Integration – Evaluating a Complex Trigonometric Integral

Question

This horrible integral (shown in the title) comes from MIT Integration Bee 2022 Final Round. I have tried to use product to sum formulae, and also Feynman's trick by introducing $$I(b)=\int^{2\pi}_0e^{b\sin(50x)\sin(51x)}\cdot\frac{\cos(2022x)\sin(10050x)\sin(10251x)}{\sin(50x)\sin(51x)}dx$$
However, it seems the approach does not work. Does anyone suggest a good solution? Thank you!
(Final answer from the official website: $6\pi$.)

Answer 1

Note that $$\frac{\sin(10050x)}{\sin(50x)} =1+2\sum_{k=1}^{100} \cos(100k x) $$ $$\frac{\sin(10251x)}{\sin(51x)} =1+2\sum_{j=1}^{100} \cos(102 j x) $$ Then \begin{align} &I=\int^{2\pi}_0\frac{\cos(2022x)\sin(10050x)\sin(10251x)}{\sin(50x)\sin(51x)}dx\\ =&\sum_{k,j=1}^{100}\int^{2\pi}_0 4\cos(2022x) \cos(100k x)\cos(102j x)dx\\ =&\sum_{k,j=1}^{100}\int^{2\pi}_0 [ \cos(1011+50k +51j)2x\\ & \hspace{20mm} + \cos(1011-50k -51j)2x\\ & \hspace{20mm} + \cos(1011+51j -50k)2x\\ & \hspace{20mm} + \cos(1011-51j +50k)2x] \ dx\\ \end{align} The surviving terms are those with \begin{align} 50k+ 51j=& 1011 \implies (k,j)=(9,11)\\ 50k- 51j=& 1011 \implies (k,j)=(60,39)\\ 51j -50k=& 1011 \implies (k,j)=(42,61)\\ \end{align} As a result $$I= \int_0^{2\pi}(1+1+1) \ dx=6\pi $$