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HERBERT GROSS: Hi, today
we start a new
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00:00:35,460 --> 00:00:37,510
segment of our course.
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00:00:37,510 --> 00:00:41,380
It could be entitled 'Techniques
of Integration'.
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00:00:41,380 --> 00:00:45,000
And a subtitle could be the
difference between 'knowing
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00:00:45,000 --> 00:00:47,810
what to do and knowing
how to do it'.
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00:00:47,810 --> 00:00:51,280
You see, essentially in this
chapter, it shall be our aim
15
00:00:51,280 --> 00:00:57,790
to develop various recipes
to evaluate the inverse
16
00:00:57,790 --> 00:00:59,010
derivative.
17
00:00:59,010 --> 00:01:00,310
So in fact, let's
call it that.
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00:01:00,310 --> 00:01:02,260
Some basic recipes.
19
00:01:02,260 --> 00:01:03,280
The idea is something
like this.
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00:01:03,280 --> 00:01:04,920
Let's start with an
easier problem.
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00:01:04,920 --> 00:01:08,820
Let's suppose we start with a
differentiable function 'G'.
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00:01:08,820 --> 00:01:11,990
Suppose, for example, that 'G'
prime is equal to 'f'.
23
00:01:11,990 --> 00:01:15,170
Then in terms of the language of
the indefinite integral, or
24
00:01:15,170 --> 00:01:19,020
the inverse derivative, what we
have is that the indefinite
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00:01:19,020 --> 00:01:22,160
integral of 'f of x' with
respect to 'x' is then 'G of
26
00:01:22,160 --> 00:01:23,780
x' plus 'c'.
27
00:01:23,780 --> 00:01:26,430
Now, you see, the harder
problem is
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00:01:26,430 --> 00:01:28,160
the inverse of this.
29
00:01:28,160 --> 00:01:32,800
What we're going to be talking
about is not do you find 'f'
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00:01:32,800 --> 00:01:36,430
once 'G' is given, but
techniques for finding 'G'
31
00:01:36,430 --> 00:01:38,200
once 'f' is given.
32
00:01:38,200 --> 00:01:42,700
And as we've seen on several
occasions already, that given
33
00:01:42,700 --> 00:01:47,110
'f', it is not that easy, in
general, to find a function
34
00:01:47,110 --> 00:01:48,860
whose derivative is 'f'.
35
00:01:48,860 --> 00:01:52,620
In fact, in many cases, we can't
do it as explicitly as
36
00:01:52,620 --> 00:01:55,450
we would like to no matter
how sharp we are.
37
00:01:55,450 --> 00:01:59,280
For example, oh, I've just
reviewed this so that you can
38
00:01:59,280 --> 00:02:00,280
think of this in one block.
39
00:02:00,280 --> 00:02:05,160
Given 'f' to require 'G' may
not exist in familiar form.
40
00:02:05,160 --> 00:02:07,740
I don't want to pin down what
familiar means any more than
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00:02:07,740 --> 00:02:08,810
that right now.
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00:02:08,810 --> 00:02:12,610
For example, the illustration
that we've used several times
43
00:02:12,610 --> 00:02:15,700
already in the course, suppose
'f of x' is 'e to
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00:02:15,700 --> 00:02:17,260
the minus 'x squared''.
45
00:02:17,260 --> 00:02:20,050
And we want a function whose
derivative is 'e to the minus
46
00:02:20,050 --> 00:02:20,760
'x squared''.
47
00:02:20,760 --> 00:02:24,100
In terms of areas, the second
fundamental theorem of
48
00:02:24,100 --> 00:02:27,640
integral calculus, we can
construct such a function in a
49
00:02:27,640 --> 00:02:30,840
rather straightforward way,
even though it may be
50
00:02:30,840 --> 00:02:34,490
difficult to name it in terms
of familiar functions.
51
00:02:34,490 --> 00:02:38,460
Namely, what we do is we take
the curve 'y' equals 'e to the
52
00:02:38,460 --> 00:02:41,520
minus 't squared''
in the yt-plane.
53
00:02:41,520 --> 00:02:45,380
Compute the area bounded above
by that curve, below by the
54
00:02:45,380 --> 00:02:49,770
t-axis, on the left by the
y-axis, and on the right by
55
00:02:49,770 --> 00:02:51,510
the line 't' equals 'x'.
56
00:02:51,510 --> 00:02:55,080
And the area under the curve
is a function of 'x'.
57
00:02:55,080 --> 00:02:58,480
And what we've already seen is
that in this particular case,
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00:02:58,480 --> 00:03:02,090
'A prime of x' is precisely 'e
to the minus 'x squared''.
59
00:03:02,090 --> 00:03:05,780
In other words, this area
function, which is a function
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00:03:05,780 --> 00:03:08,950
of 'x', does turn out to be the
function whose derivative
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00:03:08,950 --> 00:03:11,920
with respect to 'x' is 'e to
the minus 'x squared''.
62
00:03:11,920 --> 00:03:14,580
Now you see, that will not
be primarily what we're
63
00:03:14,580 --> 00:03:17,300
interested in, in
this particular
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00:03:17,300 --> 00:03:18,470
phase of our course.
65
00:03:18,470 --> 00:03:22,620
What we're interested in this
phase of the course is given
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00:03:22,620 --> 00:03:27,480
various classifications for
the function 'f', to find
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00:03:27,480 --> 00:03:31,690
recipes that will give us 'G' in
familiar form, where 'G' is
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00:03:31,690 --> 00:03:34,540
the function whose derivative
is 'f'.
69
00:03:34,540 --> 00:03:36,760
Summarize for you
to look at here.
70
00:03:36,760 --> 00:03:39,860
The objective of this block of
material is to find recipes
71
00:03:39,860 --> 00:03:44,210
for finding 'G of x' for various
types of 'f of x'.
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00:03:44,210 --> 00:03:47,110
Now the best way to illustrate
what that means is to actually
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00:03:47,110 --> 00:03:49,330
pick a few examples.
74
00:03:49,330 --> 00:03:52,970
And by the way, what you'll
notice in today's lecture is,
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00:03:52,970 --> 00:03:56,620
roughly speaking, we do nothing
at all that's new.
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00:03:56,620 --> 00:03:59,770
The new stuff will begin with
out next lecture essentially.
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00:03:59,770 --> 00:04:02,970
But for now what we'll really
be doing is giving us an
78
00:04:02,970 --> 00:04:06,620
excuse to pull together all
of the recipes that we've
79
00:04:06,620 --> 00:04:07,830
developed so far.
80
00:04:07,830 --> 00:04:10,860
All of the recipes either called
indefinite integrals or
81
00:04:10,860 --> 00:04:12,020
inverse derivatives.
82
00:04:12,020 --> 00:04:17,440
For example, the first class
of functions which we could
83
00:04:17,440 --> 00:04:21,940
handle were things of the form
integral ''u to the n' du'.
84
00:04:21,940 --> 00:04:24,930
Recall that very early in our
course, we found out that the
85
00:04:24,930 --> 00:04:29,000
integral of ''u to the n' du'
was ''u to the 'n + 1'' over
86
00:04:29,000 --> 00:04:33,960
'n plus 1' plus a constant if
'n' was unequal to minus 1.
87
00:04:33,960 --> 00:04:37,860
And then, in our last block of
material we learned to handle
88
00:04:37,860 --> 00:04:40,340
the case where 'n' was
equal to minus 1.
89
00:04:40,340 --> 00:04:43,330
Namely, when 'n' is minus
1, integral ''u to
90
00:04:43,330 --> 00:04:44,690
the minus 1' du'.
91
00:04:44,690 --> 00:04:49,010
Namely, integral 'du' over 'u'
is natural log absolute value
92
00:04:49,010 --> 00:04:51,490
of 'u' plus a constant.
93
00:04:51,490 --> 00:04:54,200
Now this then, is one of our
basic building blocks.
94
00:04:54,200 --> 00:04:57,300
What we're saying is, if an
indefinite integral can be
95
00:04:57,300 --> 00:05:00,220
written in this form, we already
have a recipe that
96
00:05:00,220 --> 00:05:03,020
allows us to evaluate
the integral.
97
00:05:03,020 --> 00:05:06,560
By the way, let us also point
out as was mentioned again,
98
00:05:06,560 --> 00:05:10,180
several times in the past, that
'u' is the generic name
99
00:05:10,180 --> 00:05:11,700
for a variable over here.
100
00:05:11,700 --> 00:05:14,930
It is not important in this
recipe that we use 'u'.
101
00:05:14,930 --> 00:05:18,050
What is important is that
whatever is being raised to
102
00:05:18,050 --> 00:05:21,530
the n-th power inside the
integrand must be precisely
103
00:05:21,530 --> 00:05:23,990
the thing that's following
the 'd' over here.
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00:05:23,990 --> 00:05:27,530
For example, instead of
'u', suppose I used
105
00:05:27,530 --> 00:05:28,960
'sine x' over here.
106
00:05:28,960 --> 00:05:30,900
In other words, suppose I
had 'sine x' to the n-th
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00:05:30,900 --> 00:05:33,810
power 'd sine x'.
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00:05:33,810 --> 00:05:37,210
See, what's being raised to
the n-th power is the same
109
00:05:37,210 --> 00:05:39,770
thing as what's following
the 'd'.
110
00:05:39,770 --> 00:05:40,770
That's all I really
care about.
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00:05:40,770 --> 00:05:44,850
In other words, if this had been
the integrand, the recipe
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00:05:44,850 --> 00:05:48,750
would have said, this is 1 over
'n + 1' times 'sine x' to
113
00:05:48,750 --> 00:05:50,170
the 'n + 1' power.
114
00:05:50,170 --> 00:05:53,170
Plus a constant if 'n' is
not equal to minus 1.
115
00:05:53,170 --> 00:05:56,620
And it's natural log absolute
value of 'sine x' plus a
116
00:05:56,620 --> 00:06:00,040
constant if 'n' is
equal to minus 1.
117
00:06:00,040 --> 00:06:03,180
Again, the important thing is
not the 'u', but the fact that
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00:06:03,180 --> 00:06:05,610
what's being raised to the
n-th power is the same as
119
00:06:05,610 --> 00:06:07,220
what's following the
'd' over here.
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00:06:07,220 --> 00:06:11,690
By the way, we can write this
in more familiar form.
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00:06:11,690 --> 00:06:16,150
You see, namely, notice that the
differential of 'sine x'
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00:06:16,150 --> 00:06:17,820
is 'cosine x dx'.
123
00:06:17,820 --> 00:06:22,370
And consequently, in this form
what our recipe says is that
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00:06:22,370 --> 00:06:27,010
integral 'sine x' to the n-th
power times 'cosine x dx' is
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00:06:27,010 --> 00:06:28,210
given by this.
126
00:06:28,210 --> 00:06:31,950
In other words, it's sine to the
'n + 1' power 'x over 'n +
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00:06:31,950 --> 00:06:35,870
1'' plus a constant if 'n'
is unequal to minus 1.
128
00:06:35,870 --> 00:06:39,510
And it's the natural log
absolute value of 'sine x'
129
00:06:39,510 --> 00:06:43,520
plus 'c' if 'n' is
equal to minus 1.
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00:06:43,520 --> 00:06:53,030
Again, notice this looks more
complicated than this.
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00:06:53,030 --> 00:06:58,390
But to get our problem, to get
into the form ''u to the n'
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00:06:58,390 --> 00:07:01,930
du', notice that the natural
substitution here is to let
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00:07:01,930 --> 00:07:03,780
'u' equal 'sine x'.
134
00:07:03,780 --> 00:07:07,030
In which case, 'du' would
be 'cosine x dx'.
135
00:07:07,030 --> 00:07:10,100
In other words, the extra factor
of 'cosine x' here
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00:07:10,100 --> 00:07:13,250
actually is to our advantage
over here.
137
00:07:13,250 --> 00:07:15,210
That when we make the
substitution, it gives us the
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00:07:15,210 --> 00:07:16,630
form that we want.
139
00:07:16,630 --> 00:07:19,290
With the factor missing, we're
in a little bit of trouble.
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00:07:19,290 --> 00:07:23,150
By the way, this is what makes
this a rather exciting topic.
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00:07:23,150 --> 00:07:27,820
That there seems to be no really
canned ways of finding
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00:07:27,820 --> 00:07:29,160
indefinite integrals.
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00:07:29,160 --> 00:07:30,610
That we have recipes.
144
00:07:30,610 --> 00:07:34,180
Frequently, the recipe that
we have doesn't cover the
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00:07:34,180 --> 00:07:35,910
situation that we have.
146
00:07:35,910 --> 00:07:38,630
In fact, I know a lot of times
in teaching a course like this
147
00:07:38,630 --> 00:07:41,280
students say, why do we have
to learn these recipes and
148
00:07:41,280 --> 00:07:41,820
techniques?
149
00:07:41,820 --> 00:07:44,030
Why can't we just look them
up in the tables?
150
00:07:44,030 --> 00:07:48,070
And the answer quite simply, is
that in many applications,
151
00:07:48,070 --> 00:07:50,790
the integral that we have
doesn't appear in
152
00:07:50,790 --> 00:07:52,110
that form in the table.
153
00:07:52,110 --> 00:07:55,490
That frequently what we must do
is a tremendous amount of
154
00:07:55,490 --> 00:07:58,700
algebraic manipulation and the
like to even get the given
155
00:07:58,700 --> 00:08:01,970
integral to resemble one that
we can find in the table.
156
00:08:01,970 --> 00:08:04,020
Oh, we're starting
off on a fairly
157
00:08:04,020 --> 00:08:05,440
straightforward level here.
158
00:08:05,440 --> 00:08:06,710
Nothing really too tricky.
159
00:08:06,710 --> 00:08:09,870
But at least enough so that we
get an idea of what kind of
160
00:08:09,870 --> 00:08:11,120
gimmicks are involved.
161
00:08:11,120 --> 00:08:15,080
For example, suppose we had
'cosine x' to the seventh
162
00:08:15,080 --> 00:08:17,030
power 'dx'.
163
00:08:17,030 --> 00:08:19,990
You see the idea here is that
when you have cosine raised to
164
00:08:19,990 --> 00:08:23,590
a power, it would have been nice
if a 'sine x' factor had
165
00:08:23,590 --> 00:08:24,380
been in here.
166
00:08:24,380 --> 00:08:26,860
Because the differential
of 'cosine x' is
167
00:08:26,860 --> 00:08:28,730
minus 'sine x dx'.
168
00:08:28,730 --> 00:08:32,120
And what I'm driving at here is
that we often have to have
169
00:08:32,120 --> 00:08:36,440
a certain kind of ingenuity that
allows us to reduce a new
170
00:08:36,440 --> 00:08:39,299
integrand to a more
familiar one.
171
00:08:39,299 --> 00:08:40,480
And, as I say, there are many
172
00:08:40,480 --> 00:08:42,159
techniques done in the textbook.
173
00:08:42,159 --> 00:08:45,060
There'll be plenty of exercises
again, that will
174
00:08:45,060 --> 00:08:46,570
allow you to experiment
on this.
175
00:08:46,570 --> 00:08:50,050
All I want to do is hit a few
highlights and create the mood
176
00:08:50,050 --> 00:08:51,360
as to what's going on.
177
00:08:51,360 --> 00:08:54,440
For example, you see frequently
what happens is, if
178
00:08:54,440 --> 00:08:57,690
we have 'cosine x' raised
to an odd power, a very
179
00:08:57,690 --> 00:09:01,520
convenient way of handling this
is to split off one of
180
00:09:01,520 --> 00:09:03,870
the factors of 'cosine
x' separately.
181
00:09:03,870 --> 00:09:06,780
In other words, write this as
'cosine x' to the sixth power
182
00:09:06,780 --> 00:09:08,700
times 'cosine x dx'.
183
00:09:08,700 --> 00:09:13,180
The idea being that cosine
squared can be written as '1
184
00:09:13,180 --> 00:09:14,800
minus 'sine squared''.
185
00:09:14,800 --> 00:09:18,620
In other words, 'cosine x' to
the sixth power is really
186
00:09:18,620 --> 00:09:21,150
'cosine squared x' cubed.
187
00:09:21,150 --> 00:09:24,840
But 'cosine squared x' is '1
minus 'sine squared x''.
188
00:09:24,840 --> 00:09:28,140
In other words, notice again the
role of the trigonometric
189
00:09:28,140 --> 00:09:29,030
identities.
190
00:09:29,030 --> 00:09:34,160
I take this integrand, write it
this way, and now you see
191
00:09:34,160 --> 00:09:38,040
if I use the binomial theorem
and expand this, notice that
192
00:09:38,040 --> 00:09:42,160
my typical term will have 'sine
x' to a power multiplied
193
00:09:42,160 --> 00:09:44,070
by a 'cosine x'.
194
00:09:44,070 --> 00:09:47,540
And that's precisely the kind
of a term that I handled in
195
00:09:47,540 --> 00:09:48,980
the previous case.
196
00:09:48,980 --> 00:09:50,500
And again, I'm not
going to bore you
197
00:09:50,500 --> 00:09:51,650
with the details here.
198
00:09:51,650 --> 00:09:53,670
That is the easiest part
of the problem.
199
00:09:53,670 --> 00:09:56,850
The hard part is getting to
understand why one would want
200
00:09:56,850 --> 00:09:59,750
to use this particular
kind of form.
201
00:09:59,750 --> 00:10:02,670
Another variation is, what
happens if the cosine is
202
00:10:02,670 --> 00:10:05,790
raised to an even power rather
than to an odd power?
203
00:10:05,790 --> 00:10:07,500
How do we handle that?
204
00:10:07,500 --> 00:10:11,010
Again, the same kind of use of
trigonometric identities.
205
00:10:11,010 --> 00:10:15,370
For example, it would not be to
our advantage in this case,
206
00:10:15,370 --> 00:10:19,040
to replace cosine squared
by 1 minus sine squared.
207
00:10:19,040 --> 00:10:21,960
Because then we would wind
up with the same kind of
208
00:10:21,960 --> 00:10:25,320
integrands that we started with,
only using powers of
209
00:10:25,320 --> 00:10:28,400
sine instead of powers
of cosine.
210
00:10:28,400 --> 00:10:31,830
You see, the idea is that if
we have a power of sine, we
211
00:10:31,830 --> 00:10:34,080
want a cosine factor
to accompany it.
212
00:10:34,080 --> 00:10:37,120
And inversely, if we have a
power of cosine, we'd like a
213
00:10:37,120 --> 00:10:39,210
factor of sine to
accompany it.
214
00:10:39,210 --> 00:10:39,900
Why?
215
00:10:39,900 --> 00:10:42,355
So that we can get the integral
into the form ''u to
216
00:10:42,355 --> 00:10:43,260
the n' du'.
217
00:10:43,260 --> 00:10:46,670
In this particular case, a
very common device is to
218
00:10:46,670 --> 00:10:50,910
invoke the identity that cosine
squared of an angle is
219
00:10:50,910 --> 00:10:54,350
1 plus cosine twice
the angle over 2.
220
00:10:54,350 --> 00:10:57,600
You see again, this is where
all of these particular
221
00:10:57,600 --> 00:10:59,430
identities come into play.
222
00:10:59,430 --> 00:11:02,370
It's not a case of saying, let's
learn these identities
223
00:11:02,370 --> 00:11:03,910
so that we can recite them.
224
00:11:03,910 --> 00:11:07,750
It's a case of getting into a
situation where you want a
225
00:11:07,750 --> 00:11:11,270
certain answer, can't handle
it conveniently in the form
226
00:11:11,270 --> 00:11:14,940
that you're in, and hope that
you can pick off a synonym
227
00:11:14,940 --> 00:11:18,240
that allows you to tackle the
problem more successfully.
228
00:11:18,240 --> 00:11:20,890
That's where this becomes a
matter of insight, lucky
229
00:11:20,890 --> 00:11:24,260
guessing, skill, whatever
you want to call it.
230
00:11:24,260 --> 00:11:27,370
It's one thing to transform an
integral into an equivalent
231
00:11:27,370 --> 00:11:30,920
one, and another thing to have
that equivalent integral be
232
00:11:30,920 --> 00:11:32,370
something that you can handle.
233
00:11:32,370 --> 00:11:35,790
Well, at any rate, for example
in a problem of this type, a
234
00:11:35,790 --> 00:11:40,730
tendency is to write cosine
sixth theta as
235
00:11:40,730 --> 00:11:42,710
cosine squared cubed.
236
00:11:42,710 --> 00:11:46,800
Cosine squared theta is 1 plus
cosine 2 theta over 2.
237
00:11:46,800 --> 00:11:50,160
To cube this we can take
out the factor of 1/8.
238
00:11:50,160 --> 00:11:53,750
You see, we use the binomial
theorem to expand 1 plus
239
00:11:53,750 --> 00:11:55,500
cosine 2 theta cubed.
240
00:11:55,500 --> 00:11:57,750
And notice that of the
terms that we get,
241
00:11:57,750 --> 00:11:59,680
this term we can handle.
242
00:11:59,680 --> 00:12:01,640
This term we can handle,
it's just cosine
243
00:12:01,640 --> 00:12:03,000
to the first power.
244
00:12:03,000 --> 00:12:05,480
Which involves a sine term
when we integrate.
245
00:12:05,480 --> 00:12:07,240
And here's a cosine cubed.
246
00:12:07,240 --> 00:12:10,110
But we've just seen how you can
handle something to an odd
247
00:12:10,110 --> 00:12:13,180
power, so that reduces
this term to a
248
00:12:13,180 --> 00:12:14,690
type that we can handle.
249
00:12:14,690 --> 00:12:15,630
And then we see what?
250
00:12:15,630 --> 00:12:17,570
That we have another term
in here, which is
251
00:12:17,570 --> 00:12:19,350
cosine squared 2 theta.
252
00:12:19,350 --> 00:12:21,280
That's an even power
of cosine.
253
00:12:21,280 --> 00:12:22,290
Again, we do what?
254
00:12:22,290 --> 00:12:27,070
We write cosine squared 2 theta
as 1 plus cosine of
255
00:12:27,070 --> 00:12:28,260
twice this angle.
256
00:12:28,260 --> 00:12:31,930
That's 1 plus cosine
4 theta over 2.
257
00:12:31,930 --> 00:12:34,510
In other words, we keep reducing
these things, always
258
00:12:34,510 --> 00:12:38,850
hoping to break them down into
collections of problems that
259
00:12:38,850 --> 00:12:40,230
we solved before.
260
00:12:40,230 --> 00:12:43,900
As I said, these are things
that we'll drill on in the
261
00:12:43,900 --> 00:12:47,410
exercises and I think it will
become more meaningful there.
262
00:12:47,410 --> 00:12:50,400
Here, as I say, the main aim
is to create the mood that
263
00:12:50,400 --> 00:12:54,470
what we're trying to do is to
reduce unfamiliar problems to
264
00:12:54,470 --> 00:12:56,750
equivalent familiar
ones to which we
265
00:12:56,750 --> 00:12:58,520
already know the answer.
266
00:12:58,520 --> 00:13:02,520
Now again, in the way of review,
sometimes an integrand
267
00:13:02,520 --> 00:13:06,190
occurs in the form of the sum or
difference of two squares.
268
00:13:06,190 --> 00:13:09,610
We've already handled this in
terms of the inverse circular
269
00:13:09,610 --> 00:13:12,830
functions and the inverse
hyperbolic functions.
270
00:13:12,830 --> 00:13:16,340
It's covered for the first time
in our textbook at this
271
00:13:16,340 --> 00:13:18,040
particular stage
of the course.
272
00:13:18,040 --> 00:13:21,100
And I think it's a tough enough
concept so that it's
273
00:13:21,100 --> 00:13:22,950
worth emphasizing
at this stage.
274
00:13:22,950 --> 00:13:25,280
So I just separate this
particular type out.
275
00:13:25,280 --> 00:13:27,790
How do we handle sums and
differences of squares?
276
00:13:27,790 --> 00:13:30,690
And what I intend to show is,
is that by a suitable
277
00:13:30,690 --> 00:13:33,980
trigonometric function, either
circular trigonometric or
278
00:13:33,980 --> 00:13:36,870
hyperbolic trigonometric, we can
always reduce that kind of
279
00:13:36,870 --> 00:13:39,550
a problem to the type
that we can handle.
280
00:13:39,550 --> 00:13:43,180
Again, to keep computation at a
minimum, I have not tried to
281
00:13:43,180 --> 00:13:45,610
pick a rather complicated
situation.
282
00:13:45,610 --> 00:13:49,060
In fact, I've picked ones where
you can very easily look
283
00:13:49,060 --> 00:13:50,780
them up in the table
if that was the
284
00:13:50,780 --> 00:13:52,370
main aim of the exercise.
285
00:13:52,370 --> 00:13:54,510
You see here, it's not so much
that I want to get this
286
00:13:54,510 --> 00:13:58,810
answer, as much as it is that
I want to emphasize the
287
00:13:58,810 --> 00:14:00,670
technique that one uses.
288
00:14:00,670 --> 00:14:01,950
The idea is something
like this.
289
00:14:01,950 --> 00:14:03,630
Here we see the difference
of two squares.
290
00:14:03,630 --> 00:14:06,160
That's the square root of a
squared minus x squared.
291
00:14:06,160 --> 00:14:10,580
What this should do for us is
to suggest a right triangle
292
00:14:10,580 --> 00:14:14,250
whose hypotenuse is a and one
of whose sides is 'x'.
293
00:14:14,250 --> 00:14:16,920
That's the diagram I've
drawn over here.
294
00:14:16,920 --> 00:14:18,570
This is my reference triangle.
295
00:14:18,570 --> 00:14:20,830
As I say, we've done this
before, but I would like to
296
00:14:20,830 --> 00:14:23,030
reinforce this at this
particular stage.
297
00:14:23,030 --> 00:14:26,460
Especially now that we're more
familiar with both the inverse
298
00:14:26,460 --> 00:14:29,860
circular and inverse hyperbolic
functions.
299
00:14:29,860 --> 00:14:32,470
At any rate, we use this
reference triangle.
300
00:14:32,470 --> 00:14:35,600
From this triangle, the easiest
relationship to pick
301
00:14:35,600 --> 00:14:40,890
off involving 'x' is that sine
theta is 'x over a', or 'a
302
00:14:40,890 --> 00:14:42,710
'sine theta'' equals 'x'.
303
00:14:42,710 --> 00:14:45,080
From which it follows that
''a' cosine theta
304
00:14:45,080 --> 00:14:47,630
'd theta'' is 'dx'.
305
00:14:47,630 --> 00:14:49,740
The square root of ''a squared'
minus 'x squared''
306
00:14:49,740 --> 00:14:51,670
over 'a' is cosine theta.
307
00:14:51,670 --> 00:14:53,930
Therefore, the square root
of ''a squared' minus 'x
308
00:14:53,930 --> 00:14:56,020
squared'' is 'a cosine theta'.
309
00:14:56,020 --> 00:14:58,330
If we now make this substitution
in this
310
00:14:58,330 --> 00:15:02,580
particular integral, we wind up
with integral 'dx' over the
311
00:15:02,580 --> 00:15:05,830
square root of ''a squared'
minus 'x squared'' is equal to
312
00:15:05,830 --> 00:15:07,880
integral 'd theta'.
313
00:15:07,880 --> 00:15:09,930
Which is theta plus 'c'.
314
00:15:09,930 --> 00:15:15,190
And since theta is the number
whose sine is 'x over a', we
315
00:15:15,190 --> 00:15:19,250
have that the answer to this
problem is inverse 'sine 'x
316
00:15:19,250 --> 00:15:21,450
over a'' plus a constant.
317
00:15:21,450 --> 00:15:24,370
Again, just a straightforward
review, but a
318
00:15:24,370 --> 00:15:25,820
rather important technique.
319
00:15:25,820 --> 00:15:29,180
In our textbook, this is put in
a separate section, and the
320
00:15:29,180 --> 00:15:32,470
technique is called
trigonometric substitution.
321
00:15:32,470 --> 00:15:34,150
And I just wanted to show
you what motivates the
322
00:15:34,150 --> 00:15:35,620
trigonometric substitution.
323
00:15:35,620 --> 00:15:38,930
The harder part as I mentioned
in our previous lecture, is
324
00:15:38,930 --> 00:15:42,130
how you motivate trigonometric
substitutions when it's the
325
00:15:42,130 --> 00:15:44,200
hyperbolic functions
that are involved.
326
00:15:44,200 --> 00:15:46,770
You see, what I'd like to do now
is instead of dealing with
327
00:15:46,770 --> 00:15:49,110
the square root of ''a squared'
minus 'x squared'',
328
00:15:49,110 --> 00:15:51,300
let me just reverse the
order of the terms.
329
00:15:51,300 --> 00:15:55,380
And now let me take integral
'dx' over the 'square root of
330
00:15:55,380 --> 00:15:57,380
''x squared' minus
'a squared'''.
331
00:15:57,380 --> 00:15:59,690
Notice that the difference
between ''x squared' minus 'a
332
00:15:59,690 --> 00:16:02,910
squared'' and ''a squared' minus
'x squared'' is just a
333
00:16:02,910 --> 00:16:04,690
factor of minus 1.
334
00:16:04,690 --> 00:16:08,060
And the fact that that minus 1
is under the square root sign,
335
00:16:08,060 --> 00:16:10,790
really means in a certain manner
of speaking that you've
336
00:16:10,790 --> 00:16:15,970
multiplied by i, the square root
of minus 1, to transform
337
00:16:15,970 --> 00:16:18,610
the integral that we
just had into this
338
00:16:18,610 --> 00:16:19,960
particular form here.
339
00:16:19,960 --> 00:16:23,520
And in terms of our lecture on
the hyperbolic functions that
340
00:16:23,520 --> 00:16:27,400
should, in some way, give us a
preview of things to come in
341
00:16:27,400 --> 00:16:30,380
the sense that the hyperbolic
functions are very strongly
342
00:16:30,380 --> 00:16:32,550
related to the circular
functions.
343
00:16:32,550 --> 00:16:34,370
My technique for doing this--
344
00:16:34,370 --> 00:16:36,630
and again, this is highly
subjective.
345
00:16:36,630 --> 00:16:37,630
It's very simple.
346
00:16:37,630 --> 00:16:40,310
I learnt the circular functions
long before I learnt
347
00:16:40,310 --> 00:16:41,790
the hyperbolic functions.
348
00:16:41,790 --> 00:16:44,450
Consequently, I feel very
much at home with
349
00:16:44,450 --> 00:16:46,350
the circular functions.
350
00:16:46,350 --> 00:16:48,820
As a result, whenever I see the
sum or difference of two
351
00:16:48,820 --> 00:16:51,550
squares, I always think in
terms of a triangle.
352
00:16:51,550 --> 00:16:54,310
I make a circular trigonometric
substitution.
353
00:16:54,310 --> 00:16:56,260
If that works for me, fine.
354
00:16:56,260 --> 00:16:58,240
I haven't lost anything.
355
00:16:58,240 --> 00:17:02,040
If it doesn't work for me, it
gives me the hint that I can't
356
00:17:02,040 --> 00:17:05,950
seem to get in my own mind
without this hint as to which
357
00:17:05,950 --> 00:17:07,900
hyperbolic function
I should use.
358
00:17:07,900 --> 00:17:10,950
Let me show you this in terms
of this particular example.
359
00:17:10,950 --> 00:17:14,290
If I saw this problem from
scratch and didn't have any
360
00:17:14,290 --> 00:17:16,910
previous knowledge of how to
handle this I'd say, this
361
00:17:16,910 --> 00:17:19,319
seems like I should think of
a right triangle whose
362
00:17:19,319 --> 00:17:22,760
hypotenuse is 'x' and one
of whose sides is 'a'.
363
00:17:22,760 --> 00:17:25,319
So I think of this particular
reference triangle.
364
00:17:25,319 --> 00:17:28,010
Now, thinking of this triangle,
and in fact, even
365
00:17:28,010 --> 00:17:31,650
look at this triangle, I very
easily pick off that 'x' is
366
00:17:31,650 --> 00:17:34,550
equal to 'a cosecant theta'.
367
00:17:34,550 --> 00:17:38,210
See, 'x over a' is 1
over sine theta.
368
00:17:38,210 --> 00:17:40,240
That's cosecant theta.
369
00:17:40,240 --> 00:17:43,610
Therefore and remembering how to
differentiate the cosecant,
370
00:17:43,610 --> 00:17:46,310
I take the differential of both
sides and I find that
371
00:17:46,310 --> 00:17:49,840
minus 'a cosecant theta
cotangent theta
372
00:17:49,840 --> 00:17:52,000
'd theta'' is 'dx'.
373
00:17:52,000 --> 00:17:55,530
Again, from this diagram I see
that the square root of ''x
374
00:17:55,530 --> 00:17:58,920
squared' minus 'a
squared'' is--
375
00:17:58,920 --> 00:18:00,080
well, let's see.
376
00:18:00,080 --> 00:18:02,020
The square root of ''x squared'
minus 'a squared''
377
00:18:02,020 --> 00:18:04,450
over 'a' would be
cotangent theta.
378
00:18:04,450 --> 00:18:06,710
So the square root of ''x
squared' minus 'a squared''
379
00:18:06,710 --> 00:18:09,720
itself is 'a cotangent theta'.
380
00:18:09,720 --> 00:18:13,090
Making the substitution in the
integral 'dx' over the square
381
00:18:13,090 --> 00:18:16,270
root of ''x squared' minus 'a
squared'', notice that the
382
00:18:16,270 --> 00:18:20,310
'a's cancel, the cotangent's
cancel, and I'm left with
383
00:18:20,310 --> 00:18:22,940
'cosecant theta 'd theta''
and a minus
384
00:18:22,940 --> 00:18:24,690
sign inside my integral.
385
00:18:24,690 --> 00:18:27,830
In other words, somehow or
other, this circular
386
00:18:27,830 --> 00:18:31,420
trigonometric function replaces
the integral 'dx'
387
00:18:31,420 --> 00:18:34,230
over the 'square root of ''x
squared' minus 'a squared'''
388
00:18:34,230 --> 00:18:37,510
by integral minus 'cosecant
theta 'd theta''.
389
00:18:37,510 --> 00:18:39,710
And the point I'd like to
emphasize at this particular
390
00:18:39,710 --> 00:18:42,470
stage is that we don't
say that the
391
00:18:42,470 --> 00:18:44,000
substitution has failed.
392
00:18:44,000 --> 00:18:45,580
The substitution
has been made.
393
00:18:45,580 --> 00:18:48,390
We have replaced an integral
involving 'x' by
394
00:18:48,390 --> 00:18:50,030
one involving theta.
395
00:18:50,030 --> 00:18:52,450
It's just that in the same way
that it's difficult to
396
00:18:52,450 --> 00:18:54,125
integrate secant theta--
397
00:18:54,125 --> 00:18:57,380
in fact, that will be later
in this block of
398
00:18:57,380 --> 00:18:58,320
material we'll do that.
399
00:18:58,320 --> 00:19:01,060
But the point is that secant
and cosecant are rather
400
00:19:01,060 --> 00:19:02,780
difficult functions
to integrate.
401
00:19:02,780 --> 00:19:05,850
In fact, at this stage of our
development, we do not know a
402
00:19:05,850 --> 00:19:09,870
function whose derivative
is cosecant theta.
403
00:19:09,870 --> 00:19:12,630
Consequently, we don't know one
whose derivative is minus
404
00:19:12,630 --> 00:19:13,830
cosecant theta.
405
00:19:13,830 --> 00:19:16,250
We seemed to have arrived
at an impasse here.
406
00:19:16,250 --> 00:19:18,890
We have successfully made the
substitution, but the key
407
00:19:18,890 --> 00:19:23,140
point is that the new integral
is no easier for us to handle
408
00:19:23,140 --> 00:19:24,610
than the old integral.
409
00:19:24,610 --> 00:19:28,780
Now, here's how I visualize the
hyperbolic substitutions.
410
00:19:28,780 --> 00:19:30,330
We did this in the
last lecture.
411
00:19:30,330 --> 00:19:32,340
I think it's tough enough, so
I'd like to do it a second
412
00:19:32,340 --> 00:19:35,760
time and give you a chance of
seeing how easy this is once
413
00:19:35,760 --> 00:19:37,800
you see the structural form.
414
00:19:37,800 --> 00:19:41,600
Notice that the substitution I
made here was 'x' equals 'a
415
00:19:41,600 --> 00:19:42,890
cosecant theta'.
416
00:19:42,890 --> 00:19:45,600
The cosecant was the
key step here.
417
00:19:45,600 --> 00:19:48,750
The reference triangle, knowing
how to draw this
418
00:19:48,750 --> 00:19:51,930
geometrically, allowed me not
to consciously have to think
419
00:19:51,930 --> 00:19:53,670
of the trigonometric
identities.
420
00:19:53,670 --> 00:19:56,850
But what I really did in
evaluating this problem, if I
421
00:19:56,850 --> 00:19:59,870
left the diagram out and wanted
to do it analytically,
422
00:19:59,870 --> 00:20:03,370
the identity that I made use
of was the one that said
423
00:20:03,370 --> 00:20:07,390
cosecant squared theta minus
cotangent squared theta is
424
00:20:07,390 --> 00:20:08,620
identically 1.
425
00:20:08,620 --> 00:20:11,590
In other words, I would like an
identity for the hyperbolic
426
00:20:11,590 --> 00:20:14,660
functions that has the structure
that the difference
427
00:20:14,660 --> 00:20:17,230
of two squares is
identically 1.
428
00:20:17,230 --> 00:20:20,480
The easiest one I can think of
is the one that says cosh
429
00:20:20,480 --> 00:20:23,690
squared theta minus sinh
squared theta is 1.
430
00:20:23,690 --> 00:20:26,680
You see, structurally, these
two are equivalent.
431
00:20:26,680 --> 00:20:30,360
In other words, sinh and cosh
are related hyperbolically,
432
00:20:30,360 --> 00:20:34,310
the same way that cotangent and
cosecant are in terms of
433
00:20:34,310 --> 00:20:35,650
circular functions.
434
00:20:35,650 --> 00:20:37,210
The identification is this.
435
00:20:37,210 --> 00:20:41,460
Since in the original problem I
tried 'x' equals 'a cosecant
436
00:20:41,460 --> 00:20:45,300
theta', and since the
identification here is that
437
00:20:45,300 --> 00:20:50,260
cosecant and cosh are matched
up, what I try next-- see once
438
00:20:50,260 --> 00:20:53,610
this has failed, I very quickly
now say, OK, what I'll
439
00:20:53,610 --> 00:20:57,590
do is instead of 'x' equals 'a
cosecant theta', I'll try 'x'
440
00:20:57,590 --> 00:21:01,160
equals 'a cosh theta'.
441
00:21:01,160 --> 00:21:04,060
That's exactly what
I did over here.
442
00:21:04,060 --> 00:21:06,110
'a cosh theta' equals 'x'.
443
00:21:06,110 --> 00:21:09,130
From which it follows
that 'dx' is 'a
444
00:21:09,130 --> 00:21:11,090
sinh theta 'd theta''.
445
00:21:11,090 --> 00:21:14,570
The square root of ''x squared'
minus 'a squared'' is
446
00:21:14,570 --> 00:21:16,720
just ''a squared' cosh
squared theta'.
447
00:21:16,720 --> 00:21:17,850
That's 'x squared'.
448
00:21:17,850 --> 00:21:19,220
Minus 'a squared'.
449
00:21:19,220 --> 00:21:22,930
That can be written as the
square root of 'a squared'
450
00:21:22,930 --> 00:21:27,040
times the quantity 'cosh
squared theta minus 1'.
451
00:21:27,040 --> 00:21:29,910
Now, the interesting point is
the 'cosh squared theta minus
452
00:21:29,910 --> 00:21:33,190
1' turns out to be sinh
squared theta.
453
00:21:33,190 --> 00:21:35,140
In fact, if we want to just look
back here for a moment,
454
00:21:35,140 --> 00:21:36,890
recall that's exactly what
we have over here.
455
00:21:36,890 --> 00:21:39,880
That cosh squared theta is 1.
456
00:21:39,880 --> 00:21:44,680
Well, cosh squared theta is
1 plus sinh squared theta.
457
00:21:44,680 --> 00:21:48,680
cosh squared theta minus 1
is sinh squared theta.
458
00:21:48,680 --> 00:21:50,660
And you might, say wasn't
that terrific?
459
00:21:50,660 --> 00:21:55,160
What a lucky break it was that
this complicated expression
460
00:21:55,160 --> 00:21:57,490
just happened to be sinh
squared theta.
461
00:21:57,490 --> 00:22:01,320
And now taking the square
root I get a sinh theta.
462
00:22:01,320 --> 00:22:03,850
The thing I'd like to point
out is that the element of
463
00:22:03,850 --> 00:22:08,750
luck was removed from this at
the instant that I recognized
464
00:22:08,750 --> 00:22:12,580
the fact that cosh and sinh
were related by the same
465
00:22:12,580 --> 00:22:15,510
identity that cosecant
and cotangent were.
466
00:22:15,510 --> 00:22:19,110
In other words, I rigged this
thing so that when I finally
467
00:22:19,110 --> 00:22:22,220
had to take this particular
difference, I had to wind up
468
00:22:22,220 --> 00:22:24,750
with an easy square
root to extract.
469
00:22:24,750 --> 00:22:26,870
You see, I had the right
structure to begin with.
470
00:22:26,870 --> 00:22:30,140
At any rate, to make a long
story short here, 'dx' is 'a
471
00:22:30,140 --> 00:22:32,190
sinh theta 'd theta''.
472
00:22:32,190 --> 00:22:35,200
The square root of ''x squared'
minus 'a squared'' is
473
00:22:35,200 --> 00:22:36,780
'a sinh theta'.
474
00:22:36,780 --> 00:22:39,640
Therefore, 'dx' over the square
root of ''x squared'
475
00:22:39,640 --> 00:22:40,580
minus 'a squared''.
476
00:22:40,580 --> 00:22:42,450
Well, look how nicely
this works out.
477
00:22:42,450 --> 00:22:45,950
The a sinh theta cancels from
both the numerator and the
478
00:22:45,950 --> 00:22:46,960
denominator.
479
00:22:46,960 --> 00:22:49,690
And it turns out that the
integral that I'm looking for,
480
00:22:49,690 --> 00:22:52,770
'dx' over the square root
of ''x squared' minus 'a
481
00:22:52,770 --> 00:22:55,730
squared'' is just theta
plus a constant.
482
00:22:55,730 --> 00:22:59,570
But now, recalling that a
cosh theta equals 'x'.
483
00:22:59,570 --> 00:23:01,970
In other words, cosh theta
is 'x over a'.
484
00:23:01,970 --> 00:23:06,100
And therefore, theta is 'inverse
cosh 'x over a'', I
485
00:23:06,100 --> 00:23:10,970
arrive at the result that this
integral is 'inverse cosh 'x
486
00:23:10,970 --> 00:23:12,850
over a'' plus a constant.
487
00:23:12,850 --> 00:23:14,920
In other words, what I hope that
these two little examples
488
00:23:14,920 --> 00:23:19,520
show is that when I have the
sum or difference of two
489
00:23:19,520 --> 00:23:24,200
squares, the trigonometric
substitutions will usually
490
00:23:24,200 --> 00:23:27,700
give me a crack at getting an
equivalent integral that will
491
00:23:27,700 --> 00:23:29,410
be easier to handle.
492
00:23:29,410 --> 00:23:32,120
What may happen is that the
circular functions won't help
493
00:23:32,120 --> 00:23:34,120
me, but the hyperbolic
ones will.
494
00:23:34,120 --> 00:23:37,330
Or vice versa, and I hope that
you see how these are so
495
00:23:37,330 --> 00:23:39,180
delicately connected.
496
00:23:39,180 --> 00:23:43,260
Well, there is one final type of
technique that I would like
497
00:23:43,260 --> 00:23:46,190
to show for today's lesson
that's related to what we've
498
00:23:46,190 --> 00:23:47,180
already done.
499
00:23:47,180 --> 00:23:50,080
And with that, we'll conclude
today's lesson and go into
500
00:23:50,080 --> 00:23:52,910
some more elaborate
recipes next time.
501
00:23:52,910 --> 00:23:58,020
But this is a take-off on our
old high school construction
502
00:23:58,020 --> 00:23:59,620
of completing a square.
503
00:23:59,620 --> 00:24:03,240
In other words, let's suppose
we have an expression of the
504
00:24:03,240 --> 00:24:06,550
form 'ax squared' plus
'bx' plus 'c'.
505
00:24:06,550 --> 00:24:08,540
And this may sound
old hat to you.
506
00:24:08,540 --> 00:24:11,430
You may remember this as looking
pretty much like the
507
00:24:11,430 --> 00:24:14,650
development of the quadratic
equation.
508
00:24:14,650 --> 00:24:16,200
The idea works something
like this.
509
00:24:16,200 --> 00:24:17,970
First of all, I can factor
an a out of here.
510
00:24:17,970 --> 00:24:19,600
I'm assuming that a is not 0.
511
00:24:19,600 --> 00:24:21,830
In fact, if a were 0, I
really wouldn't have a
512
00:24:21,830 --> 00:24:23,150
square term in here.
513
00:24:23,150 --> 00:24:26,470
I factor out an 'a' and I'm
left with 'a' times 'x
514
00:24:26,470 --> 00:24:30,730
squared' plus ''b over
a'x' plus 'c over a'.
515
00:24:30,730 --> 00:24:34,050
Now it turns out that whenever
you have something of the form
516
00:24:34,050 --> 00:24:37,610
'x squared' plus something times
'x', to convert that
517
00:24:37,610 --> 00:24:38,900
into a perfect square.
518
00:24:38,900 --> 00:24:41,970
And again, we'll have ample
opportunity to practice this
519
00:24:41,970 --> 00:24:43,350
in the exercises.
520
00:24:43,350 --> 00:24:46,960
You take half the coefficient
of 'x' and square it.
521
00:24:46,960 --> 00:24:50,010
Half the coefficient of 'x'
here is 'b over 2a'.
522
00:24:50,010 --> 00:24:53,910
If I square that it's 'b
squared' over ''4a' squared'.
523
00:24:53,910 --> 00:24:57,060
Now to keep the identity intact
here, if I add in 'b
524
00:24:57,060 --> 00:25:00,460
squared' over ''4a' squared',
I must subtract that.
525
00:25:00,460 --> 00:25:03,440
In other words, I just add and
subtract this term that will
526
00:25:03,440 --> 00:25:05,190
make this a perfect square.
527
00:25:05,190 --> 00:25:09,150
In fact, what is 'x squared'
plus ''b over a'x' plus ''b
528
00:25:09,150 --> 00:25:11,720
squared' over '4a' squared''?
529
00:25:11,720 --> 00:25:13,320
You'll notice that
this is just ''x'
530
00:25:13,320 --> 00:25:17,460
plus 'b over 2a' squared'.
531
00:25:17,460 --> 00:25:18,800
'x squared'.
532
00:25:18,800 --> 00:25:20,170
Just multiply this thing out.
533
00:25:20,170 --> 00:25:21,540
You see how this particular
thing works.
534
00:25:21,540 --> 00:25:24,950
In other words, what I've done
is I can write this as 'a'
535
00:25:24,950 --> 00:25:27,020
times a perfect square.
536
00:25:27,020 --> 00:25:29,670
And by the way, what's left over
here if I multiply these
537
00:25:29,670 --> 00:25:31,910
two terms through by 'a'?
538
00:25:31,910 --> 00:25:35,980
This is 'c' minus 'b
squared over 4a'.
539
00:25:35,980 --> 00:25:39,770
In other words, recognizing that
this is ''x' plus 'b over
540
00:25:39,770 --> 00:25:41,310
2a' squared'.
541
00:25:41,310 --> 00:25:44,330
And multiplying these two terms
through by 'a', I wind
542
00:25:44,330 --> 00:25:48,460
up with the fact that ''ax'
squared' plus 'bx' plus 'c'
543
00:25:48,460 --> 00:25:51,330
can always be written in
this particular form.
544
00:25:51,330 --> 00:25:53,340
This is called completing
the square.
545
00:25:53,340 --> 00:25:55,450
And notice that this is now a
perfect square and this is
546
00:25:55,450 --> 00:25:56,530
some constant.
547
00:25:56,530 --> 00:25:59,480
And the question that comes up
is, what does this have to do
548
00:25:59,480 --> 00:26:01,260
with integral calculus?
549
00:26:01,260 --> 00:26:03,630
In other words, with finding
antiderivatives.
550
00:26:03,630 --> 00:26:06,720
And again, as is so often
the case, and granted
551
00:26:06,720 --> 00:26:08,440
that as this course--
552
00:26:08,440 --> 00:26:12,250
especially this chapter gets
more complicated, the degree
553
00:26:12,250 --> 00:26:14,750
of sophistication will
become much greater.
554
00:26:14,750 --> 00:26:17,640
But the basic idea will always
remain the same.
555
00:26:17,640 --> 00:26:21,180
Somehow or other, we will always
try to reduce a new
556
00:26:21,180 --> 00:26:23,810
situation to a more
familiar one.
557
00:26:23,810 --> 00:26:27,100
In other words, let's suppose
now we don't have an integrand
558
00:26:27,100 --> 00:26:29,560
that's the sum and difference of
two squares, but we have an
559
00:26:29,560 --> 00:26:32,680
integrand that involves a
quadratic expression.
560
00:26:32,680 --> 00:26:36,630
Say, for example, we have
'dx' over ''ax' squared'
561
00:26:36,630 --> 00:26:38,210
plus 'bx' plus 'c'.
562
00:26:38,210 --> 00:26:41,780
Well, you see, using hindsight
rather than foresight, I
563
00:26:41,780 --> 00:26:45,020
already took care of this
problem before we started.
564
00:26:45,020 --> 00:26:47,740
Namely, we just finished
completing
565
00:26:47,740 --> 00:26:49,070
the square over here.
566
00:26:49,070 --> 00:26:53,230
Namely, what is ''ax' squared'
plus 'bx' plus 'c'?
567
00:26:53,230 --> 00:26:54,780
Notice it can be written
in what form?
568
00:26:54,780 --> 00:26:56,720
Let's come back here
and take a look.
569
00:26:56,720 --> 00:26:59,820
It's 'a' times ''x'
plus 'b over 2a'
570
00:26:59,820 --> 00:27:01,910
squared' plus some constant.
571
00:27:01,910 --> 00:27:03,600
I don't care what it
happens to be here.
572
00:27:03,600 --> 00:27:06,240
I'll just call that constant
'c sub 1'.
573
00:27:06,240 --> 00:27:09,500
And by the way, notice depending
upon the relative
574
00:27:09,500 --> 00:27:12,740
sizes of 'a', 'b', and 'c', this
constant can be either
575
00:27:12,740 --> 00:27:14,330
positive or negative.
576
00:27:14,330 --> 00:27:16,460
Or in fact, even 0.
577
00:27:16,460 --> 00:27:19,860
So to handle that what I'll say
is I can always write this
578
00:27:19,860 --> 00:27:24,680
denominator in the form 'a'
times ''x' plus 'b over 2a'
579
00:27:24,680 --> 00:27:27,590
squared' plus or minus
some constant.
580
00:27:27,590 --> 00:27:30,010
Where I'm assuming now that
the constant is positive.
581
00:27:30,010 --> 00:27:31,900
That's why I put the plus
or minus sign in.
582
00:27:31,900 --> 00:27:34,660
It's plus some positive constant
or else it's minus
583
00:27:34,660 --> 00:27:36,350
some positive constant.
584
00:27:36,350 --> 00:27:39,840
Now what I can do is factor out
an a from the denominator.
585
00:27:39,840 --> 00:27:42,920
If I factor out an a from the
denominator here, I have 1
586
00:27:42,920 --> 00:27:44,980
over 'a' times a certain
integral.
587
00:27:44,980 --> 00:27:46,160
What integral?
588
00:27:46,160 --> 00:27:52,150
'dx' over 'x plus 'b over 2a'
squared' plus or minus still
589
00:27:52,150 --> 00:27:53,890
some constant.
590
00:27:53,890 --> 00:27:56,500
In other words, I
had 'c1' here.
591
00:27:56,500 --> 00:28:00,740
I factored out on 'a'. 'c2'
was actually 'c1 over a'.
592
00:28:00,740 --> 00:28:02,740
I just factored out an 'a'
from this thing here.
593
00:28:02,740 --> 00:28:03,900
But it doesn't make
any difference.
594
00:28:03,900 --> 00:28:06,750
The important point is I now
have this written in this
595
00:28:06,750 --> 00:28:07,960
particular form.
596
00:28:07,960 --> 00:28:11,980
Now if I assume that 'c2' is a
positive constant, notice that
597
00:28:11,980 --> 00:28:14,610
any positive number is
a perfect square.
598
00:28:14,610 --> 00:28:18,190
Namely, it's the square
of its square root.
599
00:28:18,190 --> 00:28:20,960
And again, as I say with the
exercises, this will become
600
00:28:20,960 --> 00:28:23,250
clearer when we work with
specific numbers.
601
00:28:23,250 --> 00:28:25,550
But the idea here is what?
602
00:28:25,550 --> 00:28:29,200
This is some perfect square,
which I'll call 'k'.
603
00:28:29,200 --> 00:28:31,610
And in other words, this
integrand can now be written
604
00:28:31,610 --> 00:28:36,540
in the form '1 over a', 'dx'
over 'x plus 'b over 2a'
605
00:28:36,540 --> 00:28:39,820
squared' plus or minus
'k squared'.
606
00:28:39,820 --> 00:28:41,910
Well, again, what am
I leading up to?
607
00:28:41,910 --> 00:28:46,040
If I now make the substitution
that 'u' equals 'x plus 'b
608
00:28:46,040 --> 00:28:50,760
over 2a'', notice that my
denominator simply becomes 'u
609
00:28:50,760 --> 00:28:52,880
squared' plus or minus
'k squared'.
610
00:28:52,880 --> 00:28:56,560
In other words, my denominator
now has the form of either the
611
00:28:56,560 --> 00:28:58,860
sum or difference
of two squares.
612
00:28:58,860 --> 00:29:02,780
On the other hand, if 'u'
is 'x plus 'b over
613
00:29:02,780 --> 00:29:05,730
2a'', what is 'du'?
614
00:29:05,730 --> 00:29:07,620
Remember, 'b' and 'a'
are constants.
615
00:29:07,620 --> 00:29:09,420
'b over 2a' is a constant.
616
00:29:09,420 --> 00:29:11,680
The derivative of
a constant is 0.
617
00:29:11,680 --> 00:29:14,780
So 'du' is equal to 'dx'.
618
00:29:14,780 --> 00:29:18,970
In other words, notice that if
I now make this substitution
619
00:29:18,970 --> 00:29:23,850
in here, outside I have a '1
over a', inside I have 'du'
620
00:29:23,850 --> 00:29:27,560
over ''u squared' plus or
minus 'k squared''.
621
00:29:27,560 --> 00:29:31,590
In other words, when I have a
quadratic I can write that in
622
00:29:31,590 --> 00:29:35,650
the form the sum and/or
difference of two squares,
623
00:29:35,650 --> 00:29:39,560
simply by completing
the square.
624
00:29:39,560 --> 00:29:41,310
Simply by completing square.
625
00:29:41,310 --> 00:29:46,460
In other words, I can solve the
problem of the quadratic
626
00:29:46,460 --> 00:29:50,660
by converting it back into an
integral, the type of which
627
00:29:50,660 --> 00:29:52,100
I've solved before.
628
00:29:52,100 --> 00:29:54,920
And this, by and large, becomes
the technique that we
629
00:29:54,920 --> 00:29:57,080
will use throughout
this chapter.
630
00:29:57,080 --> 00:29:59,820
Well, at any rate, I won't
go into that in any more
631
00:29:59,820 --> 00:30:00,930
detail right now.
632
00:30:00,930 --> 00:30:02,070
We will pick up additional
633
00:30:02,070 --> 00:30:03,455
techniques in our next lecture.
634
00:30:03,455 --> 00:30:05,460
And until the next
lecture, goodbye.
635
00:30:05,460 --> 00:30:08,200
636
00:30:08,200 --> 00:30:10,740
ANNOUNCER: Funding for the
publication of this video was
637
00:30:10,740 --> 00:30:15,460
provided by the Gabriella and
Paul Rosenbaum Foundation.
638
00:30:15,460 --> 00:30:19,630
Help OCW continue to provide
free and open access to MIT
639
00:30:19,630 --> 00:30:23,830
courses by making a donation
at ocw.mit.edu/donate.
640
00:30:23,830 --> 00:30:28,564