Okay, I've figured out how to do this using the residue theorem, so I should probably post an answer to my own question in case anyone happens upon it in the future. How satisfying it is to answer a question that I've had for such a long time!
Consider the function
$$f(z)=\frac{1}{(z-\cos z)^2-\pi^2}$$
SINGULARITIES: The poles of $f$ occur at the zeroes of
$$(z-\cos z)^2-\pi^2=(z-\cos z+\pi)(z-\cos z-\pi)$$
Let $z_k^+$ be the poles of $f$ that are zeroes of $z-\cos z+\pi$ and $z_k^-$ be the poles of $f$ that are zeroes of $z-\cos z-\pi$. Because each $z_k^+$ satisfies
$$z_k^+-\cos z_k^+ +\pi=0$$
it follows that
$$(z_k^++2\pi)-\cos (z_k^+ +2\pi) -\pi=0$$
and so $z_k^+$ is a zero of $z-\cos z-\pi$. Thus we may establish the following correspondence:
$$z_k^++2\pi=z_k^-$$
Now let us set $z_0^+=\pi- ա$ and $z_0^-=-\pi- ա$. Note that this is the only pair of corresponding poles lying on opposite sides of the line $\Re(z)=-\pi/2$.
RESIDUES: I shall omit the algebra:
$$\text{Res}(f,z_k^+)=-\frac{1}{2\pi} \frac{1}{1+\sin z_k^+}$$
$$\text{Res}(f,z_k^-)=\frac{1}{2\pi} \frac{1}{1+\sin z_k^+}$$
Notice that
$$\text{Res}(f,z_k^+)+\text{Res}(f,z_k^-)=0$$
This will be important later.
CONTOUR: Let $C_1$ be a straight line contour from $-\pi/2+ri$ to $-\pi/2-ri$, and let $C_2$ be a semicircular arc stretching counterclockwise from $-\pi/2-ri$ to $-\pi/2+ri$, and let $\gamma = C_1 \cup C_2$.
As we let $r\to\infty$, this contour will enclose all poles of $f$ to the right of the line $\Re(z)=-\pi/2$. Because all such poles can be put in equal and opposite pairs except for $z_0^+=\pi- ա$, the sum of residues inside of $\gamma$ as $r\to\infty$ will approach
$$\text{Res}(f,\pi- ա)$$
which is equal to
$$\frac{1}{2\pi} \frac{1}{1+\sqrt{1-ա^2}}$$
EVALUATION: By the residue theorem,
$$\oint_\gamma f(z)dz=\frac{i}{1+\sqrt{1-ա^2}}$$
or
$$\int_{C_1} f(z)dz+\int_{C_2} f(z)dz=\frac{i}{1+\sqrt{1-ա^2}}$$
Since $\int_{C_2} f(z)dz$ vanishes as $r\to\infty$, we have
$$\int_{C_1} f(z)dz=\frac{i}{1+\sqrt{1-ա^2}}$$
or
$$\int_{-\pi/2+i\infty}^{-\pi/2-i\infty} \frac{dz}{(z-\cos z)^2-\pi^2}=\frac{i}{1+\sqrt{1-ա^2}}$$
Now I'm going to omit a lot of nasty algebra. After simplifying this and equating real and imaginary parts, one ends up with the result
$$\int_{-\infty}^\infty \frac{12\pi^2+16(z-\sinh z)^2}{(3\pi^2+4(z-\sinh z)^2)^2+16\pi^2(z-\sinh z)^2}dz=\frac{1}{1+\sqrt{1-ա^2}}$$
which is equivalent to
$$\color{green}{\int_{0}^\infty \frac{3\pi^2+4(z-\sinh z)^2}{(3\pi^2+4(z-\sinh z)^2)^2+16\pi^2(z-\sinh z)^2}dz=\frac{1}{8+8\sqrt{1-ա^2}}}$$
Wolfram Alpha agrees with this to a lot of digits. Whoopee!
Of course, this is a nasty integral and no-one would ever want to try and evaluate it directly, so I'm still looking for something a little nicer.
