Question

求解答，设f（x）的一个原函数为x的平方•sinx，求不定积分 如图所

Answer 1

∫缺凳f(x) dx = x^2.sinx
f(x) = x^2.cosx +2x.sinx
∫宏搭f(x)/蔽扮拿x dx
=∫ (xcosx + 2sinx ) dx
=∫ xdsinx - 2cosx
=xsinx -∫ sinx dx -2cosx
=xsinx +cosx -2cosx +C
=xsinx -cosx +C

追问

这个题目第二问到第三问是怎么来的

追答

x->0
sinx = x-(1/6)x^3+o(x^3)
sin(sinx)
=sin[x-(1/6)x^3+o(x^3)]
=[x-(1/6)x^3]-(1/6)[x-(1/6)x^3] +o(x^3)
=x-(1/3)x^3+o(x^3)
sinx -sin(sinx = (1/6)x^3+0(x^3)
lim(x->0) [ sinx -sin(sinx)]/(x^2.tanx)
=lim(x->0) (1/6)x^3/x^3
=1/6

x->0
sinx = x-(1/6)x^3+o(x^3)
sin(sinx)
=sin[x-(1/6)x^3+o(x^3)]
=[x-(1/6)x^3]-(1/6)[x-(1/6)x^3] +o(x^3)
=x-(1/3)x^3+o(x^3)
sinx -sin(sinx = (1/6)x^3+0(x^3)
lim(x->0) [ sinx -sin(sinx)]/(x^2.tanx)
=lim(x->0) (1/6)x^3/x^3
=1/6