We have the following Proposition (from Basic Complex Analysis; 3rd edition by Warden and Hoffman)
Proposition 2.1.2 If $f(z) = u(x,y) + iv(x,y)$, then $$ \int_{\gamma} f = \int_{\gamma} u(x,y)dx - v(x,y)dy + i \int_{\gamma} [u(x,y)dy + v(x,y)dx]$$
I'm using this definition of the integral from the textbook, $\int_{\gamma}f = \sum_{i=0}^{n-1} \int_{a_i}^{a_{i+1}} f(\gamma(t)) \gamma'(t)dt$.
I reached the part (in my computations) where I have, by definition,
\begin{align} f(\gamma(t))\cdot \gamma'(t) & = x'(t) u(x(t), y(t)) - y'(t)v(x(t),y(t)) + i[y'(t)u(x(t),y(t)) + x'(t)v(x(t),y(t))] \end{align}
Integrating both sides wrt to $t$, we get: $$ \int_{a_i}^{a_{i+1}}f(\gamma(t))\cdot \gamma'(t) dt = \int_{a_i}^{a_{i+1}} x'(t) u(x(t), y(t))dt - \int_{a_i}^{a_{i+1}}y'(t)v(x(t),y(t))dt ... \text{ and so forth }$$
This is where I am confused (maybe I'm missing something simple from vector calculus - like a definition or something..) but, if I may ask, how do we remove the dependence on $t$ in our equation. For example, why does $ \int_{a_i}^{a_{i+1}} x'(t) u(x(t), y(t))dt$ simplify to $$\int u(x,y)dx$$
Thanks ! in advance.
This is really just an issue of definitions: if $\gamma\colon [0,1]\to \mathbb C$ is a contour, then it can be integrated against "$1$-forms", which are expressions of the form $a(x,y)dx + b(x,y)dy$, where you either treat $dx$ and $dy$ as formal symbols for now, or learn what differential forms are (the first option is fine for the purposes of this post though!)
Using the second option, if $\gamma(t) = (x(t),y(t))$, then we simply define the integral $\int_{\gamma} adx + bdy$ to be: $$ \int_{0}^1 \big(a(\gamma(t)).x'(t)+ b(\gamma(t))y'(t)\big)\mathrm{d}t $$ In other words, the dependence on $t$ doesn't disappear, it simply gets recorded by writing $\int_{\gamma}(-)$ rather than $\int_{0}^1(-)\gamma'(t)\mathrm{d}t$.
This is then consistent with the definition $\int_{\gamma} f(z)dz = \int_{0}^1f(\gamma(t))\gamma'(t)\mathrm{d}t$ if you set $dz=dx+idy$. In this sense, you should just think of it as an extension of the definition $\int_{\gamma}f(z)\mathrm{d}z$ that allows you to integrate over $\gamma$ any expression of the form $a\mathrm{d}x+b\mathrm{d}y$ rather than ones which are of the form $f(z).dz$.
[If you do learn differential forms, then $\omega = a\mathrm{d}x+b\mathrm{d}y$ is a $1$-form on $\mathbb R^2$, and if $\gamma$ is smooth, it can be pulled back to a $1$-form $\gamma^*(\omega)$ on $[0,1]$, which can then be integrated over $[0,1]$ as usual.]