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HERBERT GROSS: Hi, today we start a new
segment of our course.
It could be entitled 'Techniques of Integration'.
And a subtitle could be the difference between 'knowing
what to do and knowing how to do it'.
You see, essentially in this chapter, it shall be our aim
to develop various recipes to evaluate the inverse
derivative.
So in fact, let's call it that.
Some basic recipes.
The idea is something like this.
Let's start with an easier problem.
Let's suppose we start with a differentiable function 'G'.
Suppose, for example, that 'G' prime is equal to 'f'.
Then in terms of the language of the indefinite integral, or
the inverse derivative, what we have is that the indefinite
integral of 'f of x' with respect to 'x' is then 'G of
x' plus 'c'.
Now, you see, the harder problem is
the inverse of this.
What we're going to be talking about is not do you find 'f'
once 'G' is given, but techniques for finding 'G'
once 'f' is given.
And as we've seen on several occasions already, that given
'f', it is not that easy, in general, to find a function
whose derivative is 'f'.
In fact, in many cases, we can't do it as explicitly as
we would like to no matter how sharp we are.
For example, oh, I've just reviewed this so that you can
think of this in one block.
Given 'f' to require 'G' may not exist in familiar form.
I don't want to pin down what familiar means any more than
that right now.
For example, the illustration that we've used several times
already in the course, suppose 'f of x' is 'e to
the minus 'x squared''.
And we want a function whose derivative is 'e to the minus
'x squared''.
In terms of areas, the second fundamental theorem of
integral calculus, we can construct such a function in a
rather straightforward way, even though it may be
difficult to name it in terms of familiar functions.
Namely, what we do is we take the curve 'y' equals 'e to the
minus 't squared'' in the yt-plane.
Compute the area bounded above by that curve, below by the
t-axis, on the left by the y-axis, and on the right by
the line 't' equals 'x'.
And the area under the curve is a function of 'x'.
And what we've already seen is that in this particular case,
'A prime of x' is precisely 'e to the minus 'x squared''.
In other words, this area function, which is a function
of 'x', does turn out to be the function whose derivative
with respect to 'x' is 'e to the minus 'x squared''.
Now you see, that will not be primarily what we're
interested in, in this particular
phase of our course.
What we're interested in this phase of the course is given
various classifications for the function 'f', to find
recipes that will give us 'G' in familiar form, where 'G' is
the function whose derivative is 'f'.
Summarize for you to look at here.
The objective of this block of material is to find recipes
for finding 'G of x' for various types of 'f of x'.
Now the best way to illustrate what that means is to actually
pick a few examples.
And by the way, what you'll notice in today's lecture is,
roughly speaking, we do nothing at all that's new.
The new stuff will begin with out next lecture essentially.
But for now what we'll really be doing is giving us an
excuse to pull together all of the recipes that we've
developed so far.
All of the recipes either called indefinite integrals or
inverse derivatives.
For example, the first class of functions which we could
handle were things of the form integral ''u to the n' du'.
Recall that very early in our course, we found out that the
integral of ''u to the n' du' was ''u to the 'n + 1'' over
'n plus 1' plus a constant if 'n' was unequal to minus 1.
And then, in our last block of material we learned to handle
the case where 'n' was equal to minus 1.
Namely, when 'n' is minus 1, integral ''u to
the minus 1' du'.
Namely, integral 'du' over 'u' is natural log absolute value
of 'u' plus a constant.
Now this then, is one of our basic building blocks.
What we're saying is, if an indefinite integral can be
written in this form, we already have a recipe that
allows us to evaluate the integral.
By the way, let us also point out as was mentioned again,
several times in the past, that 'u' is the generic name
for a variable over here.
It is not important in this recipe that we use 'u'.
What is important is that whatever is being raised to
the n-th power inside the integrand must be precisely
the thing that's following the 'd' over here.
For example, instead of 'u', suppose I used
'sine x' over here.
In other words, suppose I had 'sine x' to the n-th
power 'd sine x'.
See, what's being raised to the n-th power is the same
thing as what's following the 'd'.
That's all I really care about.
In other words, if this had been the integrand, the recipe
would have said, this is 1 over 'n + 1' times 'sine x' to
the 'n + 1' power.
Plus a constant if 'n' is not equal to minus 1.
And it's natural log absolute value of 'sine x' plus a
constant if 'n' is equal to minus 1.
Again, the important thing is not the 'u', but the fact that
what's being raised to the n-th power is the same as
what's following the 'd' over here.
By the way, we can write this in more familiar form.
You see, namely, notice that the differential of 'sine x'
is 'cosine x dx'.
And consequently, in this form what our recipe says is that
integral 'sine x' to the n-th power times 'cosine x dx' is
given by this.
In other words, it's sine to the 'n + 1' power 'x over 'n +
1'' plus a constant if 'n' is unequal to minus 1.
And it's the natural log absolute value of 'sine x'
plus 'c' if 'n' is equal to minus 1.
Again, notice this looks more complicated than this.
But to get our problem, to get into the form ''u to the n'
du', notice that the natural substitution here is to let
'u' equal 'sine x'.
In which case, 'du' would be 'cosine x dx'.
In other words, the extra factor of 'cosine x' here
actually is to our advantage over here.
That when we make the substitution, it gives us the
form that we want.
With the factor missing, we're in a little bit of trouble.
By the way, this is what makes this a rather exciting topic.
That there seems to be no really canned ways of finding
indefinite integrals.
That we have recipes.
Frequently, the recipe that we have doesn't cover the
situation that we have.
In fact, I know a lot of times in teaching a course like this
students say, why do we have to learn these recipes and
techniques?
Why can't we just look them up in the tables?
And the answer quite simply, is that in many applications,
the integral that we have doesn't appear in
that form in the table.
That frequently what we must do is a tremendous amount of
algebraic manipulation and the like to even get the given
integral to resemble one that we can find in the table.
Oh, we're starting off on a fairly
straightforward level here.
Nothing really too tricky.
But at least enough so that we get an idea of what kind of
gimmicks are involved.
For example, suppose we had 'cosine x' to the seventh
power 'dx'.
You see the idea here is that when you have cosine raised to
a power, it would have been nice if a 'sine x' factor had
been in here.
Because the differential of 'cosine x' is
minus 'sine x dx'.
And what I'm driving at here is that we often have to have
a certain kind of ingenuity that allows us to reduce a new
integrand to a more familiar one.
And, as I say, there are many
techniques done in the textbook.
There'll be plenty of exercises again, that will
allow you to experiment on this.
All I want to do is hit a few highlights and create the mood
as to what's going on.
For example, you see frequently what happens is, if
we have 'cosine x' raised to an odd power, a very
convenient way of handling this is to split off one of
the factors of 'cosine x' separately.
In other words, write this as 'cosine x' to the sixth power
times 'cosine x dx'.
The idea being that cosine squared can be written as '1
minus 'sine squared''.
In other words, 'cosine x' to the sixth power is really
'cosine squared x' cubed.
But 'cosine squared x' is '1 minus 'sine squared x''.
In other words, notice again the role of the trigonometric
identities.
I take this integrand, write it this way, and now you see
if I use the binomial theorem and expand this, notice that
my typical term will have 'sine x' to a power multiplied
by a 'cosine x'.
And that's precisely the kind of a term that I handled in
the previous case.
And again, I'm not going to bore you
with the details here.
That is the easiest part of the problem.
The hard part is getting to understand why one would want
to use this particular kind of form.
Another variation is, what happens if the cosine is
raised to an even power rather than to an odd power?
How do we handle that?
Again, the same kind of use of trigonometric identities.
For example, it would not be to our advantage in this case,
to replace cosine squared by 1 minus sine squared.
Because then we would wind up with the same kind of
integrands that we started with, only using powers of
sine instead of powers of cosine.
You see, the idea is that if we have a power of sine, we
want a cosine factor to accompany it.
And inversely, if we have a power of cosine, we'd like a
factor of sine to accompany it.
Why?
So that we can get the integral into the form ''u to
the n' du'.
In this particular case, a very common device is to
invoke the identity that cosine squared of an angle is
1 plus cosine twice the angle over 2.
You see again, this is where all of these particular
identities come into play.
It's not a case of saying, let's learn these identities
so that we can recite them.
It's a case of getting into a situation where you want a
certain answer, can't handle it conveniently in the form
that you're in, and hope that you can pick off a synonym
that allows you to tackle the problem more successfully.
That's where this becomes a matter of insight, lucky
guessing, skill, whatever you want to call it.
It's one thing to transform an integral into an equivalent
one, and another thing to have that equivalent integral be
something that you can handle.
Well, at any rate, for example in a problem of this type, a
tendency is to write cosine sixth theta as
cosine squared cubed.
Cosine squared theta is 1 plus cosine 2 theta over 2.
To cube this we can take out the factor of 1/8.
You see, we use the binomial theorem to expand 1 plus
cosine 2 theta cubed.
And notice that of the terms that we get,
this term we can handle.
This term we can handle, it's just cosine
to the first power.
Which involves a sine term when we integrate.
And here's a cosine cubed.
But we've just seen how you can handle something to an odd
power, so that reduces this term to a
type that we can handle.
And then we see what?
That we have another term in here, which is
cosine squared 2 theta.
That's an even power of cosine.
Again, we do what?
We write cosine squared 2 theta as 1 plus cosine of
twice this angle.
That's 1 plus cosine 4 theta over 2.
In other words, we keep reducing these things, always
hoping to break them down into collections of problems that
we solved before.
As I said, these are things that we'll drill on in the
exercises and I think it will become more meaningful there.
Here, as I say, the main aim is to create the mood that
what we're trying to do is to reduce unfamiliar problems to
equivalent familiar ones to which we
already know the answer.
Now again, in the way of review, sometimes an integrand
occurs in the form of the sum or difference of two squares.
We've already handled this in terms of the inverse circular
functions and the inverse hyperbolic functions.
It's covered for the first time in our textbook at this
particular stage of the course.
And I think it's a tough enough concept so that it's
worth emphasizing at this stage.
So I just separate this particular type out.
How do we handle sums and differences of squares?
And what I intend to show is, is that by a suitable
trigonometric function, either circular trigonometric or
hyperbolic trigonometric, we can always reduce that kind of
a problem to the type that we can handle.
Again, to keep computation at a minimum, I have not tried to
pick a rather complicated situation.
In fact, I've picked ones where you can very easily look
them up in the table if that was the
main aim of the exercise.
You see here, it's not so much that I want to get this
answer, as much as it is that I want to emphasize the
technique that one uses.
The idea is something like this.
Here we see the difference of two squares.
That's the square root of a squared minus x squared.
What this should do for us is to suggest a right triangle
whose hypotenuse is a and one of whose sides is 'x'.
That's the diagram I've drawn over here.
This is my reference triangle.
As I say, we've done this before, but I would like to
reinforce this at this particular stage.
Especially now that we're more familiar with both the inverse
circular and inverse hyperbolic functions.
At any rate, we use this reference triangle.
From this triangle, the easiest relationship to pick
off involving 'x' is that sine theta is 'x over a', or 'a
'sine theta'' equals 'x'.
From which it follows that ''a' cosine theta
'd theta'' is 'dx'.
The square root of ''a squared' minus 'x squared''
over 'a' is cosine theta.
Therefore, the square root of ''a squared' minus 'x
squared'' is 'a cosine theta'.
If we now make this substitution in this
particular integral, we wind up with integral 'dx' over the
square root of ''a squared' minus 'x squared'' is equal to
integral 'd theta'.
Which is theta plus 'c'.
And since theta is the number whose sine is 'x over a', we
have that the answer to this problem is inverse 'sine 'x
over a'' plus a constant.
Again, just a straightforward review, but a
rather important technique.
In our textbook, this is put in a separate section, and the
technique is called trigonometric substitution.
And I just wanted to show you what motivates the
trigonometric substitution.
The harder part as I mentioned in our previous lecture, is
how you motivate trigonometric substitutions when it's the
hyperbolic functions that are involved.
You see, what I'd like to do now is instead of dealing with
the square root of ''a squared' minus 'x squared'',
let me just reverse the order of the terms.
And now let me take integral 'dx' over the 'square root of
''x squared' minus 'a squared'''.
Notice that the difference between ''x squared' minus 'a
squared'' and ''a squared' minus 'x squared'' is just a
factor of minus 1.
And the fact that that minus 1 is under the square root sign,
really means in a certain manner of speaking that you've
multiplied by i, the square root of minus 1, to transform
the integral that we just had into this
particular form here.
And in terms of our lecture on the hyperbolic functions that
should, in some way, give us a preview of things to come in
the sense that the hyperbolic functions are very strongly
related to the circular functions.
My technique for doing this--
and again, this is highly subjective.
It's very simple.
I learnt the circular functions long before I learnt
the hyperbolic functions.
Consequently, I feel very much at home with
the circular functions.
As a result, whenever I see the sum or difference of two
squares, I always think in terms of a triangle.
I make a circular trigonometric substitution.
If that works for me, fine.
I haven't lost anything.
If it doesn't work for me, it gives me the hint that I can't
seem to get in my own mind without this hint as to which
hyperbolic function I should use.
Let me show you this in terms of this particular example.
If I saw this problem from scratch and didn't have any
previous knowledge of how to handle this I'd say, this
seems like I should think of a right triangle whose
hypotenuse is 'x' and one of whose sides is 'a'.
So I think of this particular reference triangle.
Now, thinking of this triangle, and in fact, even
look at this triangle, I very easily pick off that 'x' is
equal to 'a cosecant theta'.
See, 'x over a' is 1 over sine theta.
That's cosecant theta.
Therefore and remembering how to differentiate the cosecant,
I take the differential of both sides and I find that
minus 'a cosecant theta cotangent theta
'd theta'' is 'dx'.
Again, from this diagram I see that the square root of ''x
squared' minus 'a squared'' is--
well, let's see.
The square root of ''x squared' minus 'a squared''
over 'a' would be cotangent theta.
So the square root of ''x squared' minus 'a squared''
itself is 'a cotangent theta'.
Making the substitution in the integral 'dx' over the square
root of ''x squared' minus 'a squared'', notice that the
'a's cancel, the cotangent's cancel, and I'm left with
'cosecant theta 'd theta'' and a minus
sign inside my integral.
In other words, somehow or other, this circular
trigonometric function replaces the integral 'dx'
over the 'square root of ''x squared' minus 'a squared'''
by integral minus 'cosecant theta 'd theta''.
And the point I'd like to emphasize at this particular
stage is that we don't say that the
substitution has failed.
The substitution has been made.
We have replaced an integral involving 'x' by
one involving theta.
It's just that in the same way that it's difficult to
integrate secant theta--
in fact, that will be later in this block of
material we'll do that.
But the point is that secant and cosecant are rather
difficult functions to integrate.
In fact, at this stage of our development, we do not know a
function whose derivative is cosecant theta.
Consequently, we don't know one whose derivative is minus
cosecant theta.
We seemed to have arrived at an impasse here.
We have successfully made the substitution, but the key
point is that the new integral is no easier for us to handle
than the old integral.
Now, here's how I visualize the hyperbolic substitutions.
We did this in the last lecture.
I think it's tough enough, so I'd like to do it a second
time and give you a chance of seeing how easy this is once
you see the structural form.
Notice that the substitution I made here was 'x' equals 'a
cosecant theta'.
The cosecant was the key step here.
The reference triangle, knowing how to draw this
geometrically, allowed me not to consciously have to think
of the trigonometric identities.
But what I really did in evaluating this problem, if I
left the diagram out and wanted to do it analytically,
the identity that I made use of was the one that said
cosecant squared theta minus cotangent squared theta is
identically 1.
In other words, I would like an identity for the hyperbolic
functions that has the structure that the difference
of two squares is identically 1.
The easiest one I can think of is the one that says cosh
squared theta minus sinh squared theta is 1.
You see, structurally, these two are equivalent.
In other words, sinh and cosh are related hyperbolically,
the same way that cotangent and cosecant are in terms of
circular functions.
The identification is this.
Since in the original problem I tried 'x' equals 'a cosecant
theta', and since the identification here is that
cosecant and cosh are matched up, what I try next-- see once
this has failed, I very quickly now say, OK, what I'll
do is instead of 'x' equals 'a cosecant theta', I'll try 'x'
equals 'a cosh theta'.
That's exactly what I did over here.
'a cosh theta' equals 'x'.
From which it follows that 'dx' is 'a
sinh theta 'd theta''.
The square root of ''x squared' minus 'a squared'' is
just ''a squared' cosh squared theta'.
That's 'x squared'.
Minus 'a squared'.
That can be written as the square root of 'a squared'
times the quantity 'cosh squared theta minus 1'.
Now, the interesting point is the 'cosh squared theta minus
1' turns out to be sinh squared theta.
In fact, if we want to just look back here for a moment,
recall that's exactly what we have over here.
That cosh squared theta is 1.
Well, cosh squared theta is 1 plus sinh squared theta.
cosh squared theta minus 1 is sinh squared theta.
And you might, say wasn't that terrific?
What a lucky break it was that this complicated expression
just happened to be sinh squared theta.
And now taking the square root I get a sinh theta.
The thing I'd like to point out is that the element of
luck was removed from this at the instant that I recognized
the fact that cosh and sinh were related by the same
identity that cosecant and cotangent were.
In other words, I rigged this thing so that when I finally
had to take this particular difference, I had to wind up
with an easy square root to extract.
You see, I had the right structure to begin with.
At any rate, to make a long story short here, 'dx' is 'a
sinh theta 'd theta''.
The square root of ''x squared' minus 'a squared'' is
'a sinh theta'.
Therefore, 'dx' over the square root of ''x squared'
minus 'a squared''.
Well, look how nicely this works out.
The a sinh theta cancels from both the numerator and the
denominator.
And it turns out that the integral that I'm looking for,
'dx' over the square root of ''x squared' minus 'a
squared'' is just theta plus a constant.
But now, recalling that a cosh theta equals 'x'.
In other words, cosh theta is 'x over a'.
And therefore, theta is 'inverse cosh 'x over a'', I
arrive at the result that this integral is 'inverse cosh 'x
over a'' plus a constant.
In other words, what I hope that these two little examples
show is that when I have the sum or difference of two
squares, the trigonometric substitutions will usually
give me a crack at getting an equivalent integral that will
be easier to handle.
What may happen is that the circular functions won't help
me, but the hyperbolic ones will.
Or vice versa, and I hope that you see how these are so
delicately connected.
Well, there is one final type of technique that I would like
to show for today's lesson that's related to what we've
already done.
And with that, we'll conclude today's lesson and go into
some more elaborate recipes next time.
But this is a take-off on our old high school construction
of completing a square.
In other words, let's suppose we have an expression of the
form 'ax squared' plus 'bx' plus 'c'.
And this may sound old hat to you.
You may remember this as looking pretty much like the
development of the quadratic equation.
The idea works something like this.
First of all, I can factor an a out of here.
I'm assuming that a is not 0.
In fact, if a were 0, I really wouldn't have a
square term in here.
I factor out an 'a' and I'm left with 'a' times 'x
squared' plus ''b over a'x' plus 'c over a'.
Now it turns out that whenever you have something of the form
'x squared' plus something times 'x', to convert that
into a perfect square.
And again, we'll have ample opportunity to practice this
in the exercises.
You take half the coefficient of 'x' and square it.
Half the coefficient of 'x' here is 'b over 2a'.
If I square that it's 'b squared' over ''4a' squared'.
Now to keep the identity intact here, if I add in 'b
squared' over ''4a' squared', I must subtract that.
In other words, I just add and subtract this term that will
make this a perfect square.
In fact, what is 'x squared' plus ''b over a'x' plus ''b
squared' over '4a' squared''?
You'll notice that this is just ''x'
plus 'b over 2a' squared'.
'x squared'.
Just multiply this thing out.
You see how this particular thing works.
In other words, what I've done is I can write this as 'a'
times a perfect square.
And by the way, what's left over here if I multiply these
two terms through by 'a'?
This is 'c' minus 'b squared over 4a'.
In other words, recognizing that this is ''x' plus 'b over
2a' squared'.
And multiplying these two terms through by 'a', I wind
up with the fact that ''ax' squared' plus 'bx' plus 'c'
can always be written in this particular form.
This is called completing the square.
And notice that this is now a perfect square and this is
some constant.
And the question that comes up is, what does this have to do
with integral calculus?
In other words, with finding antiderivatives.
And again, as is so often the case, and granted
that as this course--
especially this chapter gets more complicated, the degree
of sophistication will become much greater.
But the basic idea will always remain the same.
Somehow or other, we will always try to reduce a new
situation to a more familiar one.
In other words, let's suppose now we don't have an integrand
that's the sum and difference of two squares, but we have an
integrand that involves a quadratic expression.
Say, for example, we have 'dx' over ''ax' squared'
plus 'bx' plus 'c'.
Well, you see, using hindsight rather than foresight, I
already took care of this problem before we started.
Namely, we just finished completing
the square over here.
Namely, what is ''ax' squared' plus 'bx' plus 'c'?
Notice it can be written in what form?
Let's come back here and take a look.
It's 'a' times ''x' plus 'b over 2a'
squared' plus some constant.
I don't care what it happens to be here.
I'll just call that constant 'c sub 1'.
And by the way, notice depending upon the relative
sizes of 'a', 'b', and 'c', this constant can be either
positive or negative.
Or in fact, even 0.
So to handle that what I'll say is I can always write this
denominator in the form 'a' times ''x' plus 'b over 2a'
squared' plus or minus some constant.
Where I'm assuming now that the constant is positive.
That's why I put the plus or minus sign in.
It's plus some positive constant or else it's minus
some positive constant.
Now what I can do is factor out an a from the denominator.
If I factor out an a from the denominator here, I have 1
over 'a' times a certain integral.
What integral?
'dx' over 'x plus 'b over 2a' squared' plus or minus still
some constant.
In other words, I had 'c1' here.
I factored out on 'a'. 'c2' was actually 'c1 over a'.
I just factored out an 'a' from this thing here.
But it doesn't make any difference.
The important point is I now have this written in this
particular form.
Now if I assume that 'c2' is a positive constant, notice that
any positive number is a perfect square.
Namely, it's the square of its square root.
And again, as I say with the exercises, this will become
clearer when we work with specific numbers.
But the idea here is what?
This is some perfect square, which I'll call 'k'.
And in other words, this integrand can now be written
in the form '1 over a', 'dx' over 'x plus 'b over 2a'
squared' plus or minus 'k squared'.
Well, again, what am I leading up to?
If I now make the substitution that 'u' equals 'x plus 'b
over 2a'', notice that my denominator simply becomes 'u
squared' plus or minus 'k squared'.
In other words, my denominator now has the form of either the
sum or difference of two squares.
On the other hand, if 'u' is 'x plus 'b over
2a'', what is 'du'?
Remember, 'b' and 'a' are constants.
'b over 2a' is a constant.
The derivative of a constant is 0.
So 'du' is equal to 'dx'.
In other words, notice that if I now make this substitution
in here, outside I have a '1 over a', inside I have 'du'
over ''u squared' plus or minus 'k squared''.
In other words, when I have a quadratic I can write that in
the form the sum and/or difference of two squares,
simply by completing the square.
Simply by completing square.
In other words, I can solve the problem of the quadratic
by converting it back into an integral, the type of which
I've solved before.
And this, by and large, becomes the technique that we
will use throughout this chapter.
Well, at any rate, I won't go into that in any more
detail right now.
We will pick up additional
techniques in our next lecture.
And until the next lecture, goodbye.
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