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:
$$I=\int_{0}^{\pi} \frac {(\pi - x)dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} $$
$$I=\int_{0}^{\pi} \frac {\pi dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} - \int_{0}^{\pi} \frac {x dx}{(a^2\sin^2 x+b^2\cos^2 x)^2}$$
$$I=2\pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2\cos^2 x)^2} - I$$
$$I=\pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} $$
 A: $$I=\int_{0}^{\pi} \frac{x dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(1)$$
Apply $\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx$.
$$I=\int_{0}^{\pi} \frac{(\pi-x) dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(2)$$
Add (1) and (2)
$$2I=\pi\int_{0}^{\pi} \frac{ dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(3)$$
Use $\int_{0}^{2a} f(x) dx=\int_{0}^{a} f(x) dx, ~if ~f(2a-x)=f(x)$
$$I=\pi \int_{0}^{\pi/2} \frac{dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}~~~(4)$$
Let $$J(a,b)=\int_{0}^{\pi/2} \frac{dx}{(a^2 \sin^2 x+ b^2 \cos^2x)}=\int_{0}^{\pi/2} 
\frac {\sec^2 x dx}{b^2+a^2\tan^2 x} =\frac{\pi}{2ab}~~~(5)$$
D. (5) w.r.t. $a$ to get
$$\frac{J(a,b)}{da}=\int_{0}^{\pi/2} \frac{-2a \sin^2 xdx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}=-\frac{\pi}{2a^2b}~~~~(6)$$
$$\implies \int_{0}^{\pi/2} \frac{\sin^2x dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}=\frac{\pi}{4a^3b}~~~(7)$$
Similarly by D.w.r.t. $b$, we can get
$$\implies \int_{0}^{\pi/2} \frac{\cos^2x dx}{(a^2 \sin^2 x+ b^2 \cos^2x)^2}=\frac{\pi}{4a^3b}~~~(8)$$
Adding (7) and (8), we get from (4)
$$I=\frac{\pi^2}{4}\frac{a^2+b^2}{a^3b^2}.$$
A: Continue with the substitution $t=\tan x$,
$$\begin{align}
& \pi \int_{0}^{\frac {\pi}{2}} \frac {dx}{(a^2\sin^2 x+b^2 \cos^2 x)^2} \\
 & = \pi \int_0^\infty \frac{1+t^2}{(b^2+a^2t^2)^2}dt \\
&=\frac{\pi(a^2-b^2)}{2a^2b^2} \frac t {b^2+a^2t^2}\bigg|_ 0^\infty
 + \frac{\pi(a^2+b^2)}{2a^2b^2} \int_0^\infty \frac {dt}{b^2+a^2t^2} \\
& =0+\frac{\pi(a^2+b^2)}{2a^3b^3} \tan^{-1}\frac {at}b\bigg|_0^\infty \\
&=\frac{\pi^2(a^2+b^2)}{4a^3b^3}\\
\end{align}$$
