If int_{frac{pi}{6}}^{frac{pi}{3}} sqrt{1-sin | vedclass

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(D) We know that $1 - \sin 2x = \cos^2 x + \sin^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2$.Thus,$\sqrt{1 - \sin 2x} = |\cos x - \sin x|$.In the inte

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$6$

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(D) We know that $1 - \sin 2x = \cos^2 x + \sin^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2$.Thus,$\sqrt{1 - \sin 2x} = |\cos x - \sin x|$.In the interval $[\frac{\pi}{6}, \frac{\pi}{4}]$,$\cos x \ge \sin x$,so $|\cos x - \sin x| = \cos x - \sin x$.In the interval $[\frac{\pi}{4}, \frac{\pi}{3}]$,$\sin x \ge \cos x$,so $|\cos x - \sin x| = \sin x - \cos x$.Therefore,the integral becomes:$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\cos x - \sin x) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} (\sin x - \cos x) \, dx$$I = [\sin x + \cos x]_{\frac{\pi}{6}}^{\frac{\pi}{4}} + [-\cos x - \sin x]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$$I = ((\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (\frac{1}{2} + \frac{\sqrt{3}}{2})) + ((- \frac{1}{2} - \frac{\sqrt{3}}{2}) - (- \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}))$$I = (\sqrt{2} - \frac{1}{2} - \frac{\sqrt{3}}{2}) + (\sqrt{2} - \frac{1}{2} - \frac{\sqrt{3}}{2}) = 2\sqrt{2} - 1 - \sqrt{3}$.Comparing with $\alpha + \beta \sqrt{2} + \gamma \sqrt{3}$,we get $\alpha = -1, \beta = 2, \gamma = -1$.Then $3\alpha + 4\beta - \gamma = 3(-1) + 4(2) - (-1) = -3 + 8 + 1 = 6$.

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 ..........

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