



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 
X^{2}
+ 2X or 7X^{12} 



Trigonometric 
like 
sinX or cos^{3}X 



Exponential 
like 
e^{X}
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 ... 





Example: 











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 ... 





4X^{2} =
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 ... 





Example: 





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 ... 





5X^{2} =
u e^{2X}dx
= dv 





That
leaves ... 





10Xdx
= du ½
e^{2X} = v






STOP!
TIME OUT! 





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 ^{2X}dx
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 e^{2X}dx
= dv 





5 = du
½
e^{2X} = 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 





Example: 











This
time we have an e^{X} 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 ... 





e^{X} =
u sinXdx
= dv 





e^{X}dx =
du cosX
= v 





That
gives us ... 











Simplifying
the signs a bit we get ... 











For
the second round we choose ... 





e^{X} =
u cosXdx
= dv 





e^{X}dx =
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. 





CAN
WE LEGALLY DO THAT? 





As
long as we do the same thing to each side, 


and
don't divide by zero, 


we
can do almost anything we like. 





SO
DO IT ALREADY!!! 





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 
