Đặt: $\sqrt{3\cos x+1}=t \Rightarrow 3\cos x+1=t^2 \Rightarrow -3\sin x=2tdt$
Đổi cận: $x=0 \Rightarrow t=2$
$x=\dfrac{\pi}{2} \Rightarrow t=1$
Ta có:
$\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x+\sin x}{\sqrt{3\cos x+1}}dx$
$=\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x(2\cos x+1)}{\sqrt{3\cos x+1}}dx$
$=\dfrac{-2}{3}\int\limits_2^1\dfrac{\dfrac{2(t^2-1)}{3}+1}{t}tdt$
$=\dfrac{2}{3}\int\limits_1^2\left(\dfrac{2t^2}{3}+\dfrac{1}{3}\right)dt$
$=\left(\dfrac{4t^3}{27}+\dfrac{2t}{9}\right)\left|\begin{array}{l}2\\1\end{array}\right.=\dfrac{34}{27}$