Question:
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 :
Correct Option: 3,
Solution:
$\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2 x}} d x$
$=\int \frac{\cos x-\sin x}{\sqrt{9-(\sin x+\cos x)^{2}}} d x$
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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