f'(sin^2x)=cos^2x=1-sin^2x
则f'(x)=1-x
则f(x)=x-1/2x^2+C
C是常数
f(x)=sin(π-x)sin(π/2 -x)+cos x =sinxcosx+cos x =1/2*sin2x+(1+cos2x)/2 =1/2*(sin2x+cos2x)+1/2 =√2/2*sin(2x+π/4)+1
f'(sin^2x)=cos^2x=1-sin^2x
则f'(x)=1-x
则f(x)=x-1/2x^2+C（C是常数，0<=x <=1)
f'(sin^2x)=cos^2x=1-sin^2x
f'(x)=1-x，
f(x)=x-(1/2)x^2+C
内容全部来源于网络收集，如有侵权，请联系网站删除：QQ：24596024