\[{I_6} = \int {\frac{{dx}}{{4\cos x + 3\sin x + 5}} = \int {\frac{{dx}}{{5\sin \left( {x + \alpha } \right)}}} }\]
Với $\left\{ \begin{array}{l} \sin \alpha = \dfrac{4}{5}\\ \cos \alpha = \dfrac{3}{5} \end{array} \right.$
\[{I_6} = \dfrac{1}{5}\int {\frac{{\sin \left( {x + \alpha } \right)dx}}{{{{\sin }^2}\left( {x + \alpha } \right)}} = \frac{1}{5}\int {\frac{{d\left( {\cos \left( {x + \alpha } \right)} \right)}}{{{{\cos }^2}\left( {x + \alpha } \right) - 1}} = \frac{1}{{10}}\ln \left| {\frac{{\cos \left( {x + \alpha } \right) - 1}}{{\cos \left( {x + \alpha } \right) + 1}}} \right| + C} }\]
