${\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{1}{9\cos x+12\sin x}dx=}$

$I={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{1}{9Cosx+12Sinx}dx}$ 

  $={\displaystyle \underset{0}{\overset{\pi /2}{\int }}\frac{1.\left(\frac{2dt}{1+t^2}\right)}{9\left(\frac{1-t^2}{1+t^2}\right)+12\left(\frac{2t}{1+t^2}\right)}}$

=$$\begin{gathered} {\int }_0^1\frac{2 dt}{9-9t^2+24t} \\ =\frac{2}{9}{\int }_0^1\frac{dt}{{\left({\displaystyle \frac{5}{3}}\right)}^2-{\left(t-{\displaystyle \frac{4}{3}}\right)}^2} \\ ={{\left[\frac{2}{9}\frac{1}{2\left({\displaystyle \frac{5}{3}}\right)}\log \left(\frac{{\displaystyle \frac{5}{3}}-t+{\displaystyle \frac{4}{3}}}{{\displaystyle \frac{5}{3}}+t-{\displaystyle \frac{4}{3}}}\right)\right]}_0}^1 \end{gathered}$$

  $=\frac{1}{15}\log 6$