Doesn't this integral equal zero?

[s0ft]

$\displaystyle\int_0^\pi\dfrac{x dx}{a^2sin^2(x)+b^2cos^2(x)}$
I have to prove this to be equal to $\dfrac{\pi^2}{2ab}$ but with my attempt at it this problem boils down to:
$\dfrac{\pi}{2ab}\bigg[\arctan\Big(\dfrac{atan(x)}{b}\Big)\bigg]_0^\pi$ which equals zero.

 

[D H]

Show some work. It's hard to tell where you went wrong when all you show is your final result.

It should be obvious that that integral is not zero because the integrand is positive for all x in (0,pi).