Let $M$ be a smooth manifold and a curve $\Gamma\subset M$ with $C^1$parametrizations $\gamma:[a, b]\to M$ and $\tilde{\gamma}:[c, d]\to M$. If $\omega$ is a $1$-form on $M$, prove that:
$$\int_a^b \gamma^*\omega = \int_c^d \tilde{\gamma}^*\omega$$
My intial idea was to claim that there exists an increasing diffeomorphism $\phi:[a,b]\to[c, d]$ (therefore, orientation preserving) such that $\gamma=\tilde{\gamma}\circ\phi$. And therefore:
$$\int_{[c,d]} \tilde{\gamma}^*{\omega}=\int_{\phi^{-1}([c,d])} \phi^*\circ\tilde{\gamma}^*(\omega)=\int_{[a,b]} (\tilde{\gamma}\circ\phi)^*(\omega)=\int_{[a,b]} \gamma^*\omega$$
However, I'm not sure that such a diffeomorphism always exists, since we only know that $\gamma$ and $\tilde{\gamma}$ are $C^1$, so I'm stuck with this argumentation. Any sugestions? Thanks!
You can show that any homeomorphism of between interval is monotonic. Suppose you have $x<y$ and $f(x) > f(y)$. Consider the interval $I = (f(y),f(x))$. Then $f^{-1}(I) = (y,x)$. However $x<y$, so this is a contradiction.