Lee smooth manifolds, proposition 11.38 (line integral formula) Proposition 11.38 of Lee, Introduction to smooth manifolds states if $\gamma:[a,b] \mapsto M$ is a piecewise smooth curve segment, then the line integral of $\omega$ over $\gamma$ can be written as $\int_{\gamma} \omega = \int_{a}^{b} \omega_{\gamma(t)}(\gamma'(t)) dt$. He uses coordinates chart to prove it (chop it into segements such that each image is contained in an coordinate chart, and then compute it in the coordiante to prove), but why can't we do it directly like this ? $$\int_{\gamma} \omega = \int_{[a,b]} \gamma^* (\omega) = \int_{a}^{b} \gamma_t^*(\omega) = \int_{a}^{b} \omega_{\gamma(t)} \gamma'(t) dt$$
The third equality follows from the definition of pull-backs: $F_p^*(\omega)(v) = \omega_{F(p)} dF_p(v)$, so I don't really see the point of using coordinate charts ?
 A: Probably because on page 288, the definition of a line integral is as follows (modified notation): given a one-form $\eta$ on $[a,b]$, if we let $t$ denote the coordinate on $[a,b]$ then we may express it as $\eta = f \, dt$ for some unique $f:[a,b] \to \Bbb{R}$. Then, we define $\int_{[a,b]}\eta := \int_a^b f$, where the LHS is a new symbol where we're defining the integral of a one-form over $[a,b]$, while the RHS is the standard Riemann integral of a function $f:[a,b] \to \Bbb{R}$.
Now, to prove the theorem you have apply the definition, which means you have to be able to express $\gamma^*{\omega}$ as $f\, dt$. I guess the point of introducing the chart is to somehow convince you that $f(t) = \omega_{\gamma(t)}(\gamma'(t))$? Though this is pretty straightforward even without a chart. For $t \in [a,b]$, let $e_t = \dfrac{d}{dt}\bigg|_t \in T_t\Bbb{R}$ denote the "unit tangent vector" corresponding to the element $1 \in \Bbb{R}$, but regarded as a tangent vector under the isomorphism $\Bbb{R} \cong T_t \Bbb{R}$. Then, $(dt)_t(e_t) = 1$ (here, I'm of course using $t$ to mean both the base point and also the standard identity coordinate function on $[a,b]$... hopefully it's not too confusing). So,
\begin{align}
f(t) &= f(t) \cdot (dt)_t(e_t) \\
&= (f\, dt)_t(e_t) \\
&= (\gamma^*\omega)_t(e_t) \\
&= \omega_{\gamma(t)}\left( d\gamma_t(e_t)\right) \\
&= \omega_{\gamma(t)}\left( \gamma'(t)_{\gamma(t)}\right),
\end{align}
where in the last line, I use the notation $\gamma'(t)_{\gamma(t)}$ to mean the tangent vector in $T_{\gamma(t)}\Bbb{R}^n$ corresponding to the vector $\gamma'(t) \in \Bbb{R}^n$ under the isomorphism $\Bbb{R}^n \cong T_{\gamma(t)}\Bbb{R}^n$.

Note that the argument I presented above only works for smooth $\gamma$. For piece-wise smooth $\gamma$, just chop up $[a,b]$ appropriately into finitely many pieces, and apply this argument on each piece.
