Integrate by parts ? e^-x cos 2x dx

Question

682 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-10T07:16:34

The answer is #=1/5e^-x(2sin(2x)-cos(2x))+C#

Perform the integration by parts #2# times

#intuv'=uv-intu'v#

Here,

#u=cos2x#, #=>#, #u'=-2sin2x#

#v'=e^-x#, #=>#, #v=-e^-x#

Therefore,

#inte^-xcos(2x)dx=-e^-x cos(2x)-2inte^-x sin(2x)dx#

#u=sin2x#, #=>#, #u'=2cos2x#

#v'=e^-x#, #=>#, #v=-e^-x#

#2inte^-x sin(2x)dx=-2e^-xsin2x+4inte^-xcos2x#

So,

#inte^-xcos(2x)dx=-e^-x cos(2x)-(-2e^-xsin2x+4inte^-xcos2x)#

#5inte^-xcos(2x)dx=-e^-xcos(2x)+2e^-xsin(2x)#

#inte^-xcos(2x)dx=1/5(2e^-xsin(2x)-e^-xcos(2x))+C#

#inte^-xcos(2x)dx=1/5e^-x(2sin(2x)-cos(2x))+C#