How do you integrate $int e^(2x)cosx$ by integration by parts method?

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Answer 1 · 2016-09-04T20:44:44

$I = e^(2x) (1/5 sin x + 2/5 cos x) + C$

$I = int e^(2x)cosx \ dx$

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

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

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

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

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

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

$= e^(2x) sin x - 2 ( -e^(2x) cos x + 2 I + C)$

$= e^(2x) sin x + 2 e^(2x) cos x - 4 I + C$

$5 I = e^(2x) sin x + 2 e^(2x) cos x + C$

$I = e^(2x) (1/5 sin x + 2/5 cos x) + C$

How do you integrate int e^(2x)cosxint e^(2x)cosx by integration by parts method?