Definite integration evaluation of $\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$. $$\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}~dx$$
how to proceed please help
The answer given is $\dfrac{\pi}{4a^3b}$.

Let $I(a,b)=\int_0^{\pi/2}\frac{dx}{a^2\sin^2x+b^2\cos^2x}=\frac{\pi}{2ab}$(try proving this yourself)
partially differentiating wrt $a$ and applying Leibnitz rule,
$\int_0^{\pi/2}\frac{-2a \sin^2x dx}{(a^2\sin^2x+b^2\cos^2x)^2}= \frac{- \pi}{2ba^2}$.
So $\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}dx = \frac{\pi}{4a^3b}$.
