Question

∫e的负x次方乘以sin2xdx的不定积分是什么，求详细答案

Answer 1

令A=∫ e^(-x)sin2xdx
∵A=∫ e^(-x)sin2xdx = - ∫ sin2xd(e^(-x))=-e^(-x)sin2x+∫ e^(-x)d(sin2x)
=-e^(-x)sin2x+2∫ e^(-x)cos2xdx
=-e^(-x)sin2x-2∫ cos2xd(e^(-x))
=-e^(-x)sin2x-2cos2x*e^(-x)+2∫ e^(-x)d(cos2x)
=-e^(-x)sin2x-2cos2x*e^(-x)-4∫ e^(-x)sin2xdx
=-e^(-x)sin2x-2cos2x*e^(-x)-4A
5A=-e^(-x)sin2x-2cos2x*e^(-x)
∴A=∫ e^(-x)sin2xdx
=1/5[-e^(-x)sin2x-2cos2x*e^(-x)]+C
=-1/5(sin2x+2cos2x)e^(-x) +C

Answer 2

用分步积分
I=∫e^(-x)sin2xdx

=-∫e^(-x)sin2xd(-x)
=-∫sin2xde^(-x)
=-e^(-x)sin2x+∫e^(-x)dsin2x
=-e^(-x)sin2x+2∫e^(-x)cos2xdx
=-e^(-x)sin2x-2∫cos2xde^(-x)
=-e^(-x)sin2x-2e^(-x)cos2x+2∫e^(-x)dcos2x
=-e^(-x)sin2x-2e^(-x)cos2x-4∫e^(-x)sin2xdx
移项得
I=∫e^(-x)sin2xdx=1/5[-e^(-x)sin2x-2e^(-x)cos2x]+C

Answer 3

∫e^-x *sin2xdx

=-∫e^-x *sin2xd(-x)
=-∫sin2xde^-x
=-e^-xsin2x+∫e^-x*cos2x*2dx
=-e^-xsin2x-2∫e^-x*cos2xd(-x)
=-e^-xsin2x-2∫cos2xde^-x
=-e^-xsin2x-2e^-xcos2x+2∫e^-x*(-sin2x)*2dx
=-e^-xsin2x-2e^-xcos2x-4∫e^-x*sin2xdx

∴5∫e^-x *sin2xdx=-e^-xsin2x-2e^-xcos2x
∫e^-x *sin2xdx=-e^-x *(sin2x+2cos2x )/5