做个简单点的吧,设f'(x)=[sin(x-1)]^2f(x)=∫f'(x)dx=∫[sin(x-1)]^2dx=∫[sin(x-1)]^2d(x-1)=∫{1-cos[2(x-1)]}/2*d(x-1)=(x-1)/2-1/4*sin[2(x-1)]+Cf(0)=-1/2-1/4*sin(-2)+C=0=> C=1/2-1/4*sin2f(x)=x/2-1/4*sin[2...