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 $\dfrac{\pi}{4a^3b}$.
 A: Here is my solution:
$$\int_0^\frac{\pi}{2}\frac{\sin^2 x}{(b^2\cos^2 x+a^2\sin^2 x)^2}dx=\int_0^\frac{\pi}{2}\frac{\tan^2 x\sec^2 x}{(b^2 +a^2\tan^2 x)^2}dx\overset{\tan x=t}=\int_0^\infty \frac{t^2}{(b^2+a^2 t^2)^2}dt$$
$$\overset{at=bp}=\frac{1}{a^3b}\int_0^\infty\frac{p^2}{(p^2+1)^2} dp=\frac{1}{a^3b} \left(\int_0^\infty \frac{1}{p^2+1}dp- \underbrace{\int_0^\infty \frac{1}{(p^2+1)^2}dp}_{p=\tan \theta }\right)$$
$$=\frac{1}{a^3b}\left(\arctan p\bigg|_0^\infty-\int_0^\frac{\pi}{2}\frac{1+\cos(2\theta)}{2}d\theta\right)=\frac{1}{a^3b}\left(\frac{\pi}{2}-\frac{\pi}{4}\right)=\frac{\pi}{4a^3b}$$
A: 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 $\int_0^{\pi/2} \frac{\sin^2(x)}{(b^2\cos^2(x)+a^2 \sin^2(x))^2}dx = \frac{\pi}{4a^3b}$.
A: Change variable to $t = \tan x$, we have
\begin{align}
&\quad\int_0^{\frac{\pi}{2}}\frac{\sin ^2 x}{\left(b^2\cos^2 x+a^2\sin^2 x\right)^2}\, dx\\&=\int_0^{\frac{\pi}{2}}\frac{\tan^2 x}{\left(b^2+a^2\tan^2 x\right)^2}\cdot\frac{1}{\cos^2 x}\,dx\\&=\int_0^\infty\frac{t^2}{\left(b^2+a^2t^2\right)^2}\,dt\\&=\left.-\frac{t}{2a^2}\cdot\frac{1}{b^2+a^2t^2}\,\right|_{\,0}^{+\infty} +\frac{1}{2a^2}\int_0^\infty \frac{1}{b^2+a^2t^2}\,dt\\&=\left.\frac{1}{2a^2}\cdot\frac{a}{b^3}\arctan \frac{a}{b}t\,\right|_{\,0}^{+\infty}\\&=\frac{\pi}{4a^3b}. 
\end{align}
