If \int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx = a{\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over b}} \right) + c, where c is a constant of integration, then the ordered pair (a, b) is equal to :

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Answer: B. (1, 3) 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} ight]} }}} dx = \int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x - 1} ight]} }}} dx = \int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\left( {\sin x + \cos x} ight)}^2} - 1} ight]} }}} dx = \int {{{\cos x - \sin x} \over {\sqrt {9 - {{\left( {\sin x + \cos x} ight)}^2}} }}} dx put sin x + cos x = t \Rightarrow (cos x – sin x) dx = dt = \int {{{dt} \over {\sqrt {9 - {{\left( t ight)}^2}} }}} = {\sin ^{ - 1}}\left( {{t \over 3}} ight) + C = {\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over 3}} ight) + C 	herefore a = 1 and b = 3

If $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx = a{\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over b}} \right) + c$$, where c is a constant of integration, then the ordered pair (a, b) is equal to :

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