Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 17 / c / Solution

Two applications of integration by parts lower the exponent by two. First, the sine boundary term vanishes:For $n\ge 1$, integrate the last expression once more:Thus the recurrence relation isThe starting value is $I_0={\int }_0^{\pi }\cos xdx=0$. Successive applications give $I_1=-2\pi$ and $I_2=-4{\pi }^3+24\pi$, followed by