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

Let `I = int sin(lnx) dx`

We need to decide (guess) whether to use `sin(lnx)`as `u` or `dv`. It turns out that either will work.

Method 1
Let `u = sin(lnx)` and `dv = dx`.

Then `du = 1/x cos(lnx) dx` and `v = x`

`I = uv-intvdu`

`= xsin(lnx)-intcos(lnx) dx`

Repeat with `u = cos(lnx)` and `dv = dx`,

so `du = -1/xsin(lnx)` and `v = x`.

`I = xsin(lnx)-[ xcos(lnx)- int -sin(lnx) dx ]`

so

`I = xsin(lnx)- xcos(lnx)- underbrace(int sin(lnx) dx)_I`

`2I = xsin(lnx)- xcos(lnx)`

`I = 1/2( xsin(lnx)- xcos(lnx) )`

Method 2

Let `I = int sin(lnx) dx`

In order to use `sin(lnx) dx` in `dv`, we'll need to be able to integrate `dv`.
We could use substitution if we had the derivative of `lnx` as a factor, so we'll introduce it.

`I = int x sin(lnx) 1/x dx`

Let `u = x` and `dv = sin(lnx) 1/x dx`.

Then `du =dx` and `v = -cos(lnx)`

`I = uv-intvdu`

`= -xcos(lnx)+int cos(lnx) dx`

We'll use parts again. (And we'll hope that it works. If it doesn't, we'll try something else.)

`= -xcos(lnx)+int x cos(lnx) 1/x dx`

Let `u = x` and `dv = cos(lnx) 1/x dx`.

Then `du =dx` and `v = sin(lnx)`

`I = -xcos(lnx)+[xsin(lnx)-underbrace(intsin(lnx) dx)_I]`

`2I = -xcos(lnx)+xsin(lnx)`

`I = 1/2( xsin(lnx)- xcos(lnx) )`