How do you integrate $(e^{x}) sin 4 x d x$ $(e^x)sin4x dx$ ?

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Answer 1 · 2015-04-29T08:32:29

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 $f(x)=sin4x$ and $g'(x)dx=e^xdx$ ,

$f'(x)dx=cos4x*4dx$

$g(x)=e^x$ ,

so:

$I=e^xsin4x-inte^xcos4x*4dx=$

$=e^xsin4x-4inte^xcos4xdx=(1)$ .

And now...again:

$f(x)=cos4x$ and $g'(x)dx=e^xdx$ ,

$f'(x)dx=-sin4x*4dx$

$g(x)=e^x$ .

So:

$(1)=e^xsin4x-4[e^xcos4x-inte^x(-sin4x*4)dx]=$

$=e^xsin4x-4e^xcos4x-16inte^xsin4xdx$ .

So, finally, we can write:

$I=e^xsin4x-4e^xcos4x-16IrArr$

$17I=e^xsin4x-4e^xcos4xrArr$

$I=e^x/17(sin4x-4cos4x)+c$ .

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