# How do you integrate int sin(lnx) using integration by parts?

Nov 3, 2015

$\frac{1}{2} \left(\left[x \sin \left(\ln \left(x\right)\right)\right] - \left[x \cos \left(\ln \left(x\right)\right)\right]\right)$

#### Explanation:

$\int \sin \left(\ln \left(x\right)\right) \mathrm{dx}$

Let's $u = \ln \left(x\right)$

$\mathrm{du} = \frac{\mathrm{dx}}{x}$

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

$x = {e}^{u}$

then

$\int {e}^{u} \sin \left(u\right) \mathrm{du}$

By part : dv = e^u ; v = e^u ; w = sin(u) ; dw = cos(u)

$\left[v \cdot w\right] - \int \mathrm{dw} \cdot v$

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

By part again

dv=e^u ; v = e^u ; w = cos(u) ; dw = -sin(u)

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

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

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

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

Substitute back for $u = \ln \left(x\right)$

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