$\int_0^{\pi}{x \over{a^2\cos^2 x+b^2\sin^2 x}}dx$ How to solve 
$$\int_0^{\pi}{x \over{a^2\cos^2 x+b^2\sin^2 x}}dx$$

The answer is ${\pi}^2\over{2ab}$ but I cant prove it.
$$\text{As }\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$

$$I=\int_0^\pi{x \over{a^2\cos^2 x+b^2\sin^2 x}}dx=\int_0^\pi\frac{\pi-x}{a^2\cos^2 x+b^2\sin^2 x}dx$$ as $\displaystyle\sin(\pi-x)=\sin x,\cos(\pi-x)=-\cos x$

$$I+I=\pi \int_0^\pi\frac1{a^2\cos^2 x+b^2\sin^2 x}dx$$

$$=\frac\pi{b^2} \int_0^\pi\frac{\sec^2x}{\left(\frac ab\right)^2+\tan^2 x}dx$$

Put $\tan x=u$
Set $a/b=\tan (\alpha)$ and the intgrand simplifies to the product of $x$ and $\sec (x-\alpha)$ which is readily integrable