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

Write $I=\int e^x\sin xdx$ and $J=\int e^x\cos xdx$. Two applications of integration by parts, first taking the sine and then the cosine as the differentiated factor, giveThus $I+J=e^x\sin x$ and $J-I=e^x\cos x$, in each case up to an arbitrary constant of integration. Keeping that freedom avoids identifying independently chosen antiderivatives too rigidly.

The difference and sum of the two antiderivatives in part (i) giveandTheir derivatives also verify the signs immediately.

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 $n=0$.

Use the substitution rule $u=x^2$, $du=2xdx$. The convergent improper integral is

For $n\ge 2$, write $x^n{e}^{-x^2}=x^{n-1}(x{e}^{-x^2})$ and apply integration by parts, using the antiderivative $-e^{-x^2}/2$ for the second factor:The boundary term is zero: at zero, $n-1>0$, and at infinity Gaussian decay dominates every power. Consequently the Gaussian moment recurrence is

Start with the supplied Gaussian integral $I(0)=\sqrt{\pi }/2$ and part (i)'s $I(1)=1/2$. Repeated use of the Gaussian moment recurrence givesThe even indices inherit the $\sqrt{\pi }$ factor from $I(0)$, while the odd indices inherit rational values from $I(1)$.