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

$\begin{array}{l}a{\mathrm{Sin}}^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c=\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx=\int \frac{\cos x-\sin x}{\sqrt{9-(\cos x+\sin x{)}^2}}dx\\ =\int \frac{dt}{\sqrt{9-t^2}}\text{ where }t=\cos x+\sin x={\mathrm{Sin}}^{-1}\frac{t}{3}+c={\mathrm{Sin}}^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c\\ \Rightarrow a=1,b=3\Rightarrow (a,b)=(1,3).\end{array}$