CÂU HỎI: Cho \( I = \int {\frac{{\sin 2x + \sin x}}{{\sqrt {1 + 3\cos x} }}dx} =F(x)\). Giá trị của \( F(\frac{\pi }{2}) – F(0)\)

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

\(\begin{array}{l}
\frac{{\sin 2x + \sin x}}{{\sqrt {1 + 3\cos x} }}dx = \frac{{\left( {2\cos x + 1} \right)\sin xdx}}{{\sqrt {1 + 3\cos x} }} = \frac{{\left( {2.\frac{{{t^2} – 1}}{3} + 1} \right).\frac{{ – 2tdt}}{3}}}{t} = – \frac{2}{9}\left( {2{t^2} + 1} \right)\\
\Rightarrow I = – \smallint \frac{2}{9}\left( {2{t^2} + 1} \right)dt = – \frac{2}{9}\left( {\frac{{2{t^3}}}{3} + t} \right) + C = – \frac{2}{9}\left( {\frac{{2\sqrt {{{\left( {1 + 3\cos x} \right)}^3}} }}{3} + \sqrt {1 + 3\cos x} } \right) + C\\
\Rightarrow F\left( x \right) = – \frac{2}{9}\left( {\frac{{2\sqrt {{{\left( {1 + 3\cos x} \right)}^3}} }}{3} + \sqrt {1 + 3\cos x} } \right) + C\\
\Rightarrow F\left( {\frac{\pi }{2}} \right) – F\left( 0 \right) = \frac{{34}}{{27}}
\end{array}\)