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

Nov 5, 2017

We seek:

$I = \int \setminus \sin \ln x - \cos \ln x \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \sin \ln x \setminus \mathrm{dx} - \int \setminus \cos \ln x \setminus \mathrm{dx}$

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:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

we get:

$\int \setminus \left(\cos \ln x\right) \left(1\right) \setminus \mathrm{dx} = \left(\cos \ln x\right) \left(x\right) - \int \setminus \left(x \frac{- \sin \ln x}{x}\right) \setminus \mathrm{dx}$
$\text{ } = x \cos \ln x + \int \setminus \sin \ln x \setminus \mathrm{dx} + C$

Using this result we get:

$I = \int \setminus \sin \ln x \setminus \mathrm{dx} - \left\{x \cos \ln x + \int \setminus \sin \ln x \setminus \mathrm{dx}\right\} + C$
$\setminus \setminus = - x \cos \ln x + C$