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: \( 3\alpha + 4\beta - \gamma = 6 \).
The integral can be rewritten as: \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1 - \sin 2x} \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left| \sin x - \cos x \right| \, dx \] We split the absolute value function into two parts, as \( \sin x - \cos x \) changes its sign at \( x = \frac{\pi}{4} \).
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left| \sin x - \cos x \right| \, dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx \]
We compute the two integrals separately. For the first part: \[ \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx = -1 + 2\sqrt{2} - \sqrt{3} \] For the second part: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx = \alpha + \beta \sqrt{2} + \gamma \sqrt{3} \] where we are given: \[ \alpha = -1, \quad \beta = 2, \quad \gamma = -1 \]
Now, we compute: \[ 3\alpha + 4\beta - \gamma = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6 \]
The final value is: \[ \boxed{6} \]
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