Evaluate:
${\int }_0^{\mathrm{\pi }} \frac{\mathrm{x}}{9{\sin }^2\mathrm{x}+16{\cos }^2\mathrm{x}}\mathrm{dx}$

Assume,
$\mathrm{I}={\int }_0^{\mathrm{\pi }} \frac{\mathrm{x}}{9{\sin }^2\mathrm{x}+16{\cos }^2\mathrm{x}}\mathrm{dx}\dots$
Now, recall that,
${\int }_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_{\mathrm{a}}^{\mathrm{b}} \mathrm{f}(\mathrm{a}+\mathrm{b}-\mathrm{x})\mathrm{dx}$
Using that for given integral,
$\begin{array}{l}\therefore \mathrm{I}={\int }_0^{\mathrm{\pi }} \frac{\mathrm{\pi }-\mathrm{x}}{9{\sin }^2(\mathrm{\pi }-\mathrm{x})+16{\cos }^2(\mathrm{\pi }-\mathrm{x})}\mathrm{dx}\\ \therefore \mathrm{I}={\int }_0^{\mathrm{\pi }} \frac{\mathrm{\pi }-\mathrm{x}}{9{\sin }^2\mathrm{x}+16{\cos }^2\mathrm{x}}\mathrm{dx}\dots \dots .\left(2\right)\end{array}$
Adding equation (1) and (2)
$\begin{array}{l}2\mathrm{I}={\int }_0^{\mathrm{\pi }} \frac{\mathrm{\pi }}{9{\sin }^2\mathrm{x}+16{\cos }^2\mathrm{x}}\mathrm{dx}\\ \Rightarrow \mathrm{I}=\mathrm{\pi }{\int }_0^{\frac{\mathrm{\pi }}{2}} \frac{1}{9{\sin }^2\mathrm{x}+16{\cos }^2\mathrm{x}}\mathrm{dx}\end{array}$(using property)
$\begin{array}{l}\mathrm{I}=\mathrm{\pi }\left[{\int }_0^{\mathrm{\pi }/4} \frac{{\sec }^2\mathrm{x}}{9{\tan }^2\mathrm{x}+16}\mathrm{dx}+{\left.{\int }_{\mathrm{\pi }/4}^{\mathrm{\pi }/2} \frac{{\mathrm{cosec}}^2\mathrm{x}}{9+16{\cot }^2\mathrm{x}}\mathrm{dx}\right|}_{\rfloor }=\mathrm{\pi }\left[{\mathrm{I}}_1+{\mathrm{I}}_2\right]\right.\text{ (say) }\\ =\mathrm{\pi }\left[{\int }_0^1 \frac{\mathrm{dt}}{9{\mathrm{t}}^2+16}-{\left.{\int }_1^0 \frac{\mathrm{dz}}{9+16{\mathrm{z}}^2}\right|}^{\mathrm{\top }}\right.⌋\left(\mathrm{t}=\tan x \mathrm{in}{\mathrm{I}}_1,\mathrm{z}=\cot x \mathrm{in}{\mathrm{I}}_2\right)\\ =\frac{\mathrm{\pi }}{12}\left\{\left\{{\left[{\left.{\tan }^{-1}\frac{3\mathrm{t}}{4}\right|}_{{\rfloor }_0}^1-{{\left[{\tan }^{-1}\frac{4\mathrm{z}}{3}\right]}^0}_1\right]}_1\right)\right.\\ =\frac{\mathrm{\pi }}{12}\left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{4}{3}\right)\end{array}$
or $\frac{\mathrm{\pi }}{12}\times \frac{\mathrm{\pi }}{2}=\frac{{\mathrm{\pi }}^2}{24}$
Therefore, ${\int }_0^{\mathrm{\pi }} \frac{\mathrm{x}}{9{\sin }^2\mathrm{x}+16{\cos }^2\mathrm{x}}\mathrm{dx}=\frac{\mathrm{\pi }}{12}\left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{4}{3}\right).$ Which is the required answer.