= Solution
Use equal small excisions on the two sides of each real pole. The <Cauchy principal value> is
$$
\lim_{R\to\infty}\lim_{\epsilon\downarrow0}\left(\int_{-R}^{-a-\epsilon}+\int_{-a+\epsilon}^{a-\epsilon}+\int_{a+\epsilon}^{R}\right)\frac{\cos x}{x^2-a^2}\,dx.
$$
For $e^{iz}/(z^2-a^2)$, close the <complex contour> in the upper half-plane and indent above $\pm a$, leaving the poles outside the contour. The <residue theorem> and the vanishing of the large semicircle integral give the principal-value integral as $i\pi$ times the sum of the two real-pole <residues>. Each clockwise indentation contributes $-i\pi$ times its residue. Since
$$
\operatorname{Res}_{a}\frac{e^{iz}}{z^2-a^2}=\frac{e^{ia}}{2a},\qquad \operatorname{Res}_{-a}\frac{e^{iz}}{z^2-a^2}=-\frac{e^{-ia}}{2a},
$$
the sum is $i\sin a/a$. Taking the real part yields \b[the answer]
$$
\boxed{\operatorname{PV}\int_{-\infty}^{\infty}\frac{\cos x}{x^2-a^2}\,dx=-\frac{\pi\sin a}{a}.}
$$
The decay on the large arc follows directly from $|e^{iz}|\leq1$ and the quadratic denominator.
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