Connections of complex integration, antiderivative and argument principle

Question

Let us observe usual comlex integral, say $$ \frac{1}{2 \pi i}\oint_C\frac{1}{z}dz$$ where $C$ is unit circle. Usualy we would compute it through residue theorem or directly, elemenating $e^{i t}$: $$\oint_C \frac{1}{z}\,dz = \int_0^{2\pi} \frac{1}{e^{it}} ie^{it}\,dt = i\int_0^{2\pi} 1 \, dt = i \, t\Big|_0^{2\pi} = \left(2\pi-0\right)i =2 \pi i$$ But can we use an antiderivative in order to compute an integral over a closed path? If we take the last one and don't eleminate $e^{i t}$, is it "legal" to do following: $$\oint_C \frac{1}{z}\,dz = \int_0^{2\pi} \frac{1}{e^{it}} ie^{it}\,dt = \ln(e^{i t}) \Big|_0^{2\pi}=\ln(e^{2i\pi})-ln(e^0) = \left(2\pi i-0\right) =2 \pi i$$ If we would first paste the numbers and then compute the integral, we get $0$. Next, if we try to choose anothe contour, say unit with center $5$, we cannot just open ln so easily. We could observe the problem a different way: the function $y=z$ has zero at $z=0$ and the first integral is just a using of the argumet principle for the given function. So the question is following: Is it somehow useful to know an antiderivative by using the argument principle at searching of zeroes?