Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 13

Apply integration by parts twice. Taking an antiderivative of $e^{x/2}$ to be $2e^{x/2}$ givesand the second application gives $\int x{e}^{x/2}dx=2x{e}^{x/2}-4e^{x/2}+C$. Therefore the required definite integral is

The factor $x^2$ is an even function and $\sin x$ is an odd function, so their product is an odd function. Its definite integral over the symmetric interval vanishes:Indeed the substitution rule $x=-t$ changes the integral over $[-\pi /2,0]$ into the negative of the integral over $[0,\pi /2]$.

A partial fraction decomposition givesFor the second term the substitution rule $t=x^2+1$, $dt=2xdx$ gives a logarithm. ThusThe constant in the antiderivative may be chosen independently on the two intervals separated by $0$.

Split the integral intoThe first is the standard antiderivative $\mathrm{arsinh}x=\ln (x+\sqrt{1+x^2})$ from the inverse hyperbolic functions. For the second, the substitution rule $t=1+x^2$ gives $\frac{1}{2}\int t^{-1/2}dt=\sqrt{1+x^2}$. HenceThe logarithm's argument is positive for every real $x$.

Use the tangent half-angle substitution $t=\tan (x/2)$, so $\cos x=(1-t^2)/(1+t^2)$ and $dx=2dt/(1+t^2)$. ThenThe inverse tangent antiderivative givesThis expression applies on each interval $((2j-1)\pi ,(2j+1)\pi )$ for an integer $j$. Although $5+3\cos x\ge 2$ everywhere, the chosen inverse tangent expression jumps at the endpoints because the substitution has a pole there. A global continuous antiderivative is obtained by adding $j\pi /2$ on that interval, using one common $C$, and extending by continuity at the endpoints.