How do you integrate #int e^-xsin4x# by integration by parts method?

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Answer 1 · 2016-12-24T20:10:16

The answer is #=(-e^(-x)(sin4x+4cos4x))/17+C#

#intuv'dx=uv-intu'vdx#

Here,

#u=sin4x#, #=>#, #u'=4cos4x#

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

#inte^(-x)sin4xdx=-e^(-x)sin4x-int4*-e^(-x)cos4xdx#

#=-e^(-x)sin4x+4inte^(-x)cos4xdx#

For the integral #inte^(-x)cos4xdx#, we apply the integration by parts a second time

#u=cos4x#, #=>#, #u'=-4sin4x#

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

#inte^(-x)cos4xdx=-e^(-x)cos4x-4inte^(-x)sin4xdx#

Putting it all together

#inte^(-x)sin4xdx=-e^(-x)sin4x+4(-e^(-x)cos4x-4inte^(-x)sin4xdx)#

#=-e^(-x)sin4x-4e^(-x)cos4x-16inte^(-x)sin4xdx#

Therefore,

#17inte^(-x)sin4xdx=-e^(-x)(sin4x+4cos4x)#

#inte^(-x)sin4xdx=(-e^(-x)(sin4x+4cos4x))/17+C#