Integral from MIT Integration Bee 2023 Semifinals - $\int_{0}^{\pi} \frac{2\cos(x)-\cos(2021x)-2\cos(2022x)-\cos(2023x)+2}{1-\cos(2x)}\,\textrm{d}x$ 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.
 A: 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
$$I=\int_0^\pi 2\cdot 1011\ dx =2022\pi$$
A: Noting that
$$
\cos (2021 x)+\cos (2023 x)=2 \cos (2022 x) \cos x ,
$$
we have
$$
\begin{aligned}
I & =\int_0^\pi \frac{2(\cos x+1)-2 \cos (2022 x)(\cos x+1)}{1-\cos 2 x} d x \\
& =2 \int_0^\pi \frac{(1-\cos (2022 x))(\cos x+1)}{1-\cos (2 x)} d x
\end{aligned}
$$
Letting $x\mapsto \frac{\pi}{2}-x$ yields
$$
\begin{aligned}
I= & 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1+\cos (2022 x))(\sin x+1)}{1+\cos (2 x)} d x \\
= & 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1+\cos (2022 x)) \sin x}{1+\cos (2 x)} d x+ 2\int_ {-\frac{\pi}{2}} ^{\frac{\pi}{2}} \frac{1+\cos (2022 x)}{1+\cos (2 x)} d x\\\\
\end{aligned}
$$
Since the first and second integrand are respectively odd and even, therefore using IBP gives
$$
\begin{aligned}
I= & 4 \int_0^{\frac{\pi}{2}} \frac{1+\cos (2022 x)}{1+\cos (2 x)} d x \\
= & 2 \int_0^{\frac{\pi}{2}}(1+\cos (2022 x)) d(\tan x) \\
= & 2[\tan x(1+\cos (2022 x))]_0^{\frac{\pi}{2}} +4044 \int_0^{\frac{\pi}{2}} \tan x \sin (2022 x) d x \\
= & 4044 \int_0^{\frac{\pi}{2}} \frac{\sin x \sin (2022 x)}{\cos x} d x\\=&4044 \cdot \frac{\pi}{2} \quad (*) \\=&2022\pi
\end{aligned}
$$
where (*) comes from the post.
A: \begin{aligned}
I & =\int_0^\pi \frac{2(\cos x+1)-2 \cos (2022 x)(1+\cos x)}{1-\cos 2 x} d x \\
& =2 \int_0^\pi \frac{(1-\cos (2022 x))(1+\cos x)}{1-\cos (2 x)} d x\\
&=2\int_0^\pi \frac{\sin^2(1011x)\cos^2(\frac x2))}{\sin^2x} d x\\
&=2\int_0^\pi \frac{\sin^2(1011x)}{4\sin^2(\frac x2)} d x=\frac12\int_0^{2\pi}\frac{1-\cos(2022x)}{1-\cos(x)} d x\\
&=\frac12\int_{|z|=1} \frac{z^{2022}+z^{-2022}-2}{z+z^{-1}-2} \frac{dz}{iz}\\
&=\frac1{2i}\int_{|z|=1} \frac{(1-z^{2022})^2}{(1-z)^2z^{2022}} dz=\frac1{2i}\int_{|z|=1} \frac{(1+z+z^2+z^{2021})^2}{z^{2022}} dz\\
&=\frac1{2i}\int_{|z|=1} \bigg(\frac{2\cdot1011}{z}+\cdots\bigg) dz\\
&=\frac1{2i}\cdot2\pi i\cdot2022=2022.
\end{aligned}
