题目

f`(x)=(cosx^2),求原函数f(x).题中是x的平方

提问时间：2021-03-31

答案

f'(x)=cosx^2
∫cosx^2dx=(1/2)∫cosx^2dx^2/x=(1/2)∫dsin(x^2)/x
=(1/2)sinx^2/x -(1/2)∫sinx^2d(1/x)
=(1/2)sinx^2/x+(1/4)∫sinx^2dx^2/x^3
=(1/2)sinx^2/x +(-1/4)∫dcosx^2/x^3 +(3/8)∫cosx^2dx^2/x^5
=(1/2)sinx^2/x+(-1/4)cosx^2/x^3+(3/8)sinx^2/x^5-(3*5/16)∫sinx^2dx^2/x^7
=(1/2)sinx^2/x+(-1/4)cosx^2/x^3+(1*3/8)sinx^2/x^5+(-3*5/16)cosx^2/x^7+...+F(n)
偶数项Fn=(1/2^n)*sinx^2*(1*3*..*(2n-1))
奇数项Fn=-(1/2^n)*cosx^2*(*3*..*(2n-1)