How do you integrate int sin(lnx) dx?

Aug 7, 2018

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

Explanation:

Here,

$I = \int \sin \left(\ln x\right) \mathrm{dx} \ldots \ldots \ldots \ldots \ldots . . \left(A\right)$

Subst. color(red)(lnx=u=>x=e^u=>dx=e^udu

$\therefore I = \int \sin u \cdot {e}^{u} \mathrm{du}$

Using Integration by parts:

$I = \sin u \int {e}^{u} \mathrm{du} - \int \left(\frac{d}{\mathrm{du}} \left(\sin u\right) \int {e}^{u} \mathrm{du}\right) \mathrm{du}$

$\therefore I = \sin u \times {e}^{u} - \int \cos u {e}^{u} \mathrm{du}$

Again using Integration by parts:

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

$\therefore I = {e}^{u} \sin u - {e}^{u} \cos u - \int {e}^{u} \sin u \mathrm{du} + c$

$I = {e}^{u} \sin u - {e}^{u} \cos u - I + c . \to$from$\left(A\right)$

$\therefore I + I = {e}^{u} \sin u - {e}^{u} \cos u + c$

$\therefore 2 I = {e}^{u} \sin u - {e}^{u} \cos u + c$

$\therefore I = \frac{1}{2} \left[{e}^{u} \sin u - {e}^{u} \cos u\right] + \frac{c}{2}$

Subst.back color(red)(lnx=u=>x=e^u

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

Hence ,

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