Đặt: $t=\sqrt{1+3\cos x} \Rightarrow t^2=1+3\cos x\Rightarrow 2tdt=-3\sin xdx$
Ta có:
$\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx$
$=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x(2\cos x+1)}{\sqrt{1+3\cos x}}dx$
$=-\dfrac{2}{3}\int\limits_2^1\dfrac{2\dfrac{t^2-1}{3}+1}{t}tdt$
$=\dfrac{2}{9}\int\limits_1^2(2t^2+1)dt$
$=\dfrac{2}{9}\left(\dfrac{2t^3}{3}+t\right)\left|\begin{array}{l}2\\1\end{array}\right.$
$=\dfrac{34}{27}$