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

Answer 1

Elijah Abram

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#

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Answer 2

Charles Arocha

To integrate (\int_0^1 e^x \sin(x) , dx) by parts, you can use the formula:

[\int u , dv = uv - \int v , du]

Let (u = \sin(x)) and (dv = e^x , dx). Then, differentiate (u) to find (du), and integrate (dv) to find (v).

After finding (u), (du), (v), and (dv), you can apply the integration by parts formula to find the integral.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.