We want to calculate
$ \int_0^\pi \sin x\ dx $
three days before we learn the Fundamental Theorem of Calculus, so our only tool is the limit of a Riemann sum.
So
$ \int_0^\pi \sin x\ dx\approx \sum_{n=1}^N \sin(n\pi/N)(\pi/N) $
when $ N $ is large.
Recall Euler's identity,
$ e^{i\theta}=\cos\theta + i\sin\theta. $
Hence, that Riemann sum $ R $ is the imaginary part of
$ (\pi/N)\sum_{n=1}^N e^{in\pi/N}. $
But
$ e^{in\pi/N}=\left(e^{i\pi/N}\right)^n, $
so $ R $ is just the imaginary part of a geometric sum.
The formula
$ 1+r+r^2+\dots+r^N = \frac{1-r^{N+1}}{1-r} $
lets us calculate that $ R $ is the imaginary part of
$ \frac{\pi}{N}\left(\frac{1-\left(e^{i\pi/N}\right)^{N+1}}{1-e^{i\pi/N}} -1\right). $
Now
$ \left(e^{i\pi/N}\right)^{N+1}= e^{i\pi + i\pi/N}= e^{i\pi}e^{i\pi/N}=-e^{i\pi/N} $
since
$ e^{i\pi}=\cos\pi+i\sin\pi=-1+0i=-1. $
Hence, we obtain that the Riemann sum is equal to the imaginary part of
$ \frac{\pi}{N}\left(\frac{1+e^{i\pi/N}}{1-e^{i\pi/N}}-1\right)=\frac{\pi}{N}\left(\frac{2e^{i\pi/N}}{1-e^{i\pi/N}}\right). $
Finally, use Euler's Identity and compute the imaginary part to get
$ R=\frac{\frac{\pi}{N}\sin(\pi/N)}{1-\cos(\pi/N)}, $
and then L'Hopital's rule can be used to find that the limit as N tends to infinity is equal to 2.
