# Question #1c46e

Feb 7, 2016

Trisha - integrate by parts twice ...

#### Explanation:

$u = \sin \left(\ln x\right)$

$\mathrm{du} = \frac{\cos \left(\ln x\right)}{x} \mathrm{dx}$

$\mathrm{dv} = x \mathrm{dx}$

$v = {x}^{2} / 2$

Now, using integration by parts ...

$\int x \sin \left(\ln x\right) \mathrm{dx} = \left({x}^{2} / 2\right) \left(\sin \left(\ln x\right)\right) - \int \frac{x \cos \left(\ln x\right)}{2} \mathrm{dx}$

Do it again on the integral ...

$u = \cos \frac{\ln x}{2}$

$\mathrm{du} = - \frac{\sin \left(\ln x\right)}{2 x} \mathrm{dx}$

$\mathrm{dv} = x \mathrm{dx}$

$v = {x}^{2} / 2$

Using the integration by parts formula one more time ...

$\int x \sin \left(\ln x\right) \mathrm{dx} = \left({x}^{2} / 2\right) \left(\sin \left(\ln x\right)\right) - {x}^{2} \cos \frac{\ln x}{4} - \frac{1}{4} \left(\int x \sin \left(\ln x\right)\right)$

Combine integral terms ...

$\left(\frac{5}{4}\right) \int x \sin \left(\ln x\right) \mathrm{dx} = \left({x}^{2} / 2\right) \left(\sin \left(\ln x\right)\right) - {x}^{2} \cos \frac{\ln x}{4}$

Simplify ...

$\int x \sin \left(\ln x\right) \mathrm{dx} = - {x}^{2} / 5 \left[\cos \left(\ln x\right) - 2 \sin \left(\ln x\right)\right] + C$

hope that helped