Any hint to solve given integral $\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}$? Show that for $ab>0$ $$\int_0^{2{\pi}}{{d\theta}\over{a^2\cos^2\theta+b^2\sin^2\theta}}={{2\pi}\over ab}$$
I'm not sure how to go about this. Any solutions or hints are greatly appreciated.

$\cos^2\theta,\sin^2\theta$ have period $\pi$, so the integral is $2I$ where $I$ is the integral from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$. wlog we may take $a,b$ to be positive.
We have $I=\frac{1}{a^2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sec^2\theta}{1+(\frac{b}{a})^2\tan^2\theta}\ d\theta$. Putting $x=\tan\theta$ this becomes $\frac{1}{a^2}\int_{-\infty}^{\infty}\frac{dx}{1+(\frac{b}{a})^2x^2}$. Putting $y=\frac{b}{a}x$ we get $\frac{1}{ab}\int_{-\infty}^{\infty}\frac{dy}{1+y^2}=\frac{\pi}{ab}$. [The last integral is just $\tan^{-1}y$.]
Hence the original integral is $\frac{2\pi}{ab}$ as required.
