# How do you integrate (1/x^2)(sin(2lnx)) dx ?

Apr 11, 2015

$f \left(x\right) = \int \frac{1}{x} ^ 2 \cdot \sin \left(2 \ln \left(x\right)\right) \mathrm{dx}$

$= \frac{1}{2} \int \frac{2}{x} ^ 2 \cdot \sin \left(2 \ln \left(x\right)\right) \mathrm{dx}$

$u = 2 \ln \left(x\right)$ note $u = \ln \left({x}^{2}\right)$
$\mathrm{du} = \frac{2}{x} \mathrm{dx}$

$= \frac{1}{2} \int \frac{2}{x} \cdot \frac{1}{x} \cdot \sin \left(2 \ln \left(x\right)\right) \mathrm{dx}$

$= \frac{1}{2} \int \frac{1}{x} \cdot \sin \left(u\right) \mathrm{du}$

${e}^{u} = {x}^{2}$
${e}^{\frac{1}{2} u} = x$
${e}^{- \frac{1}{2} u} = \frac{1}{x}$

$f \left(x\right) = \int \frac{1}{2} {e}^{- \frac{1}{2} u} \cdot \sin \left(u\right) \mathrm{du}$

By part :

$g \left(x\right) = \frac{1}{2} {e}^{- \frac{1}{2} u}$
$g ' \left(x\right) = - \frac{1}{4} {e}^{- \frac{1}{2} u}$

$h ' \left(x\right) = \sin \left(u\right)$
$h \left(x\right) = - \cos \left(u\right)$

$\int \frac{1}{2} {e}^{- \frac{1}{2} u} \cdot \sin \left(u\right) \mathrm{du} = - \left[\frac{1}{2} {e}^{- \frac{1}{2} u} \cos \left(u\right)\right] - \int \frac{1}{4} {e}^{- \frac{1}{2} u} \cos \left(u\right) \mathrm{du}$

by part again :

$g \left(x\right) = \frac{1}{4} {e}^{- \frac{1}{2} u}$
$g ' \left(x\right) = - \frac{1}{8} {e}^{- \frac{1}{2} u}$

$h ' \left(x\right) = \cos \left(u\right)$
$h \left(x\right) = \sin \left(u\right)$

$= \int \frac{1}{2} {e}^{- \frac{1}{2} u} \cdot \sin \left(u\right) \mathrm{du} = - \left[\frac{1}{2} {e}^{- \frac{1}{2} u} \cos \left(u\right)\right] - \left(\left[\frac{1}{4} {e}^{- \frac{1}{2} u} \sin \left(u\right)\right] + \frac{1}{4} \int \frac{1}{2} {e}^{- \frac{1}{2} u} \sin \left(u\right) \mathrm{du}\right)$

$= \int \frac{1}{2} {e}^{- \frac{1}{2} u} \cdot \sin \left(u\right) \mathrm{du} = - \left[\frac{1}{2} {e}^{- \frac{1}{2} u} \cos \left(u\right)\right] - \left[\frac{1}{4} {e}^{- \frac{1}{2} u} \sin \left(u\right)\right] - \frac{1}{4} \int \frac{1}{2} {e}^{- \frac{1}{2} u} \sin \left(u\right) \mathrm{du}$

$= 4 \int \frac{1}{2} {e}^{- \frac{1}{2} u} \cdot \sin \left(u\right) \mathrm{du} = - 4 \left[\frac{1}{2} {e}^{- \frac{1}{2} u} \cos \left(u\right)\right] - 4 \left[\frac{1}{4} {e}^{- \frac{1}{2} u} \sin \left(u\right)\right] - \int \frac{1}{2} {e}^{- \frac{1}{2} u} \sin \left(u\right) \mathrm{du}$

$= 4 \int \frac{1}{2} {e}^{- \frac{1}{2} u} \cdot \sin \left(u\right) \mathrm{du} = - 2 \left[{e}^{- \frac{1}{2} u} \cos \left(u\right)\right] - \left[{e}^{- \frac{1}{2} u} \sin \left(u\right)\right] - \int \frac{1}{2} {e}^{- \frac{1}{2} u} \sin \left(u\right) \mathrm{du}$

$= 5 \int \frac{1}{2} {e}^{- \frac{1}{2} u} \cdot \sin \left(u\right) \mathrm{du} = - 2 \left[{e}^{- \frac{1}{2} u} \cos \left(u\right)\right] - \left[{e}^{- \frac{1}{2} u} \sin \left(u\right)\right]$

$= \int \frac{1}{2} {e}^{- \frac{1}{2} u} \cdot \sin \left(u\right) \mathrm{du} = \frac{1}{5} \left(- 2 \left[{e}^{- \frac{1}{2} u} \cos \left(u\right)\right] - \left[{e}^{- \frac{1}{2} u} \sin \left(u\right)\right]\right)$

Substitute back : ${e}^{- \frac{1}{2} u} = \frac{1}{x}$

So we have :

$= \left(- \frac{1}{5 x} \left(2 \cos \left(u\right) + \sin \left(u\right)\right)\right) + C$