Cho \(I = \int {\dfrac{{\sin 2x + \sin x}}{{\sqrt {1 + 3\cos x} }}} dx = F\left( x \right) \). Giá trị của $F\left( {\dfrac{\pi }{2}} \right) - F\left( 0 \right)$ là:

Đặt \(t = \sqrt {1 + 3\cos x} \) \( \Rightarrow {t^2} - 1 = 3\cos x \Rightarrow 2tdt =  - 3\sin xdx\)

Lại có: \(\sin 2x + \sin x = 2\sin x\cos x + \sin x = \left( {2\cos x + 1} \right)\sin x\)

Do đó \(\dfrac{{\sin 2x + \sin x}}{{\sqrt {1 + 3\cos x} }}dx = \dfrac{{\left( {2\cos x + 1} \right)\sin xdx}}{{\sqrt {1 + 3\cos x} }}\) \( = \dfrac{{\left( {2.\dfrac{{{t^2} - 1}}{3} + 1} \right).\dfrac{{ - 2tdt}}{3}}}{t} =  - \dfrac{2}{9}\left( {2{t^2} + 1} \right)dt\)

$ \Rightarrow I =  - \int {\dfrac{2}{9}\left( {2{t^2} + 1} \right)dt}  =  - \dfrac{2}{9}\left( {\dfrac{{2{t^3}}}{3} + t} \right) + C$$ =  - \dfrac{2}{9}\left( {\dfrac{{2\sqrt {{{\left( {1 + 3\cos x} \right)}^3}} }}{3} + \sqrt {1 + 3\cos x} } \right) + C$

$ \Rightarrow F\left( x \right) =  - \dfrac{2}{9}\left( {\dfrac{{2\sqrt {{{\left( {1 + 3\cos x} \right)}^3}} }}{3} + \sqrt {1 + 3\cos x} } \right) + C$

$ \Rightarrow F\left( {\dfrac{\pi }{2}} \right) - F\left( 0 \right) = \dfrac{{34}}{{27}}$