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Parts is Parts
Integration by Parts


 When we need to find the integral of something

 that is the product of two different functions
 The easiest things to do are:
 1) Multiply it out
 2) Arrange one of the parts into the derivative of the other
 and "stuff it back in".
 While those tricks are nice in theory,
 they usually don't work out.
 When they don't, something you can try is called Integration by Parts.
 It usually works.
 It works really well, when the two functions are different types.
 There are three different types of functions we usually work with:
Algebraic like X2 + 2X or 7X12
Trigonometric like sinX or cos3X
Exponential like eX or ln X
 If we have to deal with something like ...

 we need the new trick.
 We are going to use a special type of substitution.
 We split the two parts up.
 One of them we call "u".
 The other one, and here's the tricky part, we call "dv".
 We say that it is ALREADY a derivative of something called "v".
 The new formula works like this ...

 The hope is that what we have at the end
 is easier to deal with.
 Sometimes we have to run through this formula a few times
 before we DO get that easier integral to deal with.
 So the deal is you take one of the two parts and integrate it,
 and you take the other part and find the derivative.
 The question is, which do you chose for which?
 Think about it.
 If you find the derivative of a trig function,
 you just get back another trig function.
 But if you take the derivative of an algebraic function,
 it gets smaller.
 If the exponents are whole numbers,
 eventually it goes away.
 So that's the basic strategy.
 If we have an algebraic part with whole number exponents,
 take derivatives of it.
 We do integrals of whatever else we have in the problem.
 OK, let's do one ...


 We want to find the derivative of the algebraic part 
 so 4X 2 gets to be "u"

 The part that's left, cosX, gets to be "dv".

 That means all together we have ...

4X2 = u       cosXdx = dv

 So that means ...

8Xdx = du       sinX = v


 So putting all of this into our formula, we have ...

 That gets us closer, 
 but we still need to work on the integral.
 Using the same strategy, we get:

 8X = u       sinXdx = dv

 So that means ...

 8dx = du       - cosX = v

 That makes the integral part that was left ...

 Substituting this back into the whole equation,
 we now have ...

 Simplifying this, we get ...

 The integral at the end is finally something that we can deal with.
 So we have ...

 Ya know we COULD have moved the 4 to the left
 at the start of the problem,
 but it wouldn't have changed much else.
 Next one ...
 If the equation has an algebraic part and an exponential part,
 we use the same strategy.
 The exponential part takes the place of the trig part.
 Say we have ...

 We choose ...

5X2 = u       e2Xdx = dv

 That leaves ...

 10Xdx = du        e2X = v

 Before you go on, take another look at dv and v,
 and make sure you see how we came up with the v.
 It's the answer to the question:
 "What has e 2Xdx as it's derivative?"
 So we have ...

 We can combine some terms and clean this up a bit to ...

 Taking another swing at the integral that's left,
 and using the same strategy, we have ...

5X = u       e2Xdx = dv


5 = du        e2X = v

 Round two gives us ...

 We can deal with that last integral directly now,
 so all together we have ...


 The last case to consider is the combination
 of a trig part and an exponential part.
 This is the most involved type,
 because neither of these two types are going to go away
 no matter how many times we take a derivative.
 The strategy for this type sounds really silly, but it works.
 What we do is go through the process a few times.
 With any luck, we wind up subtracting an integral on the end of the partial answer
 that is the same thing as the problem we started with.
 It might not make a lot of sense yet, but it will ...
 Let's try one

 This time we have an eX and a trig term.
 It doesn't really matter which term you choose for u
 and which term you chose for dv.
 So let's choose ...

eX = u       sinXdx = dv


eXdx = du       -cosX = v

 That gives us ...

 Simplifying the signs a bit we get ...

 For the second round we choose ...

eX = u       cosXdx = dv


eXdx = d u       sinX = v

 There is nothing magic about these choices,
 we can go either way we want on these.
 Applying the formula to the integral on the right, we get ...

 OK! Now we really have something!
 If you look closely at what we have,
 it is of the form ...


 We can combine terms and get ...

 Which means ...

 So our answer is ...

 Home Stretch!
 OK, the page is about over, 
 but this might be the most important part for your long term math learning.
 This is where this integration by parts comes from.
 Do you remember the derivative product rule?
 The formula looks like this:

 (F(X) G(X))' = F(X) G'(X) + F'(X) G(X)

 To get to the integration by parts formula,
 we just take the integral of both sides of this.
 As long as we do the same thing to each side,
 and don't divide by zero,
 we can do almost anything we like.
 We get ...

 Simplify the left side ...


 and rearrange the terms ...

 And there it is!
 So now we know where it came from,
 what's the big deal?
 The big deal is that now, if you don't remember the "parts" formula,
 you can work it out from the product rule.

   copyright 2005 Bruce Kirkpatrick

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