怎么求根号下xsin根号下xdx的不定积分？

Question

如题所述

Answer 1

∫√x.sin√x dx
let
y=√x
dy = [1/(2√x)] dx
dx = 2y dy

∫√x.sin√x dx
=2∫y^2. siny dy
=-2∫y^2. dcosy
=-2y^2. cosy +4∫y cosy dy
=-2y^2. cosy +4∫y dsiny
=-2y^2. cosy +4ysiny -4∫siny dy
=-2y^2. cosy +4ysiny +4cosy + C
=-2x.cos√x = 4√x.sin√x +4cos√x + C本回答被提问者和网友采纳

Answer 2

换元