Question

What is `I = int \ sinlnx-coslnx \ dx`?

Answer 1

We seek:

`I = int \ sinlnx-coslnx \ dx`
`\ \ = int \ sinlnx \ dx - int \ coslnx \ dx`

We can apply integration by Parts to the second integral

Let # { (u,=coslnx, => (du)/dx,=(-sinlnx)/x), ((dv)/dx,=1, => v,=x ) :}#

Then plugging into the IBP formula:

`int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx`

we get:

`int \ (coslnx)(1) \ dx = (coslnx)(x) - int \ (x(-sinlnx)/x) \ dx`
`" " = xcoslnx + int \ sinlnx \ dx + C`

Using this result we get:

`I = int \ sinlnx \ dx - {xcoslnx + int \ sinlnx \ dx } + C`
`\ \ = - xcoslnx + C`