$$\eqalign{ & \int {\frac{{{{\sin }^2}x}}{{\cos x\left( {3 + {{\sin }^2}x} \right)}}dx} = \int {\frac{{{{\sin }^2}x\cos x}}{{{{\cos }^2}x\left( {3 + {{\sin }^2}x} \right)}}dx} = \int {\frac{{{{\sin }^2}x}}{{{{\cos }^2}x\left( {3 + {{\sin }^2}x} \right)}}d\sin x} = \cr & \int {\frac{{{{\sin }^2}x}}{{\left( {1 - {{\sin }^2}x} \right)\left( {3 + {{\sin }^2}x} \right)}}d\sin x} = \int {\frac{{{t^2}}}{{\left( {1 - {t^2}} \right)\left( {3 + {t^2}} \right)}}dt} = \cr & \frac{1}{8}\ln \left| {\frac{{t + 1}}{{t - 1}}} \right| - \frac{{\sqrt 3 }}{4}\operatorname{arctg} \frac{t}{{\sqrt 3 }} + C = \frac{1}{8}\ln \left| {\frac{{\sin x + 1}}{{\sin x - 1}}} \right| - \frac{{\sqrt 3 }}{4}\operatorname{arctg} \frac{{\sin x}}{{\sqrt 3 }} + C \cr} $$

$%\frac{1}{8}\ln \left| {\frac{{\sin x + 1}}{{\sin x - 1}}} \right| - \frac{{\sqrt 3 }}{4}\operatorname{arctg} \frac{{\sin x}}{{\sqrt 3 }} + C$%