$\underset{0}{\overset{\pi /2}{\int }}\left(\frac{{\displaystyle 8+7\cos x}}{{\displaystyle {(7+8\cos x)}^2}}\right)dx$

$$\begin{gathered} {\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{8+7 \cos \mathrm{x}}{{(7+8 \cos \mathrm{x})}^2}\mathrm{dx} \\ ={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{8 {\mathrm{cosec}}^2\mathrm{x}+7\mathrm{cosec} \mathrm{x} \cot \mathrm{x}}{{(7 \mathrm{cosec} \mathrm{x}+8 \cot \mathrm{x})}^2} \\ ={\int }_{\infty }^7\frac{-\mathrm{dt}}{{\mathrm{t}}^2}, \mathrm{Put} 7 \mathrm{cosec} \mathrm{x}+8\cot \mathrm{x}=\mathrm{t} \\ ={\left(\frac{1}{\mathrm{t}}\right)}_{\infty }^7 \\ =\frac{1}{7}-0=\frac{1}{7} \end{gathered}$$