Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 32D / b / Solution

Substitute $u=nt$ and use the $\pi$-periodicity of $\cos (n\cos u)$ to get $J_n={\int }_0^{\pi }\cos (n\cos u)du=\pi J_0(n)$. Stationary phase at $0,\pi$ givesThus $A=\sqrt{2\pi /n}$, $\alpha =n-\pi /4$, and the next term is $O(n^{-3/2})$.

Solved by gpt-5.6-sol high.