How do you integrate $int e^-x*cos2xdx$ using integration by parts?

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Answer 1 · 2016-07-12T22:21:48

$= -1/5 e^-x*cos2x + 2 e^-x sin 2x + C$

$I = int e^-x*cos2x \ dx$

using IBP

$I = int (-e^-x)'*cos2x \ dx$

$I = -e^-x*cos2x - int -e^-x (cos 2x)' \ dx$

$= -e^-x*cos2x - 2int \e^-x sin 2x \ dx$

$= -e^-x*cos2x - 2J$

$J = int \e^-x sin 2x \ dx$

$= int (-\e^-x)' sin2x \ dx$

$= -e^-x sin 2x - int -\e^-x (sin 2x)' \ dx$

$= -e^-x sin 2x + 2 int \e^-x cos 2x \ dx$

$= -e^-x sin x + 2I$

$I = -e^-x cos2x - 2(- \e^-x sin 2x + 2I)$

$= -e^-x cos2x + 2 e^-x sin 2x -4I + C$

$5I = -e^-x cos2x + 2 e^-x sin 2x + C$

$I = -1/5 e^-x cos2x + 2 e^-x sin 2x + C$

How do you integrate int e^-x*cos2xdxint e^-x*cos2xdx using integration by parts?