How do you integrate $int e^xcos(2x)$ by parts?

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Answer 1 · 2017-01-14T19:24:10

$int e^x cos(2x) dx = 1/5(e^xcos2x+2e^xsin2x) + C$

As $d(e^x) = e^xdx$ we can integrate by parts as:

$int e^x cos(2x) dx = int cos(2x) d(e^x) = e^xcos2x + 2int e^xsin(2x)dx$

We integrate the resulting integral by parts again:

$int e^xsin(2x)dx = int sin(2x)d(e^x) = e^xsin2x -2int e^xcos(2x)dx$

So if we name:

$I = int e^x cos(2x) dx$

we get the following equation:

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

and solving for $I$ :

$I= 1/5(e^xcos2x+2e^xsin2x)$

so that:

$int e^x cos(2x) dx = 1/5(e^xcos2x+2e^xsin2x) + C$

How do you integrate int e^xcos(2x)int e^xcos(2x) by parts?