$\underset{0}{\overset{\pi }{\int }}\frac{xdx}{4{\cos }^{2}x+9{\sin }^{2}x}=$

$$\begin{gathered} I=\underset{0}{\overset{\pi }{\int }}\frac{xdx}{4{\cos }^{2}x+9{\sin }^{2}x} \\ =\underset{0}{\overset{\pi }{\int }}\frac{\left(\mathrm{\pi }-x\right)dx}{4{\cos }^{2}x+9{\sin }^{2}x} \\ =\pi \underset{0}{\overset{\pi }{\int }}\frac{dx}{4{\cos }^{2}x+9{\sin }^{2}x}-I \\ I=\frac{\pi }{2}\underset{0}{\overset{\pi }{\int }}\frac{dx}{4{\cos }^{2}x+9{\sin }^{2}x} \\ =\frac{\pi 2}{2}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{4{\cos }^{2}x+9{\sin }^{2}x} \\ =\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{se{c}^{2}xdx}{4+9ta{n}^{2}x} \\ put \tan x=t then se{c}^{2}x dx=dt \\ =\pi \underset{0}{\overset{\infty }{\int }}\frac{dt}{4+9t^2} \\ =\frac{\pi }{9}\underset{0}{\overset{\infty }{\int }}\frac{dt}{{\left({\displaystyle \frac{2}{3}}\right)}^2+t^2} \\ =\frac{\pi }{9}{\left(\frac{3}{2}{\tan }^{-1}\left(\frac{3t}{2}\right)\right)}_0^{\infty } \\ =\frac{\mathrm{\pi }}{6}\left(\frac{\mathrm{\pi }}{2}-0\right)=\frac{{\mathrm{\pi }}^2}{12} \end{gathered}$$