sinx ^2 (1+tanx)=(1-cosx ^2)(1+tanx)=1+tanx- cosx ^2 - sinx cosx
=1/2+tanx- cos2x /2 –sin2x /2
∫sin^2x(1+tanx)dx =∫ [1/2+tanx- cos2x /2 –sin2x /2]dx
=x/2 – ln| cosx | - sin2x /4 + cos2x /4 +C