If ∫ _ (π)/(6) ^ (π)/(3) 1 - 2x dx = α + β 2 + γ 3 , where α, β and...

Given,

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

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

$⇒ I = {∫}_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} | \sin x - \cos x | dx$

$⇒ I = {∫}_{\dfrac{\pi}{6}}^{\dfrac{\pi}{4}} ( \cos x - \sin x ) dx + {∫}_{\dfrac{\pi}{4}}^{\dfrac{\pi}{3}} ( \sin x - \cos x ) dx$

$⇒ I = - 1 + 2\sqrt{2} - \sqrt{3}$

Now, comparing with 

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

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

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