chanhhoang1328

Ta có:
${I_6} = \int {\frac{{dx}}{{4\cos x + 3\sin x + 5}} = \int {\frac{{dx}}{{5\sin \left( {x + \alpha } \right)}}} } $
Với:
$\left\{ \begin{array}{l}
\sin \alpha = \frac{4}{5}\\
\cos \alpha = \frac{3}{5}
\end{array} \right.$
${I_6} = \frac{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} } $

5+sin(x+a) chứ , mình ko hiểu