= Solution
Use the positively oriented unit circle and $z=e^{i\theta}$. Writing $d=\sqrt{a^2-b^2}$, the <contour integral for a sine-squared trigonometric quotient> becomes
$$
J=\oint_{|z|=1}-\frac{(z^2-1)^2}{2iz^2(bz^2+2az+b)}\,dz.
$$
There is a double pole at zero and simple poles at $r_\pm=(-a\pm d)/b$. Since $a>b>0$, $|r_+|<1$ and $|r_-|>1$. The coefficient of $z^{-1}$ at zero is $a/(ib^2)$. At $r=r_+$, the <residue> is
$$
-\frac{(r^2-1)^2}{4idr^2}=-\frac d{ib^2},
$$
because $(r-r^{-1})^2=4d^2/b^2$. The <residue theorem> consequently gives
$$
\boxed{J=\frac{2\pi}{b^2}\left(a-\sqrt{a^2-b^2}\right).}
$$
For the second integral, use $x=\cos(\theta/2)$ with $0\leq\theta\leq\pi$. The square root is $\sin(\theta/2)$, and reversing the integration limits gives
$$
\begin{aligned}
I&=\frac12\int_0^\pi\frac{\sin^2(\theta/2)\cos^2(\theta/2)}{1+\cos^2(\theta/2)}\,d\theta\\
&=\frac14\int_0^\pi\frac{\sin^2\theta}{3+\cos\theta}\,d\theta=\frac18J(3,1).
\end{aligned}
$$
The last factor uses the symmetry of the integrand about $\pi$ over a full period. Therefore
$$
\boxed{I=\frac\pi4(3-2\sqrt2).}
$$
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