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

Concept:

\(\int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C\)

Calculation:

Given, 

∫(cos x - sin x)/\(\rm \sqrt{8-\sin 2x)}\) dx = a sin-1(sin x + cos x)/b + c

Let I = ∫(cos x - sin x)/\(\rm \sqrt{8-\sin 2x)}\) dx 

Put sin x + cos x = t ⇒ (cos x -sin x ) dx = dt

Also, 1 + sin 2x = t2

⇒ I = ∫dt/√(8-(t2-1))

= ∫dt/(9-t2)

= sin-1 (t/3) + c

= sin-1(sin x + cos x)/3 + c = a sin-1(sin x + cos x)/b + c

⇒ a = 1 and b = 3

∴ The the ordered pair (a, b) is equal to (1, 3).

The correct answer is Option 2.