Question

How do you integrate `int cos(lnx)` by integration by parts method?

Answer 1

This isn't necessarily done by only integration by parts, but it's one step in the process; this actually looks like one where you simply choose your methods and see what works.

`int cos(lnx)dx = 1/2 x(sin(lnx) + cos(lnx)) + C`

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First, let us try:

`u = lnx`
`du = 1/xdx`

Then, we have:

`int cos(u)xdu = int e^ucosudu`

This looks more reasonable, and actually would be done using integration by parts, since we have a function where `(df)/(dx) = f(x)` (`e^u)`, and a function with cyclic derivatives (`cosu`).

Let:

`s = e^u`
`dt = cosudu`
`t = sinu`
`ds = e^udu`

This gives:

`int sdt = st - int tds`

`= e^usinu + int e^u(-sinu)du`

Since `sin` derivatives are cyclic, repeat the process. Let:

`s = e^u`
`dt = -sinudu`
`t = cosu`
`ds = e^udu`

Therefore:

`int e^ucosudu = int sdt`

`= e^usinu + [e^ucosu - int e^ucosudu]`

If you notice, the original integral has shown up again, so:

`2int e^ucosudu = e^usinu + e^ucosu = e^u(sinu + cosu)`

`=> int e^ucosudu = 1/2e^u(sinu + cosu)`

And now, we undo the substitution, using `u = lnx`, to get:

`int cos(lnx)dx = color(blue)(1/2 x(sin(lnx) + cos(lnx)) + C)`