Modular Integral

Let $S=(3-3^2+3^3-\cdots +3^{2019})\pmod{100}$ and $3^n \pmod{100} \in \left\{3,9,27,81,43,29,87,61,83,49,47,41,23,69,7,21,63,89,67,1 \right\}$

Notice each remainder can be paired with another to make $\pm 90$ , if we keep the signs alternated, then writing in a table looks like this,

$\begin{array}{|c|c|} \hline -1 & -89 \\ \hline 7 & 83 \\ \hline -9 & -81 \\ \hline 3 & 87 \\ \hline 27 & 63 \\ \hline 43 & 47 \\ \hline -29 & -61 \\ \hline -21 & -69 \\ \hline -41 & -49 \\ \hline 23 & 67 \\ \hline \end{array}$

As the remainders are in cycles of $20$ and each cycle, $C$ , has property $C\equiv 0 \pmod{100}$ as there are an equal amount of positive pairings as negative.

$2019\pmod{20}=19$ so only the $20^{th}$ element in the cycle, or $1$ is excluded. We have $S\equiv 1 \pmod{100}$

$\implies \displaystyle\int_0^{\pi} S \, dx = \displaystyle\int_0^{\pi} \, dx = \boxed{\pi}$