Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 1

For real $x$, the angle $\arctan x$ lies in $(-\pi /2,\pi /2)$ and has sine $x/\sqrt{1+x^2}$. Thus the principal arcsine in the expression equals the arctangent of $x$. Its derivative isThe chain rule gives the same result directly, using $1-x^2/(1+x^2)=1/(1+x^2)$ with a positive square root.

The lower-order terms cancel, so the numerator must be expanded through sixth order. The Taylor series give $\sin (\sin x)=x-x^3/3+x^5/10+O(x^7)$ and ${\sin }^{2}x=x^2-x^4/3+2x^6/45+O(x^8)$. ThereforeThe first nonzero term is $x^2/18$, with the quotient continuously extended to zero if needed.

With the change of variables $u=\tan x$, one has $dx=du/(1+u^2)$, $\sin x=u/\sqrt{1+u^2}$ and $\cos x=1/\sqrt{1+u^2}$. The transformed improper integral is integrable at its zero endpoint and becomesHence