



When we
found the derivative of a function like:






F(X) = (3X^{2}
 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(3X^{2}  2X + 5)^{4}(6X  2)^{}






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





F(x)
= u^{5}



And
the derivative of u^{ 5} is: 


F'(x)
= 5u^{4} 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: 


u^{5}






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. 





Example: 









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. 





Example: 









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. 





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? 





Example: 


This
is a very tricky one, so watch closely! 









Most
people start this one OK. That is, they set u = X1. 


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 
