Let I denote the requested integral. Symmetry and the period of sin2θ make the corresponding integral over 0≤θ≤2π equal to 4I. With z=eiθ, this becomesThe denominator factors as −2(z2−1/2)(z2−2). The two poles inside the unit circle are z=±1/2. Each residue is 1/(6i), because the derivative of the polynomial denominator there is 6z. The residue theorem gives 4I=2π/3, so
4I=∮∣z∣=1i(−2z4+5z2−2)zdz.
I=6π.
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