How do you integrate $(e^x)(cosx) dx$ ?

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Answer 1 · 2015-04-29T08:37:38

This integral is a cyclic one and has to be done two times with the theorem of the integration by parts, that says:

$intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx$

We can assume that $f(x)=cosx$ and $g'(x)dx=e^xdx$ ,

$f'(x)dx=-sinxdx$

$g(x)=e^x$ ,

so:

$I=e^xcosx-inte^x(-sinx)dx=$

$=e^xcosx+inte^xsinxdx=(1)$ .

And now...again:

$f(x)=sinx$ and $g'(x)dx=e^xdx$ ,

$f'(x)dx=cosxdx$

$g(x)=e^x$ .

So:

$(1)=e^xcosx+[e^xsinx-inte^xcosxdx]=$

$=e^xcosx+e^xsinx-inte^xcosxdx$ .

So, finally, we can write:

$I=e^xcosx+e^xsinx-IrArr$

$2I=e^xcosx+e^xsinx-IrArr$

$I=e^x/2(cosx+sinx)+c$ .

How do you integrate (e^x)(cosx) dx(e^x)(cosx) dx?