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#