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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.

Stamp
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  • The alternating $2$'s, hmm... – Kamal Saleh Feb 15 '23 at 22:38
  • Regrouping the numerator as$$[\cos(x) - \cos(2021x)] + [\cos(x) - \cos(2023x)] + 2 - 2 \cos(2022x)$$and applying these identities gives the equivalent integrand$$\frac{\sin(1010x)\sin(1011x) + \sin(1011x)\sin(1012x) + 2\sin^2(1010x)}{\sin^2(x)}$$which may be susceptible to the approach like used here – user170231 Feb 15 '23 at 23:32
  • See also this question – user170231 Feb 15 '23 at 23:41
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    The numerator simplifies to ... $$(2+2\cos(x))(1-\cos(2022x))$$ – WW1 Feb 15 '23 at 23:42
  • A possible start: apply the transformation/substitution $u = x - \pi/2$, noting that $x = u + \pi/2$ and that $$ \cos(n(u + \pi/2)) = \begin{cases} \cos(nu) & n \equiv 0 \pmod 4\ -\sin(nu) & n \equiv 1 \pmod 4\ -\cos(nu) & n \equiv 2 \pmod 4\ \sin(nu) & n \equiv 3 \pmod 4 \end{cases} $$ – Ben Grossmann Feb 16 '23 at 02:24
  • With that, the integral becomes $$ \int_{-\pi/2}^{\pi/2} \frac{-2\sin(u)+\sin(2021u)+2\cos(2022u)-\sin(2023u)+2}{1+\cos(2u)},\textrm{d}u = \ \int_{-\pi/2}^{\pi/2} \frac{-2\sin(u)+\sin(2021u)-\sin(2023u)}{1+\cos(2u)},\textrm{d}u
    • \int_{-\pi/2}^{\pi/2} \frac{2\cos(2022u)+2}{1+\cos(2u)},\textrm{d}u=\

    \int_{-\pi/2}^{\pi/2} \frac{2\cos(2022u)+2}{1+\cos(2u)},\textrm{d}u, $$ where we note that the first integral is an integral of an odd function over $[-\pi/2,\pi/2]$.

     – Ben Grossmann Feb 16 '23 at 02:24
  • @BenGrossman In the second and third integrals, how does that deal with the $\cos(2022u)$? Rather, what's the next step at least? – Stamp Feb 16 '23 at 19:31
  • @Stamp Honestly, I have no idea – Ben Grossmann Feb 16 '23 at 19:54

3 Answers3

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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 $$I=\int_0^\pi 2\cdot 1011\ dx =2022\pi$$

Ace
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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.

Lai
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\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}

xpaul
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