1. Key Concepts and Formulas

• Trigonometric Identities: The identity $1 - \sin 2x = (\sin x - \cos x)^2$ is crucial for simplifying the integrand. This identity stems from $\sin^2 x + \cos^2 x = 1$ and $\sin 2x = 2 \sin x \cos x$.
• Property of Square Roots: The fundamental property $\sqrt{A^2} = |A|$ must be applied to correctly handle the square root of a squared trigonometric expression.
• Absolute Value Integration: Integrals involving absolute values require splitting the integration interval based on the sign of the expression within the absolute value.
• Definite Integration Techniques: Evaluating definite integrals by finding antiderivatives and applying the limits of integration.

1. Step-by-Step Solution

Step 1: Simplify the Integrand We are asked to evaluate the integral I = \int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x. We use the trigonometric identities: $1 = \sin^2 x + \cos^2 x$ $\sin 2x = 2 \sin x \cos x$ Substituting these into the expression under the square root: $1 - \sin 2x = (\sin^2 x + \cos^2 x) - 2 \sin x \cos x$ This expression is a perfect square trinomial: $(\sin^2 x + \cos^2 x) - 2 \sin x \cos x = (\sin x - \cos x)^2$ So, the integrand becomes: $\sqrt{1 - \sin 2x} = \sqrt{(\sin x - \cos x)^2}$

Step 2: Apply the Absolute Value Property The square root of a squared term is the absolute value of that term: $\sqrt{(\sin x - \cos x)^2} = |\sin x - \cos x|$ Thus, the integral transforms to: I = \int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} |\sin x - \cos x| d x

Step 3: Determine the Sign of $\sin x - \cos x$ in the Interval To remove the absolute value, we need to find where $\sin x - \cos x$ is positive or negative within the interval $\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$. We set $\sin x - \cos x = 0$, which implies $\sin x = \cos x$. Dividing by $\cos x$ (which is non-zero in this interval), we get $\tan x = 1$. The principal value for this equation is $x = \frac{\pi}{4}$. Now we check the position of $\frac{\pi}{4}$ relative to the integration limits $\frac{\pi}{6}$ and $\frac{\pi}{3}$: $\frac{\pi}{6} = 30^\circ$, $\frac{\pi}{4} = 45^\circ$, $\frac{\pi}{3} = 60^\circ$. Since $\frac{\pi}{6} < \frac{\pi}{4} < \frac{\pi}{3}$, we must split the integral at $x = \frac{\pi}{4}$.

• For $x \in \left[\frac{\pi}{6}, \frac{\pi}{4}\right]$: In this interval, $x$ is less than $\frac{\pi}{4}$. For angles in the first quadrant, as $x$ increases from $0$ to $\frac{\pi}{4}$, $\sin x$ increases and $\cos x$ decreases. Thus, for $x < \frac{\pi}{4}$, $\sin x < \cos x$. This means $\sin x - \cos x < 0$. Therefore, $|\sin x - \cos x| = -(\sin x - \cos x) = \cos x - \sin x$.

• For $x \in \left[\frac{\pi}{4}, \frac{\pi}{3}\right]$: In this interval, $x$ is greater than or equal to $\frac{\pi}{4}$. For $x > \frac{\pi}{4}$, $\sin x > \cos x$. This means $\sin x - \cos x > 0$. Therefore, $|\sin x - \cos x| = \sin x - \cos x$.

Step 4: Split and Evaluate the Integral We split the integral into two parts based on the sign analysis: I = \int_\limits{\frac{\pi}{6}}^{\frac{\pi}{4}} (\cos x - \sin x) d x + \int_\limits{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) d x

Evaluate the first integral: \int_\limits{\frac{\pi}{6}}^{\frac{\pi}{4}} (\cos x - \sin x) d x = [\sin x + \cos x]_\limits{\frac{\pi}{6}}^{\frac{\pi}{4}} $= \left(\sin \frac{\pi}{4} + \cos \frac{\pi}{4}\right) - \left(\sin \frac{\pi}{6} + \cos \frac{\pi}{6}\right)$ $= \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - \left(\frac{1}{2} + \frac{\sqrt{3}}{2}\right)$ $= \frac{2}{\sqrt{2}} - \frac{1+\sqrt{3}}{2} = \sqrt{2} - \frac{1+\sqrt{3}}{2}$

Evaluate the second integral: \int_\limits{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) d x = [-\cos x - \sin x]_\limits{\frac{\pi}{4}}^{\frac{\pi}{3}} $= \left(-\cos \frac{\pi}{3} - \sin \frac{\pi}{3}\right) - \left(-\cos \frac{\pi}{4} - \sin \frac{\pi}{4}\right)$ $= \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}\right) - \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)$ $= -\frac{1+\sqrt{3}}{2} - \left(-\frac{2}{\sqrt{2}}\right) = -\frac{1+\sqrt{3}}{2} + \sqrt{2}$

Step 5: Combine the Results Add the values of the two integrals: $I = \left(\sqrt{2} - \frac{1+\sqrt{3}}{2}\right) + \left(\sqrt{2} - \frac{1+\sqrt{3}}{2}\right)$ $I = \sqrt{2} - \frac{1}{2} - \frac{\sqrt{3}}{2} + \sqrt{2} - \frac{1}{2} - \frac{\sqrt{3}}{2}$ Combine like terms: $I = (\sqrt{2} + \sqrt{2}) + \left(-\frac{1}{2} - \frac{1}{2}\right) + \left(-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right)$ $I = 2\sqrt{2} - 1 - \sqrt{3}$

Step 6: Identify $\alpha, \beta, \gamma$ The problem states that $I = \alpha+\beta \sqrt{2}+\gamma \sqrt{3}$, where $\alpha, \beta, \gamma$ are rational numbers. Comparing $I = -1 + 2\sqrt{2} - \sqrt{3}$ with the given form: $\alpha = -1$ $\beta = 2$ $\gamma = -1$ These are indeed rational numbers.

Step 7: Calculate the Final Expression We need to find the value of $3 \alpha+4 \beta-\gamma$. Substitute the determined values: $3 \alpha+4 \beta-\gamma = 3(-1) + 4(2) - (-1)$ $= -3 + 8 + 1$ $= 5 + 1$ $= 6$

1. Common Mistakes & Tips

• Forgetting Absolute Value: The most common error is to treat $\sqrt{A^2}$ as $A$ instead of $|A|$. This leads to incorrect sign analysis and an incorrect final answer.
• Incorrect Interval Sign Analysis: Double-check the sign of $\sin x - \cos x$ in each sub-interval. Visualizing the graphs of $\sin x$ and $\cos x$ or considering the properties of $\tan x$ can help.
• Arithmetic Errors: Be extremely careful with fractions, square roots, and signs during the evaluation of the definite integrals and the final calculation.

1. Summary The integral \int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x was evaluated by first simplifying the integrand to $|\sin x - \cos x|$ using trigonometric identities. The integral was then split at $x=\frac{\pi}{4}$ to handle the absolute value. After evaluating the two definite integrals and combining the results, we obtained $I = -1 + 2\sqrt{2} - \sqrt{3}$. By comparing this with the given form $\alpha+\beta \sqrt{2}+\gamma \sqrt{3}$, we found $\alpha=-1$, $\beta=2$, and $\gamma=-1$. Finally, the expression $3 \alpha+4 \beta-\gamma$ was calculated to be $3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6$.

The final answer is \boxed{6}.