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

We need to evaluate $$\int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} \, dx$$.

Let $$u = \sin x + \cos x$$. Then $$du = (\cos x - \sin x) \, dx$$, which is exactly the numerator.

Also, $$u^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x$$, so $$\sin 2x = u^2 - 1$$.

Substituting: $$8 - \sin 2x = 8 - (u^2 - 1) = 9 - u^2$$.

The integral becomes $$\int \frac{du}{\sqrt{9 - u^2}}$$.

This is a standard form: $$\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}\left(\frac{u}{a}\right) + c$$, with $$a = 3$$.

So the integral is $$\sin^{-1}\left(\frac{\sin x + \cos x}{3}\right) + c$$.

Comparing with $$a\sin^{-1}\left(\frac{\sin x + \cos x}{b}\right) + c$$, we get $$a = 1$$ and $$b = 3$$.

Hence, the correct answer is Option D.