tìm nguyên hàm của I

$\begin{array}{l}
I = \int {\dfrac{{dx}}{{\sin x\left( {{{\cos }^3}x - 1} \right)}}} \\
 = \int {\dfrac{{ - \sin xdx}}{{ - {{\sin }^2}x\left( {{{\cos }^3}x - 1} \right)}}} \\
 = \int {\dfrac{{ - \sin xdx}}{{\left( {{{\cos }^2}x - 1} \right)\left( {{{\cos }^3}x - 1} \right)}}} \\
 = \int {\dfrac{{d\left( {\cos x} \right)}}{{{{\left( {\cos x - 1} \right)}^2}\left( {\cos x + 1} \right)\left( {{{\cos }^2}x + \cos x + 1} \right)}}} \\
\cos x = t \Rightarrow d\left( {\cos x} \right) = dt\\
 = \int {\left( {\dfrac{1}{{3\left( {{t^2} + t + 1} \right)}} + \dfrac{1}{{4\left( {t + 1} \right)}} - \dfrac{1}{{4\left( {t - 1} \right)}} + \dfrac{1}{{6{{\left( {t - 1} \right)}^2}}}} \right)dt} \\
 = \dfrac{1}{3}\int {\dfrac{1}{{{t^2} + t + 1}}dt + \int {\dfrac{1}{{4\left( {t + 1} \right)}}dt - \dfrac{1}{4}\int {\dfrac{1}{{t - 1}}dt + \dfrac{1}{6}\int {\dfrac{1}{{{{\left( {t - 1} \right)}^2}}}} dt} } } \\
 = \dfrac{1}{3}{I_1} + \dfrac{1}{4}\ln \left| {t + 1} \right| - \dfrac{1}{4}\ln \left| {t - 1} \right| - \dfrac{1}{6}.\dfrac{1}{{t - 1}} + C\\
 = \dfrac{1}{3}{I_1} + \dfrac{1}{4}\ln \left| {\dfrac{{t + 1}}{{t - 1}}} \right| - \dfrac{1}{6}.\dfrac{1}{{t - 1}} + C\\
{I_1} = \int {\dfrac{{dt}}{{{t^2} + t + 1}} = \int {\dfrac{{dt}}{{{{\left( {t + \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}}} } \\
t + \dfrac{1}{2} = \dfrac{{\sqrt 3 }}{2}\tan u \Rightarrow dt = \dfrac{{\sqrt 3 }}{2}\left( {{{\tan }^2}u + 1} \right)du\\
 \Rightarrow {I_1} = \int {\dfrac{{\dfrac{{\sqrt 3 }}{2}\left( {{{\tan }^2}u + 1} \right)du}}{{\dfrac{3}{4}\left( {{{\tan }^2}u + 1} \right)}} = \int {\dfrac{{2\sqrt 3 }}{3}du = \dfrac{{2\sqrt 3 }}{3}u + {C_1} = \dfrac{{2\sqrt 3 }}{3}.\arctan \left( {\dfrac{{t + \dfrac{1}{2}}}{{\dfrac{{\sqrt 3 }}{2}}}} \right)}  + {C_1}} \\
 \Rightarrow I = \dfrac{{2\sqrt 3 }}{9}\arctan \left( {\dfrac{{\cos x + \dfrac{1}{2}}}{{\dfrac{{\sqrt 3 }}{2}}}} \right) + \dfrac{1}{4}\ln \left| {\dfrac{{\cos x + 1}}{{\cos x - 1}}} \right| - \dfrac{1}{6}.\dfrac{1}{{\cos x - 1}} + C
\end{array}$