How do you integrate $e^(-x) * cos(2x) dx$ ?

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Answer 1 · 2018-04-03T11:43:13

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

$intuv'=uv-intu'v$

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

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

Therefore,

$inte^-xcos2xdx=-e^-xcos2x-int2e^-xsin2xdx$

$u=2sin2x$ , $=>$ , $u'=4cos2x$

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

$int2e^-xsin2xdx=-2e^-xsin2x+int4e^-xcos2x$

So,

$inte^-xcos2xdx=-e^-xcos2x-(-2e^-xsin2x+int4e^-xcos2x)$

$=-e^-xcos2x+2e^-xsin2x-4inte^-xcos2x$

$5inte^-xcos2xd=-e^-xcos2x+2e^-xsin2x$

$inte^-xcos2xd=1/5(2e^-xsin2x-e^-xcos2x)+C$

$=e^-x/5(2sin2x-cos2x)+C$

How do you integrate e^(-x) * cos(2x) dxe^(-x) * cos(2x) dx ?