\[ \int_{[-\pi, \pi]} {1\over 1 + \sin^2(\theta)} \dtheta .\]
Solution
Set \(z=e^{i\theta}\), so \(\sin(\theta) = {z-\inverseof{z}\over 2i}\) and \(\sin^2(\theta) = -{1\over 4}(z^{-2}-2+z^2)\). Then \[ I\da \int_{[-\pi, \pi]} {1\over 1 + \sin^2(\theta)} \dtheta &= \int_{S^1} {1\over 1 - {1\over 4}(z^{-2} -2 + z^2) } {1\over iz}\dz \\ &= \int_{S^1} {-4i \over z(4-(z^{-2} -2 + z^2))}\dz \\ &= \int_{S^1} {-4iz \over z^2(6-z^{-2} - z^2)}\dz \\ &= \int_{S^1} {4iz \over z^4-6z^2+1}\dz \\ &= 2\pi i \sum_{z_k\in \DD} \Res_{z=z_k} {4iz\over z^4-6z^2+1} .\]
Factoring the denominator, with \(w=z^2\): \[ w^2 - 6w + 1 =0 &\implies w^2 - 6w + \qty{-6\over 2}^2 - \qty{-6\over 2}^2 + 1 =0 \\ &\implies (w - 3)^2 = -1 + 9 = 8 \\ &\implies w-3 = \zeta_2^k \sqrt{8},\, k=0, 1 \\ &\implies w = 3 \pm \sqrt{8} \\ &\implies z= \pm\sqrt{3\pm \sqrt 8} .\] Write these roots as
\(z_1 \da \sqrt{3-\sqrt 8}\)
\(z_2 \da -\sqrt{3-\sqrt 8}\)
\(z_3 \da \sqrt{3+\sqrt 8}\)
\(z_4 \da -\sqrt{3 + \sqrt 8}\)
Their moduli:
\(3 + \sqrt{8} > 1\), so \(\abs{ \pm \sqrt{3+\sqrt 8}}>1\).
\(0<3-\sqrt{8} < 1\) since \(2<\sqrt 8<3\), so \(\abs{ \pm \sqrt{3-\sqrt 8} } < 1\).
So it suffices to compute the residues at \(z_1, z_2 = \pm \sqrt{3-\sqrt 8}\): At \(z_1\): \[ \Res_{z = z_1} &= \lim_{z\to z_1} {(z-z_1) 4i z \over z^4-6z^2+1 } \\ &\equalsbecause{\text{LH}} \lim_{z\to z_1} { 4iz + 4i(z-z_1) \over 4z^3 -12z} \\ &= {4iz_1 \over 4z_1^3 - 12z_1} \\ &= {i\over z_1^2 - 3} \\ &= {i\over (3-\sqrt 8) - 3} \\ &= -{i\over \sqrt 8} .\] At \(z_2\): \[ \Res_{z=z_2} &= \lim_{z\to z_2} {(z-z_2) 4i z \over z^4-6z^2+1 } \\ &= {4iz_2\over 4z_2^3 - 12z_2}\\ &= {i \over z_2^2 - 3} \\ &= -{i\over \sqrt 8} .\]
Thus \[ I = 2\pi i \cdot -{2i\over \sqrt 8} = {4\pi \over \sqrt{8}} .\]