How can I show $\int _0^{2π} \frac{d\theta}{\alpha^2\cos^2\theta + \beta^2\sin^2\theta}=\frac{2\pi}{\alpha\beta}$ where $\alpha, \beta \in R^+$
Calculate the integral $\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt$, by deformation theorem.