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 / iii

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 $\pm 1$.