If _( )/(6)^( )/(3)√1- 2xdx= + √2+ √3, where , and are rational numbers, then 3 +4 - is eq…

Question

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 \_\_\_\_\_\_\_\_\_.

Answer: 6

Accepted Answer

$\begin{gathered} =\int _{\frac{\pi }{6}}^{\frac{\pi }{3}}\sqrt{1-\sin 2x}dx \ =\int _{\frac{\pi }{6}}^{\frac{\pi }{3}}|\sin x-\cos x|dx \ =\int _{\frac{\pi }{6}}^{\frac{\pi }{4}}(\cos x-\sin x)dx+\int _{\frac{\pi }{4}}^{\frac{\pi }{3}}(\sin x-\cos x)dx \ =-1+2\sqrt{2}-\sqrt{3} \ =\alpha +\beta \sqrt{2}+\gamma \sqrt{3} \ \alpha =-1,\beta =2,\gamma =-1 \ 3\alpha +4\beta -\gamma =6 \end{gathered}$

Solutions · 1