Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 17

Use integration by substitution $u=\cos x$, $du=-\sin xdx$. The denominator factorizes as $(u-2)(u-3)$, and partial fraction decomposition givesIntegrating and restoring $x$ yieldsFor real $x$, both $2-\cos x$ and $3-\cos x$ are positive, so no absolute-value signs are needed in this real logarithm. The denominator never vanishes.

For integration by parts, choose $u=\ln x$ and $dv=x^{-4}dx$. Then $du=dx/x$ and $v=-1/(3x^3)$, givingHere $C$ is an arbitrary constant of integration.

Choose the trigonometric substitution $x=\sin u$ with $-\pi /2\le u\le \pi /2$. The cosine is nonnegative there, so $\sqrt{1-x^2}=\cos u$ and $dx=\cos udu$. The double-angle formula then givesSince $u=\arcsin x$ and $\sin 2u=2x\sqrt{1-x^2}$,The formula is an antiderivative on $(-1,1)$ and has continuous endpoint values at $\pm 1$.

The product in the integrand simplifies to $\sin x{\cos }^{3}x$ below its upper endpoint. With the integration by substitution $t=\cos x$, the limits become $c$ and zero and $dt=-\sin xdx$. ThereforeAlthough the tangent function itself is undefined at the upper endpoint, the product has a zero limit there, so its continuous extension, or its improper integral, gives this same result.

To obtain $J$ directly from a Riemann sum, divide $[0,c]$ into $n$ equal pieces of width $c/n$ and use right endpoints $t_k=kc/n$. The sum isThe integrand is continuous, so the limit of these Riemann sums is its definite integral, namely $c^4/4$, agreeing with the antiderivative calculation.