= Solution
Write the integral's multiplier in the <Chebyshev polynomial> basis:
$$
(1-x^2)^{-1/2}+(1-x^2)^{1/2}
=w(2-x^2)=w\left(\frac32y_0-\frac12y_2\right).
$$
At $\gamma^2=5$, part (i) gives
$$
Z=\left(\frac15+\frac{\varepsilon^2}{20}\right)y_0
+\frac\varepsilon4y_1+\frac{\varepsilon^2}{4}y_2+O(\varepsilon^3).
$$
By <orthogonality>, the $y_1$ term and all cross terms vanish. Using $I_0=\pi$ and $I_2=\pi/2$, the integral is
$$
\frac32\pi\left(\frac15+\frac{\varepsilon^2}{20}\right)
-\frac12\frac\pi2\frac{\varepsilon^2}{4}+O(\varepsilon^3)
=\boxed{\frac{3\pi}{10}+\frac{\pi}{80}\varepsilon^2+O(\varepsilon^3)}.
$$
The remainder can be integrated against this integrable weight in the regular expansion, so it retains the stated order.
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