# Integration by parts question.

1. Oct 31, 2013

### braceman

Hi guys,

Stuck on an integration by parts question....Not going to post the question as I want to work it out myself, but as I'm a bit of a novice on diff/integration I'm stuck on what we do at a certain step of the process....anyway..

I know integration by parts we end up using ∫udv = uv - ∫vdu

where obviously we assign u,v,du,dv as parts of our equation..

Now what I'm stuck on is what happens if we have say 1/2 ∫ udv = uv - ∫vdu

how does the 1/2 effect how it's processed?

does it end up as

1/2 ∫ udv = uv - 1/2∫vdu

or something like

1/2 ∫ udv = 1/2 uv - 1/2∫vdu

Anyone able to explain (reasonably simply) how it ends up and why??

2. Oct 31, 2013

### arildno

Where does the integration by parts formula come from in the first place?
Why is it correct?

3. Oct 31, 2013

### Staff: Mentor

This...
It's pretty simple. Since ∫ udv = uv - ∫vdu, then (1/2) ∫ udv = (1/2)[uv - ∫ vdu]

4. Oct 31, 2013

### economicsnerd

To elaborate on what arildno said:

"Integration by parts" isn't really its own concept. (It's always mystified me why every calculus class teaches it as its own thing... and the same goes for substitution). Our only technique for analytically integrating things is to find anti-derivatives.
- You know the product rule for differentiation. If you take an integral of both sides, you get something familiar (what some people call "integration by parts").
- You know the chain rule for differentiation. If you take an integral of both sides, you get something familiar (what some people call "substitution").

So these rules are nothing new.

5. Nov 1, 2013

### braceman

Cool......pretty obvious when re-thinking about it, but I just wanted to check..thanks for the replies guys.