Question

f'(cos^2x)=sin^2x ,f(0)=0,求f(x)

Answer 1

f'(cos^2x)=sin^2x=1-cos^2x

所以f'(x)=1-x

所以f(x)=∫(1-x)dx=x-x^2/2+C

f(0)=0-0+C=0

所以f(x)=x-x^2/2