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#