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$\int \frac{(\cos x - \sin x)}{\sqrt{8 - \sin 2 x}} d x = a \sin^{- 1} (\frac{\sin x + \cos x}{b}) + c$ where ‘c’ is a constant of integration then the ordered pair(a,b) equal to

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a

(1,-3)

b

(-1,3)

c

(3,1)

d

(1,3)

answer is D.

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Detailed Solution

$\begin{array}{l} \sin x + \cos x = t \\ 1 + \sin 2 x = t^{2} \end{array} (\cos x - \sin x) d x = d t I = \int \frac{d t}{\sqrt{8 - (t^{2} - 1)}} = \int \frac{d t}{\sqrt{9 - t^{2}}} \Rightarrow \sin^{- 1} (t / 3) + c \Rightarrow \sin^{- 1} (\frac{\sin x + \cos x}{3}) + c$

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