= Solution
Let $I$ denote the requested <integral>. Symmetry and the period of $\sin^2\theta$ make the corresponding integral over $0\leq\theta\leq2\pi$ equal to $4I$. With $z=e^{i\theta}$, this becomes
$$
4I=\oint_{|z|=1}\frac{z\,dz}{i(-2z^4+5z^2-2)}.
$$
The denominator factors as $-2(z^2-1/2)(z^2-2)$. The two poles inside the <unit circle> are $z=\pm1/\sqrt2$. Each <residue> is $1/(6i)$, because the derivative of the polynomial denominator there is $6z$. The <residue theorem> gives $4I=2\pi/3$, so
$$
\boxed{I=\frac\pi6.}
$$
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