= Solution
Choose the <square root of one minus z squared> with $f(0)=1$: explicitly $f(z)=\sqrt{1-z}\sqrt{1+z}$ with principal <square roots>. It is single valued and <holomorphic> away from the two specified exterior real <branch cuts>. If $q(x)=\sqrt{x^2-1}>0$ for $|x|>1$, its boundary values are
$$
\boxed{\begin{array}{c|cc}&f(x+i0)&f(x-i0)\\\hline x>1&-iq(x)&+iq(x)\\x<-1&+iq(x)&-iq(x)\end{array}.}
$$
For example, above the right cut $1-z$ approaches the negative real axis from below, selecting $-i\sqrt{x-1}$; above the left cut $1+z$ approaches it from above, selecting $+i\sqrt{-x-1}$. Continuing once around either <branch point> changes the sign, which explains the need for cuts.
Apply the <residue theorem> to $F(z)=1/[zf(z)]$ on a large disk slit along both cuts, with small arcs excluding $\pm1$. The only enclosed <pole> is at zero, and its <residue> is $1/f(0)=1$. On the positively oriented boundary of the slit domain, the upper right lip runs from one to $R$, and the lower right lip returns from $R$ to one. The jump in $F$ gives their combined <integral> as $2iI_R$, where $I_R=\int_1^Rdx/[x\sqrt{x^2-1}]$. The left lips similarly give $2iI_R$ after the substitution $x=-t$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-2-two-cut-contour.png]
{title=A contour around the two exterior square-root cuts}
{description=Original schematic. The two slit boundaries are traversed with the retained domain on the left; each contributes twice the same real integral multiplied by i. The sole enclosed pole is at zero.}
The small arcs contribute $O(\sqrt\varepsilon)$ because the branch singularities have order $-1/2$. The large arc contributes $O(R^{-1})$ because $F(z)=O(|z|^{-2})$. Taking <limits> therefore yields $4iI=2\pi i$, and hence
$$
\boxed{I=\frac\pi2.}
$$
The real substitution $x=\sec\theta$ independently reduces the <integral> to $\int_0^{\pi/2}d\theta$, confirming its value and sign.
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