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$$ \int_{-\pi}^{\pi} \frac{d\theta}{1+\sin^2(\theta)} = \pi\sqrt{2} $$

  • $\begingroup$ Hint: use the t-formula. No complex analysis is necessary. $\endgroup$ – Millardo Peacecraft Apr 5 '14 at 23:48
  • $\begingroup$ Although complex analysis makes this question a lot easier :P ... in my opinion $\endgroup$ – InsigMath Apr 6 '14 at 0:20
  • $\begingroup$ The trick really boils down to using the fact that $\sin(\theta) = \frac{z-1/z}{2i}$ where $z = e^{i\theta}$ gives |z| = 1$. This is essentially what Git Gud does. $\endgroup$ – Chris K Apr 6 '14 at 0:23

Hint: Let $\gamma \colon[-\pi,\pi]\to \mathbb C,\theta \mapsto e^{i\theta}$ and use the residue theorem after proving that$$\displaystyle \int \limits_{-\pi}^{\pi} \frac{1}{1+\sin^2(\theta)}\mathrm d\theta = 4i\int_\gamma\dfrac{z}{z^4-6z^2+1}\mathrm dz=4i\int_\gamma\dfrac{z}{(z^2-2z-1)(z^2+2z-1)}\mathrm dz.$$


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