This question is from the MIT Integration Bee 2023 Semifinal #1. This integral should be solved within three minutes, and the goal is to show $$\int_{0}^{\pi} \frac{2\cos(x)-\cos(2021x)-2\cos(2022x)-\cos(2023x)+2}{1-\cos(2x)}\,\textrm{d}x = 2022\pi$$
One of the things I tried was to use the difference of cosines identity, but that didn't help me. I also looked at dividing each term individually after applying the double-angle identity to the denominator, but $\cos(ax)/\cos^2(x)$ becomes very hard to integrate for sufficiently large $a$. Finally, I tried to apply the transformation $x \mapsto \pi/2-x$ in an attempt to see if there was symmetry I could take advantage of, but it simply resulted in the same problem as before.
I'm not sure how to approach this question from here.
Best Answer
Note that $$\frac{2\cos x-\cos2021x-2\cos2022x-\cos2023x+2}{1-\cos2x}\\=\frac{2(1+\cos x)(1-\cos 2022x)}{1-\cos 2x}$$ and $$\frac{1-\cos 2022x}{1-\cos 2x}= 1011+2020\cos2x+2018\cos4x+\cdots+2\cos2020x$$
Then, all the individual terms vanish upon integration except $$I=\int_0^\pi 2\cdot 1011\ dx =2022\pi$$