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

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Answer 1 · 2017-11-05T14:18:31

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:

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

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