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.$$=\dfrac{34}{27}$
Đặt: $t=\sqrt{1+3\cos x} \Rightarrow t^2=1+3\cos x\Rightarrow 2tdt=-3\sin xdx$Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x+\sin x}{\sqrt{1+3\cos x}}dx$$=\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{44}{45}$
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}$