 Solution

Let $I$ denote the requested integral. Symmetry and the period of ${\sin }^2\theta$ make the corresponding integral over $0\le \theta \le 2\pi$ equal to $4I$. With $z=e^{i\theta }$, this becomesThe denominator factors as $-2(z^2-1/2)(z^2-2)$. The two poles inside the unit circle are $z=\pm 1/\sqrt{2}$. Each residue is $1/(6i)$, because the derivative of the polynomial denominator there is $6z$. The residue theorem gives $4I=2\pi /3$, so