Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 17 / b / Solution

Let $I_n=\int e^{nx}\sin xdx$ and $J_n=\int e^{nx}\cos xdx$. Integrating the trigonometric functions and differentiating $e^{nx}$ in integration by parts gives, modulo constants,Substitution yields $(1+n^2)I_n=e^{nx}(n\sin x-\cos x)$ and $(1+n^2)J_n=e^{nx}(n\cos x+\sin x)$. ThereforeNo division by $n$ was used, so this formula also applies at $n=0$.