How do I evaluate $inte^(2x)cosx dx$ by parts?

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Answer 1 · 2015-02-01T00:31:13

I´ll start by integrating $e^(2x)$ and leaving $cos(x)$ as it is and then derive it leaving the integrated part as it is.

$e^(2x)/2*cos(x)-inte^(2x)/2*(-sin(x))dx=$
$=e^(2x)/2*cos(x)+1/2inte^(2x)*(sin(x))dx=$

by parts again:
$=e^(2x)/2*cos(x)+1/2[e^(2x)/2*(sin(x))-inte^(2x)/2*cos(x)dx]=$
$=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))-1/4inte^(2x)*cos(x)dx$

So your integral is:
$inte^(2x)cos(x)dx=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))-1/4inte^(2x)*cos(x)dx$

Now I can take to the left the integral: $-1/4inte^(2x)*cos(x)dx$
Giving:

$inte^(2x)cos(x)dx+1/4inte^(2x)*cos(x)dx=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))$

$5/4inte^(2x)*cos(x)dx=e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))$

and finally:
$inte^(2x)*cos(x)dx=4/5[e^(2x)/2*cos(x)+e^(2x)/4*(sin(x))]+c$

How do I evaluate inte^(2x)cosx dxinte^(2x)cosx dx by parts?