Question

If $\int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} dx = a \sin^{-1} \left( \frac{\sin x + \cos x}{b} \right) + c$, find (a, b) :

• A. (3, 1)
• B. (1, 3)
• C. (-1, 3)
• D. (1, -3)

Hint:

For integrals involving $(\sin x \pm \cos x)$ and $\sin 2x$, use the substitution $(\sin x \mp \cos x)$ to simplify the radical.

Answer 1

The Correct Option is B

Step 1: Substitute $t = \sin x + \cos x \Rightarrow dt = (\cos x - \sin x) dx$.
Step 2: $\sin 2x = t^2 - 1$.
Step 3: $\int \frac{dt}{\sqrt{8 - (t^2-1)}} = \int \frac{dt}{\sqrt{9-t^2}} = \sin^{-1}(t/3) + c$.
Step 4: Comparing with the format, $a=1, b=3$.