Integration by Parts

**Preston019** · September 6th 2012, 12:27 PM

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?

**MaxJasper** · September 6th 2012, 12:35 PM

Use $u=\text{ln}(2x)$

**Preston019** · September 6th 2012, 12:44 PM

Originally Posted by MaxJasper

Use $u=\text{ln}(2x)$

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

**skeeter** · September 6th 2012, 12:53 PM

Originally Posted by Preston019

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

$u = \ln(2x)$

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

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

substitute ...

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

now perform the integration by parts ...

**Preston019** · September 6th 2012, 01:10 PM

Originally Posted by skeeter

$u = \ln(2x)$

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

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

substitute ...

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

now perform the integration by parts ...

Oooooooohhh! Okay, that helped so much. Thanks!

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