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

To solve the integral \(\int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} \, dx\) in terms of the given expression \(a \sin^{-1} \left( \frac{\sin x + \cos x}{b} \right) + c\), we follow these steps:

1. Recognize that the key to solving this integral is transforming it in terms of a function whose derivative can be easily integrated.
2. Recall the trigonometric identity \(\sin 2x = 2 \sin x \cos x\). Therefore, the integral becomes: \(\int \frac{\cos x - \sin x}{\sqrt{8 - 2 \sin x \cos x}} \, dx\)
3. Notice that \((\sin x + \cos x)^2 = \sin^2 x + 2 \sin x \cos x + \cos^2 x = 1 + \sin 2x\). Hence: \(\sin x + \cos x = \sqrt{1 + \sin 2 x}\)
4. Rearrange to find b such that the term inside the inverse sine is \(\frac{\sin x + \cos x}{b}\). We have: \(\frac{\sin x + \cos x}{\sqrt{3}}\) since \(\left( \frac{\sin x + \cos x}{\sqrt{3}} \right)^2 = \frac{1 + \sin 2x}{3}\)
5. For the integral to match the term \(a \sin^{-1} \left( \frac{\sin x + \cos x}{b} \right) + c\), compare and find \(a = 1\) and \(b = 3\).
6. Thus, the values for a and b that satisfy the given condition are (1, 3).

Therefore, the correct answer is (1, 3).