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

  1. (1) $(1,-3)$

  2. (2) $(1,3)$

  3. (3) $(-1,3)$

  4. (4) $(3,1)$

Correct Option: , 2


Put $\sin x+\cos x=t$

$\Rightarrow 1+\sin 2 x=t^{2}$

$\Rightarrow(\cos x-\sin x) d x=d t$

$\therefore I=\int \frac{d t}{\sqrt{8-\left(t^{2}-1\right)}}=\int \frac{d t}{\sqrt{9-t^{2}}}$

$=\sin ^{-1}\left(\frac{t}{3}\right)+C=\sin ^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c$

$\Rightarrow a=1$ and $b=3$

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