth Manifolds will prove helpful:

Let $M$ be an oriented smooth $n$-manifold w/ or w/o boundary, and let $\omega$ be a compactly supported $n$-form on $M$. Suppose $D_1,\ldots,D_k$ are open domains of integration in $\mathbb{R}^n$ (bounded subset with boundary measure zero), and for $i=1,\ldots,k$ we are given smooth maps $F_i:\overline{D_i}\rightarrow M$ satisfying

1. $F_i$ restricts to an orientation-preserving diffeomorphism from $D_i$ onto an open subset $W_i\subset M$
2. $W_i\cap W_j=\emptyset$ when $i\neq j$
3. $\operatorname{supp}\omega\subset\overline{W_1}\cup\dots\cup\overline{W_k}$

Then \begin{equation} \int_M\omega=\sum_{i=1}^k\int_{D_i}F_i^*\omega. \end{equation}

At first glance, assuming the orientations are the standard ones, the $F$'s and $D'$s we pick should be quite simple.