Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 2 / 18

Write $I_n=\int {\sin }^{n}xdx$. Apply integration by parts with ${\sin }^{n-1}x$ and $\sin xdx$:Using ${\cos }^{2}x=1-{\sin }^{2}x$ on the right gives $I_n=-\cos x{\sin }^{n-1}x+(n-1)(I_{n-2}-I_n)$. Rearranging provesFor $J_n=\int {\cos }^{n}xdx$, the analogous integration by parts uses ${\cos }^{n-1}x$ and $\cos xdx$, giving $J_n=\sin x{\cos }^{n-1}x+(n-1)(J_{n-2}-J_n)$. HenceAn arbitrary constant of integration absorbs the choices in the lower-order antiderivatives.

Starting from $I_0=J_0=x$, successive use of $n=2,4,6$ givesThese expressions follow directly from the requested recurrence, rather than requiring a separate trigonometric expansion.

Evaluate the recurrence boundary terms at zero and $\pi /2$. For $n>1$, both $\cos x{\sin }^{n-1}x$ and $\sin x{\cos }^{n-1}x$ vanish at both endpoints. Therefore the Wallis integrals obeyThe starting value for both is ${\int }_0^{\pi /2}1dx=\pi /2$. Thus