Question #34cc6

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Answer 1 · 2017-02-13T08:15:53

The answer is $=1/2(e^xsinx-e^xcosx)+C$

We integrate by parts 2 times

$intu'vdx=uv-intuv'dx$

For the first time

$v=sinx$ , $=>$ , $v'=cosx$

$u'=e^x$ , $=>$ , $u=e^x$

So,

$inte^xsinxdx=e^xsinx-inte^xcosxdx$

For the second time,

$v=cosx$ , $=>$ , $v'=-sinx$

$u'=e^x$ , $=>$ , $u=e^x$

$inte^xcosxdx=e^xcosx+inte^2sinxdx$

Therefore,

$inte^xsinxdx=e^xsinx-(e^xcosx+inte^2sinxdx)$

$inte^xsinxdx=e^xsinx-e^xcosx-inte^2sinxdx$

$2inte^xsinxdx=e^xsinx-e^xcosx$

$inte^xsinxdx=1/2(e^xsinx-e^xcosx)+C$