Solution (final part)

Let $ h $ be the function we are integrating. Let $ R_k $ denote the region for which $ |h(t)| \geq 1 $ and that contains the value of $ t $ where $ \cos(t) = \frac{k\pi}{r} $. Then we note that $ |\int_{R_k} h(x) dx| > |\int_{R_{k-1}} h(x) dx| $ and the sines of the integral alternate, so we can just calculate the first one and everything else will be bounded (in particular by $ \frac{\pi}{2} $). With a bit of Taylor approximation, we can bound the size of each $ R_k $ by $ \frac{4\sqrt[4]{a-1}}{\sqrt{r}} $, and noting that $ h $ is always positive for $ r \leq \frac{\pi}{2} $, we can replace the $ \sqrt{r} $ with $ 1 $ and then bound $ h $ by $ \frac{1}{a-1} $. This gives us the bound we claimed above and we are done.

Jacob Steinhardt

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