Đặt \(t = 1 + 4\sin x \Rightarrow dt = 4\cos xdx\) \( \Rightarrow \cos xdx = \dfrac{{dt}}{4}\) và \(\sin x = \dfrac{{t - 1}}{4}\)
\(\begin{array}{l} \Rightarrow I = \int {\dfrac{{2\sin x\cos xdx}}{{1 + 4\sin x}}} = \int {\dfrac{{2.\dfrac{{t - 1}}{4}.\dfrac{{dt}}{4}}}{t}} \\ = \dfrac{1}{8}\int {\dfrac{{t - 1}}{t}dt} = \dfrac{1}{8}\int {\left( {1 - \dfrac{1}{t}} \right)dt} = \dfrac{1}{8}\left( {t - \ln \left| t \right|} \right) + C\\ = \dfrac{1}{8}\left( {1 + 4\sin x - \ln \left| {1 + 4\sin x} \right|} \right) + C\end{array}\)
