Question

x^2sinx定积分用换元法怎么求

Answer 1

=-x^2cosx+2xsinx+2cosx+C

解题过程如下：

运用分部积没启分孙贺法

∫x^2sinxdx

=-∫x^2dcosx

=-x^2cosx+∫cosx*2xdx

=-x^2cosx+2∫xdsinx

=-x^2cosx+2xsinx-2∫则察派sinxdx

=-x^2cosx+2xsinx+2cosx+C

Answer 2

设u=sinx，则侍山坦du=cosxdx
原式=∫x^2sinxdx=∫x^2udu
=1/3x^3u+1/唯绝2∫x^2du
=1/3x^3sinx+1/老桐2x^3+C
=1/3x^3sinx+1/2x^3+C