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PROFESSOR: Hi.
The aim of our lecture today is to present what is perhaps
the singly most powerful recipe for evaluating
indefinite or, for that matter, definite integrals.
It's called integration by parts, and therefore, today's
lecture is called 'Integration by Parts'.
And what I want to emphasize is essentially that this is a
technique which shows us how to find indefinite integrals,
but then in theory, it changes nothing over
what we've had before.
This is essentially, as we've mentioned, a recipe for being
able to simplify or reduce indefinite integrals to things
that we can handle.
Well, it centers around the rule for
differentiating a product.
And to show you what I mean by that, let's just barge right
into this thing.
In differential notation, we call the 'u' and 'v' our
differentiable functions of 'x'.
The differential of 'uv' is 'u' times the differential of
'v', plus 'v' times the differential of 'u'.
And solving for 'udv', just transposing, we have that
'udv' is the differential of 'u' times 'v' minus 'vdu'.
Now, what we do is we integrate both
sides of this equality.
On the left-hand side, we have interval udv.
On the right-hand side, recall that the integral of the
derivative just gives us back the original function.
In other words, if we integrate the differential of
'uv' up to an arbitrary constant, we get
back 'u' times 'v'.
The integral of 'vdu' we'll just denote by integral 'vdu'.
And observe that there's really an arbitrary constant
coming from this term, an arbitrary constant coming from
this term, an arbitrary constant
coming from this term.
But since the sum and difference of arbitrary
constants is still an arbitrary constant, we can
neglected all the constants here, and just lump them
together, and at the end write plus 'c'.
Or we can leave the 'c' off, making it implied that the
arbitrary constant is contained in the symbol
integral 'vdu'.
All right, now we've done this, but a much more
important question would be-- or exclamation, for that
matter-- would be so what?
See, who cares whether this happens to be true or not?
The answer is we care.
That's why we're making up a lecture on it.
And the reason that we care is this.
That frequently, for a given 'u' and 'v', 'udv' may have a
completely different nature about it than does 'vdu'
insofar as being able to find the integral.
Now again, the best way to show what we mean by this is
to choose a particular example.
Let me start, just arbitrarily here, picking a couple of
simple functions for us to differentiate or to integrate.
Let's pick 'u' to equal 'x' and 'v' to equal 'sine x'.
Now, it's simple to see from this example that in this
case, if 'u' equals 'x' and 'v' is 'sine x', then 'dv' is
''cosine x' dx'.
In fact, let's write that down so we can see it here.
Notice that in this particular case, integral 'udv' is
integral of 'x 'cosine x' dx'--
'udv'.
On the other hand, what is 'vdu'?
'v' is 'sine x'.
Then since 'u' equals 'x', 'du' is 'dx'.
Therefore, 'vdu' is just integral ''sine x' dx'.
Now, look at these two integrals.
This one we can integrate just by looking at it.
It's minus 'cosine x'.
This one, well, offhand, we don't know a function whose
derivative with respect to 'x' is 'x 'cosine x''.
However, by this recipe, we now have a way of showing how
to 'udv' once we know integral 'vdu'.
In other words, in this particular problem, letting
'u' equal 'x' and 'dv' being ''cosine x' dx', we can
proceed as follows.
Namely, by this recipe, we get what?
Integral 'x 'cosine x' dx' is equal to-- this is 'udv'
equals 'u' times 'v'.
Recall that 'u' was 'x', 'v' is 'sine x', so 'u' times 'v'
is 'x sine x', minus integral 'vdu'.
But 'v' is 'sine x', 'du' is 'dx', so we wind up with this.
But now we know that the integral of 'sine x' with
respect to 'x' is minus the 'cosine x'.
The minus and the minus cancel to give me a plus.
And I wind up with that the integral of 'x 'cosine x' dx'
is 'x sine x' plus 'cosine x' plus an
arbitrary constant, OK.
That's all integration by parts really is.
All we do here is we evaluate one integral by knowing
another integral.
And by the way, what I meant here when I said that one did
not have to change any of the theory, what does the integral
mean here anyway?
What we mean by this is what?
Find all functions whose derivative with respect to 'x'
is 'x cosine x'.
Now, what integration by parts did for us was it showed us
how to get this result.
But if we didn't know how to get this result, observe that
we check the answer the same way as always.
Namely, one thing that hopefully we can do is
differentiate this thing.
And if we differentiate it, let's see.
This is a product.
It's the first factor times the derivative of the second.
That's 'x cosine x'.
Plus the second factor times the derivative of the first.
That's just 'sine x'.
The derivative of 'cosine x' is minus 'sine x'.
The derivative of a constant is 0.
And if we now sum up the pieces that we have here, we
wind up with 'x cosine x'.
Exactly what we were supposed to get, OK?
Now, you see again, nothing has happened in theory.
All we have now is a technique that allows us to explicitly
solve certain problems that we couldn't have solved
otherwise, even though we could've stated them before.
For example, let's go back to a problem that technically
speaking we could have presented before we got to
this unit of our material.
Let's support that the region 'R' is the region bounded
above by 'y' equals 'cosine x', below by the x-axis
between the y-axis at 'x' equals pi/2.
What we want to do is take this region 'R' and revolve it
about the y-axis.
And what we'd like to find is what volume is generated when
the region 'R' is revolved about the y-axis.
Now, when we talked about volumes and cylindrical
shells, we already had the recipe.
Recall that to rotate this about the y-axis, just to
refresh memories here, if we call this 'x' and we call this
'y', the element of volume was what?
Well, you swung out a circumference '2 pi x' times
height 'y' times thickness 'dx'.
In other words, that the volume when this was rotated
about the y-axis was just 2 pi integral 'x 'cosine x' dx' as
'x' goes from 0 to pi/2.
You see, I rigged this particular problem so we
wouldn't have to waste the fact that we've already found
out what the indefinite integral of 'x
'cosine x' dx' is.
In other words, we get down to this step here.
Now, by the first fundamental theorem of integral calculus,
we know that whatever this integral is, it's also equal
to 'G of pi/2' minus 'G of 0', where 'G' is any function of
'x' whose derivative with respect to 'x' is the
integrand here, which is 'x cosine x'.
Now, you see, this part, or up to here, we could have got
this result without today's lecture.
The only thing that we learned to do today that we couldn't
do before was the idea of how we find a function whose
derivative is 'x cosine x'.
In fact, what we found was what?
A function whose derivative is 'x cosine x' is just 'x sine
x' plus 'cosine x'.
So by the first fundamental theorem, the answer to our
volume problem is that it's 2 pi times the quantity 'x sine
x' plus 'cosine x' evaluated between 0 and pi/2.
Putting in 'x' equals pi/2, this term is pi/2.
This term is 0.
Putting in the lower limit when 'x' is 0, this term is 0.
This term is 1.
We subtract the lower limit.
That gives us a minus 1.
And therefore, the volume that we get is simply 2 pi times
pi/2 minus 1.
Again, notice then, that the only new thing that we had to
do here was devise a way for finding a function whose
derivative with respect to 'x' was 'x cosine x'.
By the way, I think you'll observe over here that I
started in kind of a rigged fashion, namely, I started
with a specific 'u' and 'v'.
It might be interesting to see how would one tackle this
problem if one had just been given the 'x 'cosine x' dx'
and had to work from there.
In other words, instead of working backwards now, let's
present the problem as it would come up in a real
problem, namely, all of a sudden, we find that we have
to integrate 'x cosine x'.
And the idea go something like this.
You see, we say lookit.
If this 'x' weren't in here, this would be
pretty easy to handle.
So why don't we do this?
Why don't we call this piece here 'u' and call
this piece here 'dv'.
Then, you see, 'du' would just be 'dx'.
'v' would just be 'sine x'.
In other words, the differential of 'sine x' is
'cosine x dx'.
See, if 'u' is 'x', 'du' is 'dx'.
If 'dv' is 'cosine x dx', 'v' is equal to 'sine x'.
I suppose, technically speaking here, I should put in
plus a constant, but I'll show later why we can disregard
this constant.
Roughly speaking, it just hinges on the fact that we
want only one solution.
But we'll talk about that later.
For the time being, I'll just leave out the fact that
there's a whole family of functions.
All I want is one function whose differential
is 'cosine x dx'.
At any rate, I now apply my recipe, namely, integral 'udv'
is equal to 'u' times 'v'--
that's 'x sine x'--
minus integral 'v', which is 'sine x', times 'du',
which is just 'dx'.
In other words, integral 'x 'cosine x' dx' is just 'x sine
x' minus 'sine x dx'.
But I know how to integrate 'sine x'.
It's just minus 'cosine x'.
Therefore, replacing integral 'sine x dx' by minus 'cosine
x' plus a constant.
The minus cancels here.
I wind up with the same result as before, only now hopefully
in terms of a smooth flow not that didn't
presuppose the answer.
Namely, this integral is 'x sine x' plus
'cosine x' plus a constant.
And by the way, I should mention here that this
technique does require a certain amount of ingenuity.
In other words, in a given problem, where you're not told
what to call 'u' and what to call 'v', where you have the
choice of picking what's going to be 'u' and what's going to
be 'dv', there is a certain element of insight that one
needs to guess wisely.
In this particular problem, it's kind of obvious
which way to guess.
But let me, just to give you the experience, show you what
happens when you guess incorrectly.
For example, given the same problem, why couldn't I have
said gee, I think I'll let 'cosine x' be 'u', and what's
left over, namely, 'xdx'.
That will be 'dv'.
Now, notice that mathematically, no harm is
done this way.
Namely, if 'u' is 'cosine x', 'du' is minus 'sine x dx'.
If 'dv' is 'xdx', 'v' is 1/2 'x squared'.
And now, applying the recipe, integral 'udv' is what?
It's 'uv'.
That's 1/2 'x squared' times 'cosine x' minus
the integral 'vdu'.
But 'vdu' is already minus 1/2 'x squared' 'sine x dx', so
minus that is plus 1/2 integral ''x
squared' 'sine x' dx'.
And I wind up with the result that the integral 'x 'cosine
x' dx' is 1/2 'x squared' 'cosine x' plus 1/2 integral
''x squared' 'sine x' dx'.
And there is absolutely nothing at all wrong with
this, except that the chances are if I didn't know how to
find the function whose derivative was 'x cosine x',
what is the likelihood that I know a function whose
derivative is ''x squared' 'sine x' dx'?
In other words, here's a case where replacing 'udv' by 'vdu'
gave me an equivalent integral to evaluate which was more
cumbersome than the first.
In fact, somehow or other, it would seem that by multiplying
through here by 2 and solving for ''x squared' 'sine x' dx',
that it would seem that somehow or other this would be
a way of simplifying the integral ''x
squared' 'sine x' dx'.
In other words, if you had started with this, I think
you'd accept the fact, well, this still isn't as nice as
I'd like it, but it does seem less complicated than this.
You see, at least the power of 'x' is only to the first power
here, whereas here it's the second power.
In fact, this leads to a technique
called a reduction formula.
Again, using hindsight rather than foresight, having
motivated the integral ''x squared' 'sine x' dx', let's
see how I could have solved this particular problem if I
didn't know what the answer was going to be in advance.
You see, I say to myself, you know, this x squared here is
bothering me.
If that 'x squared' weren't here, this would be pretty
easy to handle.
Well, the best way to knock the 'x squared' down is to let
'u' equal 'x squared', in which case,
'du' will be '2xdx'.
In other words, your second power will now be replaced by
a first power.
It's not perfect, but at least there's hope here.
At any rate, what I'm saying is motivated into calling 'u'
'x squared' and therefore letting 'dv' be the rest of
this integrand, which is 'sine x dx', we wind up with the
fact that 'du' is '2xdx' and 'v' is minus 'cosine x'.
Well, you see now, remembering that 'u' is 'x squared', 'dv'
is 'sine x dx', the recipe tells us that integral 'udv'
is 'u' times 'v'.
That's minus 'x cosine x' minus integral 'vdu'.
But integral 'dvu' is integral twice 'x 'cosine x' dx' with a
minus sign.
The minus sign in the recipe makes it plus.
And we therefore wind up with that integral ''x squared'
'sine x' dx' is equal to minus ''x squared' 'cosine x'' plus
twice integral 'x 'cosine x' dx'.
By the way, just to refresh your memories on this, if we
come back and compare this with what we had over here,
it's exactly the same thing.
The only difference now is that we arrived
at this thing directly.
Now you see the point is, having reduced this to this, I
can now use integration by parts again to simplify this.
In other words, what I would do next is let 'u' equal 'x'
and 'dv' be 'cosine x dx'.
But let's face it, I've already done this problem.
Time is getting short.
Why do it again?
The point is that integral 'x cosine x' using integration by
parts again is going to come out to be 'x sine x' plus
'cosine x' plus 'c' because that's what we got before when
we used integration by parts to do this.
At any rate, what I would now do is replace integral 'x
cosine x' in here by this, OK?
And if I do that, I have what?
Integral ''x squared' 'sine x' dx' is equal to what?
Minus ''x squared' 'cosine x'' plus twice times this, which
is twice this.
That's what?
2 'x sine x' plus 2 'cosine x' plus a constant.
And by the way, again notice, if I wanted to check whether
this was right, the one thing I can do if somebody gives me
this is what?
I can differentiate it with respect to 'x'.
When I differentiate it with respect to 'x', if I get ''x
squared' 'sine x'' as my result, this is correct.
Otherwise, I've made a mistake.
But the idea is keep in mind that these can always be
checked because they're inverse derivatives.
They can always be checked by actual differentiation.
And again, notice the beauty of this technique.
If somebody said construct the function whose derivative with
respect to 'x' is ''x squared' 'sine x'', notice how
difficult it would be to try to piece this thing together.
See, notice the difference between being able to check
that this is a right answer and how one would find this.
This is the technique of a power of integration by parts.
In fact, in making up exam questions, this is exactly how
professor sometimes do this.
That what you start with is you start with this, then give
a person this problem and tell them to differentiate this.
And that whole mess simply comes out to be ''x
squared' 'sine x''.
And that's why a lot of times when you have a messy problem
to differentiate and you look at the answer in the back of
the book, you become amazed to see that the thing really
simplified.
Again, notice the idea.
The integrand here is relatively simple.
What we have to differentiate to get that relatively simple
integrand is not quite so simple.
But enough about that.
Let me just go on with a few more examples.
See, I have plenty of examples in the exercises to keep you
busy and to give you drill on this.
All I'm trying to do now is to motivate places where one
would want to use integration by parts.
Well, let me show you one place it might be used.
One place that it's always worth trying is when you look
at the integrand, haven't the vaguest notion as to how to
integrate it, but you are able to differentiate it.
You see, you've got nothing to lose.
The worst that will happen is that the resulting integral
vdu will be no easier to handle than the original udv,
but at least there's a chance that something will work.
But let me give you a for instance.
I see I've written this rather badly.
This is not 'Stan'.
This is the integral of inverse 'tangent x dx', OK?
Integral inverse 'tangent x dx'.
What function has its derivative equal to inverse
'tangent x'?
Well, you see, the thing I do know hopefully is what the
derivative of inverse 'tangent x' is.
The derivative of inverse 'tangent x' with respect to
'x' is '1 over '1 plus x squared''.
And by the way, if you don't know this, there's no hope of
even beginning the integration by parts. in other words, we
have to assume that we can do something in a
problem like this.
At any rate, knowing this, what do I do next?
I say, well, since I know how to differentiate the inverse
tangent, that's a very natural thing to call 'u' because I
can certainly find 'du'.
Moreover, the nice thing is that calling 'dv' 'dx', if I
can integrate anything, it's certainly 'dx'.
Namely, what I'm saying is if 'u' equals inverse 'tan x' and
'dv' is 'dx', then 'du' is 'dx' over '1 plus x squared',
and 'v' is just equal to 'x'.
Now, therefore, integral inverse 'tan x dx', in other
words, integral 'udv', is just what?
It's still just 'u' times 'v'.
That's 'x' times inverse 'tan x' minus 'vdu'.
That's minus integral 'x' times 'dx'
over '1 plus x squared'.
Now, look what happens in this case.
Obviously, with time being short and having other places
to show you my failures, namely in the exercises, I
picked one that worked nicely.
Here's the case, you see, that the integral inverse 'tan x
dx', by using parts, becomes replaced by integral 'xdx'
over '1 plus x squared'.
This is one I can handle from before.
This is one where the substitution 'u' equals '1
plus x squared' works very nicely for me.
Or if you're getting glib with these things, just observe
that this integrand is 1/2 natural
log '1 plus x squared'.
Because how would you differentiate natural log '1
plus x squared'?
It would just be what? '1 over '1 plus x squared'' times the
derivative of this, which is '2x'.
The 2 and the 1/2 would cancel, and we just have left
'x over '1 plus x squared''.
In other words, again, notice that the answer here is what?
This is the function that one would have to differentiate
with respect to 'x' to wind up with inverse 'tan x'.
I'll show you another example in a few minutes.
What I did just remember is that I mentioned why we can
drop the arbitrary constant.
Let me just take a minute here and show how this problem
should have really been worked.
If we wanted to be precise here, we'd say OK, 'u' equals
inverse 'tan x'.
Therefore, 'du' is 'dx over '1 plus x squared''.
That part's fine.
Then we say, OK, 'dv' is 'dx'.
Now, if 'dv' is 'dx', what must 'v' be?
Well, 'v' could be 'x', but it must be something of the form
'x' plus some constant, which I'll call 'c1'.
Could I have chosen 'v' to be 'x plus c1'?
Why not?
In fact, what would have happened if I did that?
Notice that what would have happened in this case is that
'u' times 'v' now would simply have been 'x plus c1' times
inverse 'tan x', whereas integral 'vdu', instead of
being 'xdx' over '1 plus x squared' would have been ''x
plus c1' dx' over '1 plus x squared'.
In other words, leaving the constant in, I would get this.
And now all I want you to see here, and I've done this just
for this particular example, and in the exercises we'll
work it out in general.
The whole idea is this.
This, of course, can be written as two terms, one of
which is 'x inverse tan x', and the other of which is 'c1
inverse tan x'.
This also can be written as two integrals.
Namely, with the minus sign kept aside here, this is
integral of 'xdx' over '1 plus x squared' and plus integral
'c1 dx' over '1 plus x squared', the minus making
this minus.
But we know that the integral 'dx' over '1 plus x squared'
is inverse 'tan x'.
In other words, if we now separate these into
pieces, we have what?
'x inverse tan x' minus integral 'xdx'
over '1 plus x squared'.
And notice that the 'c1' term drops out because we have 'c1
inverse tan x' occurring twice.
It's this term here, and it's also
this part of our integral.
And see, with the minus sign here, they just cancel out.
And that's why we really don't have to keep the arbitrary
constant in there, but if we wish to, we may.
At any rate, let me summarize with just one more problem to
make sure that we have the technique down fairly pat.
Let's suppose we were given integral 'log x dx'.
What we certainly do know I hope by this time is that the
derivative of 'log x' with respect to 'x' is '1/x'.
Since I can differentiate 'log x', it becomes very meaningful
to try letting 'u' equal 'log x'.
If I do this, 'u' is 'log x'.
'dv' is 'dx'.
'du' then becomes ''1 over x' dx', and 'v' is 'x'.
And now I put in plus 'c' in parentheses here to indicate
that if you wanted to use it, you could.
At any rate, what happens now?
Integral udv, integral 'log x dx' in this case is 'u' times
'v', which is 'x log x', minus integral 'vdu'.
Well, you see the 'x' times the '1/x' is just 1.
Therefore, 'vdu' is just 'dx', so minus that
is just minus 'dx'.
In other words, minus the integral 'vdu' is just minus
integral 'dx'.
And now, you see I have what? 'x log x'.
And this is very easy to integrate.
It just happens to be minus 'x', then I tack on the
arbitrary constant.
Again, to check whether this was the right answer, all I
have to do is differentiate 'x log x' minus 'x' with respect
to 'x' and see if I get 'log x', and I will, OK?
Now, how would I use a problem like this, say,
in a practical situation?
Well, for example, suppose I wanted to find the area of the
region 'R' where 'R' was the region bounded above by the
curve 'y' equals 'natural log x', below by the x-axis, on
the left by 'x' equals 1, and on the right by 'x' equals
'b', where 'b' can be any number you want right now as
long as it's bigger than 1 in this diagram.
Notice that the definite integral says that the area of
the region 'R' is the definite integral from 1
to 'b', 'log x dx'.
And the first fundamental theorem of
calculus says lookit.
If you happen to know a function whose derivative with
respect to 'x' is 'log x', you can evaluate
this thing very quickly.
Namely, it's just 'G of b' minus 'G of 1' where 'G prime
of x' is 'log x'.
Well, you see, we do know a function now whose derivative
with respect to 'x' is 'log x', namely, 'x
log x' minus 'x'.
Consequently, to solve this problem, all I have to do is
evaluate this between the limits of 1 and 'b', which I
won't bother doing right now, OK?
Now, here's the idea.
What we've really learned here is one more powerful technique
for evaluating integrals.
How we arrive at integrals in no way changes in this
particular lecture.
All we have now is what?
A new technique whereby we evaluate udv by being able to
evaluate equivalently vdu.
At any rate, that finishes up what we had in mind for today,
and so until next time, goodbye.
NARRATOR: Funding for the publication of this video was
provided by the Gabriella and Paul Rosenbaum Foundation.
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