I'm trying to understand why for a differentiable arc $\Gamma:[a,b]\to\Omega$ and a $1$-form $h=fdx+gdy$, then $ \int_\Gamma\varphi^*h=\int_{\varphi\circ\Gamma}h? $
For background, $\Omega$ is an open set in $\mathbb{C}$, and $\varphi:\Omega\to\mathbb{C}$ a smooth map. For a function $f$, I have the definition $\varphi^*f=f\circ\phi$, (when this makes sense for $f$ of course).
I also have the definitions $ \varphi^*\,dx=\frac{\partial u}{\partial x}\,dx+\frac{\partial u}{\partial y}\,dy, \qquad \varphi^*dy=\frac{\partial v}{\partial x}\,dx+\frac{\partial v}{\partial y}\,dy, $ where $u$ is the $x$ component of $\varphi$ and $v$ is the $y$ component. For a $1$-form $h=f\,dx+g\,dy$, $ \varphi^*h=(\varphi^*f)\varphi^*\,dx+(\varphi^*g)\varphi^*\,dy. $
I calculate \begin{align*} \int_\Gamma \varphi^*h &= \int_\Gamma(\varphi^*f)\varphi^*dx+\int_\Gamma (\varphi^*g)\varphi^*dy\\ &= \int_\Gamma(f\circ\varphi)\frac{\partial u}{\partial x}dx+ \int_\Gamma(f\circ\varphi)\frac{\partial u}{\partial y}dy+ \int_\Gamma(g\circ\varphi)\frac{\partial v}{\partial x}dx+ \int_\Gamma(f\circ\varphi)\frac{\partial v}{\partial y}dy\\ &= \int_\Gamma\left((f\circ\varphi)\frac{\partial u}{\partial x}+(g\circ\varphi)\frac{\partial v}{\partial x}\right)dx+\int_\Gamma\left((f\circ\varphi)\frac{\partial u}{\partial y}+(g\circ\varphi)\frac{\partial v}{\partial y}\right)dy \end{align*}
but I don't see if this fits into the form $\int_{\varphi\circ\Gamma}h=\int_{\varphi\circ\Gamma}fdx+\int_{\varphi\circ\Gamma}gdy$? Can it be made to fit? Thanks.