# Integration by Parts

• Sep 6th 2012, 12:27 PM
Preston019
Integration by Parts
I'm sure how to go about solving this integral:

∫cos(ln(2x))dx

Could someone explain how to set it up to use integration by parts?
• Sep 6th 2012, 12:35 PM
MaxJasper
Re: Integration by Parts
Use $\displaystyle u=\text{ln}(2x)$
• Sep 6th 2012, 12:44 PM
Preston019
Re: Integration by Parts
Quote:

Originally Posted by MaxJasper
Use $\displaystyle u=\text{ln}(2x)$

Okay, that'll give me du = 1/x dx which I cannot substitute back into the original equation.
• Sep 6th 2012, 12:53 PM
skeeter
Re: Integration by Parts
Quote:

Originally Posted by Preston019
Okay, that'll give me du = 1/x dx which I cannot substitute back into the original equation.

$\displaystyle u = \ln(2x)$

$\displaystyle x = \frac{e^u}{2}$

$\displaystyle dx = \frac{e^u}{2} \, du$

substitute ...

$\displaystyle \int \cos(u) \cdot \frac{e^u}{2} \, du$

now perform the integration by parts ...
• Sep 6th 2012, 01:10 PM
Preston019
Re: Integration by Parts
Quote:

Originally Posted by skeeter
$\displaystyle u = \ln(2x)$

$\displaystyle x = \frac{e^u}{2}$

$\displaystyle dx = \frac{e^u}{2} \, du$

substitute ...

$\displaystyle \int \cos(u) \cdot \frac{e^u}{2} \, du$

now perform the integration by parts ...

Oooooooohhh! Okay, that helped so much. Thanks!