$\underset{0}{\overset{\pi }{\int }}\frac{xdx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x},whena=bis:$

$I=\underset{0}{\overset{\pi }{\int }}\frac{x dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

$2I=\pi \underset{0}{\overset{\pi }{\int }}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

$2I=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

$I=2\pi \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

$=2\pi \underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{{\sec }^{2}xdx}{a^2+b^2{\tan }^{2}x}$

$=2\pi \underset{0}{\overset{1}{\int }}\frac{dt}{a^2+b^2{t}^2}$ where $\tan x=t\Rightarrow {\sec }^{2}x dx=dt$

$=\frac{2\pi }{b^2}\times \frac{b}{a}{\tan }^{-1}{\left(\frac{bt}{a}\right)}_0^1$

$=\frac{2\pi }{ab}\times {\tan }^{-1}{\left(t\right)}_0^1$, since $a=b$

$=\frac{2\pi }{ab}\times \frac{\pi }{4}=\frac{{\pi }^2}{2ab}$