展开全部
∫[e^(-2x)]sin(x/2)dx=-1/2∫sin(x/2)de^(-2x)
=-1/2e^(-2x)sin(x/2)+1/4∫[e^(-2x)]cos(x/2)dx
=-1/2e^(-2x)sin(x/2)-1/8∫cos(x/2)de^(-2x)
=-1/2e^(-2x)sin(x/2)-1/8e^(-2x)cos(x/2)dx-1/16∫e^(-2x)sin(x/2)dx
故∫e^(-2x)sin(x/2)dx
=16/17[-1/2e^(-2x)sin(x/2)-1/8e^(-2x)cos(x/2)dx]+C
=-1/2e^(-2x)sin(x/2)+1/4∫[e^(-2x)]cos(x/2)dx
=-1/2e^(-2x)sin(x/2)-1/8∫cos(x/2)de^(-2x)
=-1/2e^(-2x)sin(x/2)-1/8e^(-2x)cos(x/2)dx-1/16∫e^(-2x)sin(x/2)dx
故∫e^(-2x)sin(x/2)dx
=16/17[-1/2e^(-2x)sin(x/2)-1/8e^(-2x)cos(x/2)dx]+C
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询