Solve this following


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. $(-1,3)$

  2. $(3,1)$

  3. $(1,3)$

  4. $(1,-3)$

Correct Option: 3,


$\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x$

$=\int \frac{\cos x-\sin x}{\sqrt{9-(\sin x+\cos x)^{2}}} d x$

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

$\int \frac{\mathrm{dt}}{\sqrt{9-\mathrm{t}^{2}}}=\sin ^{-1} \frac{\mathrm{t}}{3}+\mathrm{c}$

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

So $a=1, b=3$


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