Definite integration of a trigonometric function 
How to integrate $$\int_0^{\pi/2}\!\dfrac{2a \sin^2 x}{a^2
\sin^2 x +b^2 \cos^2 x}\,dx $$

my first step is $$\frac{2}{a} \int_0^{\pi/2}\!\dfrac{a^2 \sin^2 x}{a^2 +(b^2 - a^2) \cos^2 x}\, dx $$
I would kind of want to do some sort of $u=\cos x$ substitution, to get at $\arctan u $ but no idea what to do with the sine in the numerator.

$$\displaystyle \int_0^{\dfrac{\pi}{2}} \dfrac{2a \sin^2 x}{a^2\sin^2 x +b^2 cos^2 x} dx$$
$$\displaystyle 2a\int_0^{\dfrac{\pi}{2}} \dfrac{\csc^2x }{a^2\csc^2 x +b^2 \cot^2 x \csc^2 x} dx$$
$$\displaystyle 2a\int_0^{\dfrac{\pi}{2}} \dfrac{\csc^2x }{(\cot^2 x+1)(a^2+b^2\cot^2 x)} dx$$
Let $u=\cot x$
$$\displaystyle 2a\int_0^{\infty} \dfrac{du }{(u^2 +1)(a^2+b^2u^2)} $$
$$\displaystyle 2a\int_0^{\infty} \left(\dfrac{1}{(u^2+1)(a^2-b^2)}-\dfrac{b^2 }{(a^2 -b^2)(a^2+b^2u^2)}\right) du $$
$$\displaystyle \frac{2a}{a^2-b^2}\int_0^{\infty} \left(\dfrac{1}{(u^2+1)}-\dfrac{b^2 }{(a^2+b^2u^2)}\right) du $$
And I hope you'll take it from here...
