9. (i) Show that the substitution \(u = \sin x\) transforms the integral $$\int \frac { 6 } { \cos x ( 2 - \sin x ) } d x$$ into the integral $$\int \frac { 6 } { \left( 1 - u ^ { 2 } \right) ( 2 - u ) } \mathrm { d } u$$ (ii) Express \(\frac { 6 } { \left( 1 - u ^ { 2 } \right) ( 2 - u ) }\) in partial fractions.
(iii) Hence, evaluate $$\int _ { 0 } ^ { \frac { \pi } { 6 } } \frac { 6 } { \cos x ( 2 - \sin x ) } d x$$ giving your answer in the form \(a \ln 2 + b \ln 3\), where \(a\) and \(b\) are integers.