Question

If\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1 - \sin 2x} \, dx = \alpha + \beta \sqrt{2} + \gamma \sqrt{3},\]where \( \alpha \), \( \beta \), and \( \gamma \) are rational numbers, then \( 3\alpha + 4\beta - \gamma \) is equal to ______.

Answer 1

Correct Answer: 6

Step 1. Evaluate the integral:
  \(\int_{\frac{\pi}{8}}^{\frac{\pi}{3}} \sqrt{1 - \sin 2x} \, dx\)
 
Step 2. Rewrite \( \sqrt{1 - \sin 2x} \) using trigonometric identities:
  \(\int_{\frac{\pi}{8}}^{\frac{\pi}{3}} |\sin x - \cos x| \, dx\)

Step 3. Split the integral based on the intervals where \( \sin x - \cos x \) changes sign:

  \(= \int_{\frac{\pi}{8}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx\)

Step 4. Solve each integral:
 \(= -1 + 2\sqrt{2} - \sqrt{3}\)
 Thus, we have:  
\(\alpha + \beta \sqrt{2} + \gamma \sqrt{3} = -1 + 2\sqrt{2} - \sqrt{3}\)

where \( \alpha = -1 \), \( \beta = 2 \), and \( \gamma = -1 \).

Step 5. Calculate \( 3\alpha + 4\beta - \gamma \):
 \(3\alpha + 4\beta - \gamma = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6\)
  
The Correct Answer is: \( 3\alpha + 4\beta - \gamma = 6 \).