Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / 17 / b

Set $x=2\sin \theta$ with $-\pi /2<\theta <\pi /2$, so $dx=2\cos \theta d\theta$ and $\sqrt{4-x^2}=2\cos \theta$. The integration by substitution givesThis uses the principal inverse sine; the chosen angle interval keeps the square root positive.

Complete the square in the denominator and split the numerator into a multiple of its derivative plus a constant:ThusThe natural logarithm needs no absolute-value correction because its quadratic argument is strictly positive for every real $x$. The second term is an inverse tangent antiderivative.