Mathematics - Definite Integration Question with Solution | TestHub

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, $3\alpha +4\beta -\gamma =-3+8+1=6$