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

(1, - 3)

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(1, 3)

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(- 1, 3)

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(3, 1)

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Solution

The correct option is B $(1, 3)$
Let $sin x + cos x = t$
then $1 + sin 2 x = t^{2}$
$\Rightarrow (cos x - sin x) d x = d t$
$\Rightarrow I = \int \frac{d t}{\sqrt{8 - (t^{2} - 1)}}$
$= \int \frac{d t}{\sqrt{9 - t^{2}}}$
$= {sin}^{- 1} (\frac{t}{3}) + c = (\frac{sin x + cos x}{3}) + c$
$\Rightarrow a = 1$ and $b = 3$

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