If ∫ π 6 π 3 1 - sin 2 x d x = α + β 2 + γ 3 , where α , β and γ are rational numbers, then 3 α + 4 β - γ is…

Given,

${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\sqrt{1-\sin 2x}dx=\alpha +\beta \sqrt{2}+\gamma \sqrt{3}$

Now, let $I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\sqrt{1-\sin 2x}dx$

$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}|\sin x-\cos x|dx$

$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}(\cos x-\sin x)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}(\sin x-\cos x)dx$

$\Rightarrow I=-1+2\sqrt{2}-\sqrt{3}$

Now, comparing with 

$I=\alpha +\beta \sqrt{2}+\gamma \sqrt{3}$

We get, $\alpha =-1,\beta =2,\gamma =-1$

Hence, $3\alpha +4\beta -\gamma =-3+8+1=6$