# Integrating by parts

• Mar 8th 2013, 11:21 AM
Steelers72
Integrating by parts
Simple question, but marked me wrong. Am I forgetting something?

To use the integration-by-parts formula
integral udv= uv - integral vdu

we must choose one part of
 8x cos 5x dx http://www.webassign.net/wastatic/wa...g/integral.gif

to be u, with the rest becoming dv.

Since the goal is to produce a simpler integral, we will choose u = 8x.
This means that dv = cos5xdx <----says this is wrong)
• Mar 8th 2013, 11:33 AM
veileen
Re: Integrating by parts
Uhm,

$\displaystyle \int 8x \cos {5x}\, dx = \int 8x \left ( \frac{1}{5}\sin {5x} \right )'\, dx =8x\cdot \frac{1}{5}\sin {5x} -\int (8x)'\cdot \frac{1}{5}\sin {5x}\, dx$
• Mar 8th 2013, 11:36 AM
Shakarri
Re: Integrating by parts
Veileen that's not very helpful.

You made the correct substitution, nothing is wrong with that.
• Mar 8th 2013, 01:58 PM
Steelers72
Re: Integrating by parts
so dv= 1/5 sin 5x ?

PS: wait...that would be the antiderivative of dv...dv would be cos5xdx but Im not sure why I'm marked wrong

Never mind, I needed the parenthesis.....stupid online hw.