# How do you integrate int cos(lne(x))  using integration by parts?

May 19, 2018

$\int \cos \left(\ln \left(x\right)\right) \mathrm{dx} = x \frac{\cos \left(\ln \left(x\right)\right) - \sin \left(\ln \left(x\right)\right)}{2} + C$, $C \in \mathbb{R}$

#### Explanation:

$\int \cos \left(\ln \left(x\right)\right) \mathrm{dx}$
let : $X = \ln \left(x\right)$
$x = {e}^{X}$
$\mathrm{dx} = {e}^{X} \mathrm{dX}$
so :
$\int \cos \left(\ln \left(x\right)\right) \mathrm{dx} = \int \cos \left(X\right) {e}^{X} \mathrm{dX}$
Using integration by parts :
$\int u v ' \mathrm{dX} = u v - \int u ' v \mathrm{dX}$
there :
$u = {e}^{X}$ $v ' = \cos \left(X\right)$

$u ' = {e}^{X}$ $v = \sin \left(X\right)$

So : $\int \cos \left(X\right) {e}^{X} \mathrm{dX} = \left(- \sin \left(X\right) {e}^{X}\right) - \int \left(\sin \left(X\right) {e}^{X}\right) \mathrm{dx}$

$= - \sin \left(X\right) {e}^{X} - \int \sin \left(X\right) {e}^{X} \mathrm{dX}$

Using integration by parts again :

intsin(X)e^XdX=-cos(X)e^X+int(cos(X)e^XdX

So :

intcos(X)e^XdX=-sin(X)e^X+cos(X)e^X-int(cos(X)e^XdX

$2 \int \cos \left(X\right) {e}^{X} \mathrm{dX} = {e}^{X} \left(\cos \left(X\right) - \sin \left(X\right)\right)$

$\int \cos \left(X\right) {e}^{X} \mathrm{dX} = \frac{{e}^{X} \left(\cos \left(X\right) - \sin \left(X\right)\right)}{2} + C$, $C \in \mathbb{R}$

$\int \cos \left(\ln \left(x\right)\right) \mathrm{dx} = {\cancel{{e}^{\ln \left(x\right)}}}^{= x} \frac{\cos \left(\ln \left(x\right)\right) - \sin \left(\ln \left(x\right)\right)}{2} + C$, $C \in \mathbb{R}$

$\int \cos \left(\ln \left(x\right)\right) \mathrm{dx} = x \cdot \frac{\cos \left(\ln \left(x\right)\right) - \sin \left(\ln \left(x\right)\right)}{2} + C$, $C \in \mathbb{R}$