How do you integrate #int cos2x*e^-x# by integration by parts method?

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Answer 1 · 2016-11-24T09:02:25

The answer is #=(e^(-x)(2sin2x-cos2x))/5+C#

#intu'v=uv-intuv'#

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

#u'=e^(-x)#, #=>#, #u=-e^(-x)#

#inte^(-x)*cos2xdx=-e^(-x)*cos2x-2inte^(-x)sin2xdx#

We apply once more the integration by parts

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

#u'=e^(-x)#,#=>##u=-e^(-x)#

so, #2inte^(-x)sin2xdx=-2e^(-x)sin2x+4inte^(-x)cos2xdx#

So,

#inte^(-x)*cos2xdx=-e^(-x)*cos2x-(-2e^(-x)sin2x+4inte^(-x)cos2xdx)#

#inte^(-x)*cos2xdx=#

#-e^(-x)*cos2x+2e^(-x)sin2x-4inte^(-x)cos2xdx#

#5inte^(-x)*cos2xdx=-e^(-x)*cos2x+2e^(-x)sin2x#

#inte^(-x)*cos2xdx=(e^(-x)(2sin2x-cos2x))/5+C#