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

1 Answer
Nov 5, 2017

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#