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

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