Official solution
Given \[\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx\]
Write sin2x = 1 + sin2x - 1
= \[\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {1 + \sin 2x - 1} \right]} }}} dx\]
= \[\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x - 1} \right]} }}} dx\]
= \[\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\left( {\sin x + \cos x} \right)}^2} - 1} \right]} }}} dx\]
= \[\int {{{\cos x - \sin x} \over {\sqrt {9 - {{\left( {\sin x + \cos x} \right)}^2}} }}} dx\]
put sin x + cos x = t
\[ \Rightarrow \] (cos x -- sin x) dx = dt
= \[\int {{{dt} \over {\sqrt {9 - {{\left( t \right)}^2}} }}} \]
= \[{\sin ^{ - 1}}\left( {{t \over 3}} \right)\] + C
= \[{\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over 3}} \right) + C\]
\[ \therefore \] a = 1 and b = 3