Show $\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+ b^2\cos^2 x)^{2}}=\frac {\pi^2 (a^2+b^2)}{4a^3b^3}$ Show that
$$\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+ b^2\cos^2 x)^{2}}=\frac {\pi^2 (a^2+b^2)}{4a^3b^3}$$
My Attempt:
Let $$I=\int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} $$
Using $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x)dx$ we can write:
$$I=\int_{0}^{\pi} \frac {(\pi - x)dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} $$
$$I=\int_{0}^{\pi} \frac {\pi dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} - \int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+b^2\cos^2 x)^2}$$
$$I=2\pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} - I$$
$$I=\pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} $$

Continue with the substitution $t=\tan x$,
$$\begin{align}
& \pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} \\
 & = \pi \int_0^\infty \frac{1+t^2}{(b^2+a^2t^2)^2}dt \\
&=\frac{\pi(a^2-b^2)}{2a^2b^2} \frac t {b^2+a^2t^2}\bigg|_ 0^\infty
 + \frac{\pi(a^2+b^2)}{2a^2b^2} \int_0^\infty \frac {dt}{b^2+a^2t^2} \\
& =0+\frac{\pi(a^2+b^2)}{2a^3b^3} \tan^{-1}\frac {at}b\bigg|_0^\infty \\
&=\frac{\pi^2(a^2+b^2)}{4a^3b^3}\\
\end{align}$$
