Đặt: $t=\sin^2x \Rightarrow dt=\sin2xdx$
Ta có:
$\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x}{\sqrt{\cos^2x+4\sin^2x}}dx$
$=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x}{\sqrt{1+3\sin^2x}}dx$
$=\int\limits_0^1\dfrac{dt}{\sqrt{1+3t}}$
$=\dfrac{1}{3}\int\limits_0^1(1+3t)^{\frac{-1}{2}}d(1+3t)$
$=\dfrac{2}{3}(1+3t)^{\frac{1}{2}}\left|\begin{array}{l}1\\0\end{array}\right.$
$=\dfrac{2}{3}$