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

Expand hyperbolic sine in exponential functions:Integrating each term gives the indefinite integral

For integration by parts, take $u=\arctan x$ and $dv=x^{-2}dx$. Then $du=dx/(1+x^2)$ and $v=-1/x$, soUse partial fraction decomposition $1/[x(1+x^2)]=1/x-x/(1+x^2)$. ThereforeThe constant of integration may be chosen independently on the positive and negative intervals.

Use integration by substitution $u=\ln x$, $du=dx/x$. The bounds become $3$ and $4$, and the remaining denominator factorizes as $(u-1)(u-2)$. The partial fraction decomposition isConsequently the definite integral isAll logarithm arguments here are positive on the integration interval.

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