\[\int_{0}^{\frac{\pi }{3}}\frac{2sin2x+3sinx}{\sqrt{6cosx-2}}dx\]
\[2sin2x + 3sinx = sinx(4cosx + 3)\]
Đặt
\[t = \sqrt{6cosx - 2} => dt = \frac{-3sinx}{\sqrt{6cosx - 2}}dx\]
\[ => \frac{sinx}{\sqrt{6cosx - 2}}dx = \frac{-dt}{3}\]
\[cosx = \frac{t^2 + 2}{6} => 4cosx + 3 = \frac{2t^2+13}{3}\]
\[x = 0 => t = 2\]
\[x = \frac{\pi}{3} => t = 1\]
\[I = \frac{-1}{9}\int_{2}^{1}(2t^2+13})dt = ... = \frac{-53}{27}\]