Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 11 / b / Solution

First take $a>0$ and $|x|<a$, and put $u=\arcsin (x/a)$, $q=\sqrt{a^2-x^2}$. Then $u^{\prime }=1/q$. The chain rule givesConsequentlywhich proves the identity on the open domain of differentiability. The squared inverse sine is unchanged when $a$ is replaced by $-a$, so the identity also holds for negative $a$ with $|x|<|a|$. The value $a=0$ is excluded.