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,

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

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

How do you integrate int e^-xsin4xint e^-xsin4x by integration by parts method?