题目

求不定积分［e的（-2x）次幂］*sin(x/2)

提问时间：2021-12-27

答案

I=∫e^(-2x)sin(x/2)dx
=2∫e^(-2x)dcos(x/2)
=2e^(-2x)cos(x/2)+4∫e^(-2x)cos(x/2)dx
=省略+8∫e^(-2x)dsin(x/2)
=省略+8e^(-2x)sin(x/2)+16∫e^(-2x)sin(x/2)dx
-17I=.
I= -(2/17)[4sin(x/2)+cos(x/2)]e^(-2x)