已知f´(sin^x)=cos2x+tan^x,当0

已知f´(sin^x)=cos2x+tan^x,当0

f'(sin^2x)=1-2sin^2x+sin^2x/(1-sin^2x)
f'(x)=1-2x+x/(1-x)
f(x)=∫[1-2x+x/(1-x)]dx
=x-x^2+∫x/(1-x)dx
=x-x^2-∫-x/(1-x)dx
=x-x^2-∫[1+1/(x-1)]dx
=x-x^2-x-ln(1-x)+C
=-x^2-ln(1-x)+C