Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 17

Use integration by substitution $t=\tan x$, so $dt=dx/{\cos }^{2}x$. The indefinite integral becomes $\int dt/[t(t^2-1)]$. Its partial fraction decomposition isIntegrating gives $-\ln |t|+\frac{1}{2}\ln |t^2-1|+C$. HenceThe formula applies on each interval where the printed integrand is defined: $\cos x\ne 0$ and $\tan x\notin \{0,1,-1\}$. The constant of integration may differ between such intervals.

For $n>1$, integration by parts with $u=x^{n-1}$ and $dv=e^{-x}dx$ yieldsThe boundary term vanishes at zero because $n-1>0$, and at infinity because exponential decay dominates every fixed power. The initial improper integral is $I_1={\int }_0^{\infty }{e}^{-x}dx=1$. Iterating the recurrence relation therefore givesThis is the positive-integer case of the Gamma function identity $\Gamma (n)=(n-1)!$.

Differentiate $x/(1+x^2{)}^n$ and integrate between zero and infinity. Its boundary values vanish for every integer $n\ge 1$, givingSince $x^2/(1+x^2{)}^{n+1}=(1+x^2{)}^{-n}-(1+x^2{)}^{-n-1}$, this becomes $0=E_n-2n(E_n-E_{n+1})$. Hence the recurrence relation isThe starting value is $E_1=[\arctan x{]}_0^{\infty }=\pi /2$. Therefore $E_2=\pi /4$, $E_3=3\pi /16$, andThese are also Wallis integrals after the integration by substitution $x=\tan t$.