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

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 $\pm 1$.