$I=\int\limits_{0}^{\pi/2}\frac{sìn 2x}{\sqrt{cos^2x+4sin^2x}}dx=\int\limits_{0}^{\pi/2}\frac{sìn2x}{\sqrt{\left ( \frac{cos2x+1}{2} \right )+4\left ( \frac{1-cos2x}{2} \right )}}dx$$=\int\limits_{0}^{\pi/2}\frac{sìn2x}{\sqrt{-\frac{3}{2}cos2x+\frac{5}{2}}}dx$
Đặt $t=\sqrt{-\frac{3}{2}cos2x+\frac{5}{2}}\Rightarrow t^2=-\frac{3}{2}cos2x+\frac{5}{2}\Rightarrow 2tdt=3sin2xdx\Rightarrow sin2xdx=\frac{2}{3}tdt$
Đổi cận: $x=0\Rightarrow t=1,x=\pi/2\Rightarrow t=2$
$I=\int\limits_{1}^{2}\frac{2}{3}t\frac{1}{t}dt=\int\limits_{2}^{3}dt=\frac{2}{3}$