The value of  $I={\int }_0^{\pi } \frac{x\mathrm{d}x}{4{\cos }^2 x+9{\sin }^2 x}$ is 

Using Property 9, we have

$\begin{array}{l}I={\int }_0^{\pi } \frac{(\pi -x)\mathrm{d}x}{4{\cos }^2(\pi -x)+9{\sin }^2(\pi -x)}\\ =\pi {\int }_0^{\pi } \frac{\mathrm{d}x}{4{\cos }^{2}x+9{\sin }^{2}x}-{\int }_0^{\pi } \frac{x\mathrm{d}x}{4{\cos }^{2}x+9{\sin }^{2}x}\\ 2I=\pi {\int }_0^{\pi } \frac{\mathrm{d}x}{4{\cos }^{2}x+9{\sin }^{2}x}\\ =2\pi {\int }_0^{\pi /2} \frac{\mathrm{d}x}{4{\cos }^{2}x+9{\sin }^{2}x}\end{array}$

$\begin{array}{l}=2\pi {\int }_0^{\pi /2} \frac{{\sec }^{2}x\mathrm{d}x}{4+9{\tan }^{2}x}\\ =2\pi {\int }_0^{\mathrm{\infty }} \frac{\mathrm{d}t}{4+9t^2}(t=\tan x)\\ =\frac{2\pi }{9}{\int }_0^{\mathrm{\infty }} \frac{\mathrm{d}t}{t^2+4/9}={\left.\frac{2\pi }{9}\cdot \frac{3}{2}{\tan }^{-1}\frac{3}{2}t\right|}_0^{\mathrm{\infty }}=\frac{{\pi }^2}{6}\end{array}$

Hence $I={\pi }^2/12$