Calculus Substitution
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Do You du?


 When we found the derivative of a function like:


 F(X) = (3X2 - 2X + 5)5

 We had an outside function ( stuff ) 5
 and an inside function 3X 2 - 2X + 5 to deal with.
 We wind up with:

 F'(X) = 5(3X2 - 2X + 5)4(6X - 2)

 If we substituted the letter "u" for the inside part, we would have had:

  F(x) = u5

 And the derivative of u 5 is:

 F'(x) = 5u4 du

 The du part was created when we found the derivative.
 Since "u" stood for some complicated thing,
 the derivative of u (du) is probably not equal to 1.
 When we do an integral it's like we were given a derivative,
 and we have to find the function that it came from.
 If we are finding the integral of something 
 that looks like an outside/inside type function,
 we have to arrange it so the derivative of the inside part (the du)
 is sitting next to the outside part.
 That is, the (  ) n part.
 If we were asked to find the integral of something like:

 (3X - 4)5

 The first thing we do is say: 
"WOW! That looks like one of those outside/inside type things.
 We need a "u," and we need a "du."
 It looks like 3X - 4 should be the "u."
 That makes the expression:


 And here we hit the snag.
 We've got no du lying around.
 Since we said that 3X - 4 was our "u," then du should be the derivative of this.
 That would be 3dX.
 dX is equal to 1, so that's not a problem. But we don't have a 3 handy.
 Where do we get one?
 The rules say that you can only multiply something by 1 and not change the value.
 So here our fancy name for 1 is going to be 1/3 x 3 x dX.


 Writing the expression as a function so that we can work with it,
 and multiplying by this fancy 1:



 So we got the three we wanted, but we also had to take a 1/3 that we didn't want.
 What do we do with the 1/3?
 We put it in a place where it won't be in the way.
 Like to the left of the integral sign.


 So now we've got what we needed.
 We have the u (3X - 4),
 and we have the du (3dX).
 We have:


 Working out the integral, we get:


 Now we put back the (3X - 4) in place of the u ...


 and simplify.


 The du or dX or dWhatever, is created when we find the derivative.
 So it is used up, destroyed, absorbed, or whatever when we find the integral.
 In any case, it goes away.
 And as usual, don't forget the "+ C" part!
 We can't always find a du lying around the equation.
 We can change constants around, like we did in the last example.
 But if we need an X or even an extra power of X, we're stopped.
 For now anyway, but stay tuned ... 
 So if we have:


 The u part can be X 4 - 3X 2 + 2, so the du part needs to be 4X 3 - 6X.
 We don't have this term in the equation, 
 and we can't get it without changing the value of the equation.
 So we can't do this one with the tricks we have so far.
 Here are some we can solve now.
 These examples will take some studying.
 You may not get all or any of them right off.
 Remember that we can live with extra constants in the problems.


 So here we tried the obvious first, the 3X.
 That made du equal to 3dx. 
 We only had dx, but we can deal with differences in multiplied constants.
 Since 3X probably needs to be u here, what else is there? 
 3dX has to be du.
 It's the first step to really mastering these
 to see that you multiply du = 3dX by 1/3
 to keep du equal to 3dX and turn the 3dX into what you need 
 to substitute into the equation.


 The tricky part here was realizing that p is just a number,
 so it can be moved to the left and dealt with later.
 From there we have the same "fix the coefficient on the du"
 like the last example.
 Another fix the constants on the du deal.
 The equation was just nastier.
 This fix the constants on the du looks like a common thing, eh?
 This is a very tricky one, so watch closely!
 Most people start this one OK. That is, they set u = X-1.
 The problem is that when they stop there, 
 they notice that they are going to have an extra X in the problem.
 The trick is to do the second step on the right.
 That gives you SOMETHING to substitute for that first X.
 Then just multiply everything out and do the integral term by term.
 These problems are really like big puzzles.
 The only way to get good at them is to do a bunch of them
 and learn as many tricks to move the numbers around as you can.

   copyright 2005 Bruce Kirkpatrick

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