Question

f'(2+cosx)=(sinx)^2+(tanx)^2,求f(x)

Answer 1

令t=2+cosx,则cosx=t-2,t的取值范围是[1,3]且t≠2

(cosx)^2=(t-2)^2,(sinx)^2=1-(cosx)^2=1-(t-2)^2

带入f'表达式整理得：

f'(t)=-(t-2)^2+1/(t-2)^2=-t^2+4t-4+1/(t-2)^2

所以f(t)=-(1/3)t^3+2t^2-4t-1/(t-2)

即f(x)=-(1/3)x^3+2x^2-4x-1/(x-2)

x的取值范围是[1,3]且x≠2