\[{I_7} = \int\limits_0^{\dfrac{\pi }{2}} {\frac{{2\sin x\cos x + \sin x}}{{\sqrt {1 + 3\cos x} }}dx = \int\limits_0^{\dfrac{\pi }{2}} {\frac{{\sin x\left( {2\cos x + 1} \right)}}{{\sqrt {1 + 3\cos x} }}} } dx\]
Đặt $\left\{ \begin{array}{l} u = 2\cos x + 1\\ dv = \dfrac{{\sin x}}{{\sqrt {1 + 3\cos x} }}dx = - \dfrac{{d\left( {1 + 3\cos x} \right)}}{{3\sqrt {1 + 3\cos x} }} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = - 2\sin xdx\\ v = - \dfrac{2}{3}\sqrt {1 + 3\cos x} \end{array} \right.$
Khi đó:
\[{I_7} = \left. { - \frac{2}{3}\left( {2\cos x + 1} \right)\sqrt {1 + 3\cos x} } \right|_0^{\dfrac{\pi }{2}} - \dfrac{4}{3}\int\limits_0^{\dfrac{\pi }{2}} {\sin x\sqrt {1 + 3\cos x} } dx\]
\[= \frac{2}{3} + \frac{4}{9}\int\limits_0^{\dfrac{\pi }{2}} {\sqrt {1 + 3\cos x} } d\left( {1 + 3\cos x} \right) = \frac{2}{3} + \left. {\dfrac{8}{{27}}\sqrt {{{\left( {1 + 3\cos x} \right)}^3}} } \right|_0^{\dfrac{\pi }{2}} = \dfrac{{34}}{{27}}\]
