If ∫ _ π 6^ π 3√1- sin2xdx=α +β √2+γ √3, where α ,β and γ are rational numbers, then 3α +4β -γ is equal to .

Given,

$${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\sqrt{1-\mathrm{sin}2x}dx=\alpha +\beta \sqrt{2}+\gamma \sqrt{3}$$∫6π​3π​​1−sin2x​dx=α+β2​+γ3​

Now, let $\displaystyle I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\sqrt{1-\mathrm{sin}2x}dx$I=∫6π​3π​​1−sin2x​dx

$$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}|\mathrm{sin}x-\mathrm{cos}x|dx$$⇒I=∫6π​3π​​∣sinx−cosx∣dx

$$\Rightarrow I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{4}}(\mathrm{cos}x-\mathrm{sin}x)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{3}}(\mathrm{sin}x-\mathrm{cos}x)dx$$⇒I=∫6π​4π​​(cosx−sinx)dx+∫4π​3π​​(sinx−cosx)dx

$$\Rightarrow I=-1+2\sqrt{2}-\sqrt{3}$$⇒I=−1+22​−3​

Now, comparing with

$$I=\alpha +\beta \sqrt{2}+\gamma \sqrt{3}$$I=α+β2​+γ3​

We get, $\alpha =-1,\beta =2,\gamma =-1$α=−1,β=2,γ=−1

Hence, $3\alpha +4\beta -\gamma =-3+8+1=6$3α+4β−γ=−3+8+1=6