Đặt $t=\sin^2x+2\cos^2x=1+\cos^2x \Rightarrow dt=-2\sin x\cos xdx=-\sin 2xdx$
Đổi cận: $x=0 \Rightarrow t=2$
$x=\dfrac{\pi}{2} \Rightarrow t=1$
Ta có:
$\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin 2x}{\sin^2x+2\cos^2x}dx$
$=\int\limits_2^1\dfrac{-dt}{t}$
$=\int\limits_1^2\dfrac{dt}{t}$
$=\ln t \left|\begin{array}{l}2\\1\end{array}\right.=\ln 2$