Question

How do you integrate `int e^-xcosx` by parts from `[0,2]`?

Answer 1

`int_0^2e^-xcosxdx=1/2(e^-xsinx-e^-xcosx)|_0^2=1/2(e^-2sin2-e^-2cos2+1)`

Explanation:

Let's pick `u, dv` and solve for `v, du:`

`u=e^-x`

`du=-e^-xdx`

`dv=cosxdx`

`v=sinx`

Thus, we apply `intudv=uv-intvdu`:

`inte^-xcosxdx=e^-xsinx+inte^-xsinxdx`

This didn't yield anything solvable, let's integrate by parts once more, for `inte^-xsinxdx`:

`u=e^-x`

`du=-e^-xdx`

`dv=sinxdx`

`v=-cosx`

Applying the integration by parts formula, we get

`inte^-xsinxdx=-e^-xcosx-inte^-xcosxdx`

There's nothing here we can solve, but note how our original integral showed up again.

Now, we said

`inte^-xcosxdx=e^-xsinx+inte^-xsinxdx`

So, let's plug in what we got for `inte^-xsinxdx` in:

`inte^-xcosxdx=e^-xsinx-e^-xcosx-inte^-xcosxdx`

Solve for `inte^-xcosxdx:`

`2inte^-xcosxdx=e^-xsinx-e^-xcosx`

`inte^-xcosxdx=1/2(e^-xsinx-e^-xcosx)`

Note I have not put in a constant of integration because we're going to be taking a definite integral:

`int_0^2e^-xcosxdx=1/2(e^-xsinx-e^-xcosx)|_0^2=1/2(e^-2sin2-e^-2cos2+1)`