How do you integrate $int e^xsinx$ by parts from $[0,1]$ ?

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Answer 1 · 2016-12-19T06:31:44

The answer is $=1/2e^x(sinx-cosx)+C$

The integration by parts is applied 2 times

$intuv'=uv-intu'v$

Here,

$u=sinx$ , $=>$ , $u'=cosx$

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

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

We do the integration by parts a second time

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

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

$inte^xcosxdx=e^xcosx-inte^x-sinxdx$

$=e^xcosx+inte^xsinxdx$

Therefore,

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

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

So,

$2*inte^xsinx=e^x(sinx-cosx)$

$inte^xsinxdx=1/2e^x(sinx-cosx)+C$

How do you integrate int e^xsinxint e^xsinx by parts from [0,1][0,1]?