1
00:00:00,000 --> 00:00:01,940
FEMALE VOICE: The following
content is provided under a
2
00:00:01,940 --> 00:00:03,690
Creative Commons license.
3
00:00:03,690 --> 00:00:06,630
Your support will help MIT
OpenCourseWare continue to
4
00:00:06,630 --> 00:00:09,990
offer high quality educational
resources for free.
5
00:00:09,990 --> 00:00:12,830
To make a donation, or to view
additional materials from
6
00:00:12,830 --> 00:00:16,760
hundreds of MIT courses, visit
MIT OpenCourseWare at
7
00:00:16,760 --> 00:00:18,010
ocw.mit.edu.
8
00:00:33,010 --> 00:00:33,980
PROFESSOR: Hi.
9
00:00:33,980 --> 00:00:37,220
Today, we're going to learn a
rather powerful technique,
10
00:00:37,220 --> 00:00:40,020
called the technique of partial
fractions, that's
11
00:00:40,020 --> 00:00:44,680
particularly applicable for one
special type of integrand.
12
00:00:44,680 --> 00:00:48,950
In particular, it's going to be
applicable to the situation
13
00:00:48,950 --> 00:00:52,650
where the integral has the form,
the quotient, of two
14
00:00:52,650 --> 00:00:53,400
polynomials.
15
00:00:53,400 --> 00:00:57,140
In other words, suppose we have
integral ''P of x' dx'
16
00:00:57,140 --> 00:01:00,490
over 'Q of x', where 'P'
and 'Q' are both
17
00:01:00,490 --> 00:01:02,740
polynomials in 'x'.
18
00:01:02,740 --> 00:01:05,080
And, by the way, for reasons
that will become apparent
19
00:01:05,080 --> 00:01:10,200
later, we'll assume that the
degree of 'P' is less than the
20
00:01:10,200 --> 00:01:11,300
degree of 'Q'.
21
00:01:11,300 --> 00:01:13,940
In other words, that the
highest exponent in the
22
00:01:13,940 --> 00:01:17,110
numerator is less than the
highest exponent in the
23
00:01:17,110 --> 00:01:17,920
denominator.
24
00:01:17,920 --> 00:01:21,470
By the way, if that's not the
case, we can always carry out
25
00:01:21,470 --> 00:01:26,850
long division first and carry
out enough terms until we get
26
00:01:26,850 --> 00:01:29,800
a term in which this is
the particular case.
27
00:01:29,800 --> 00:01:32,420
You see, I'll make that clearer
as we go along.
28
00:01:32,420 --> 00:01:35,230
But, for the time being, all we
really care about in terms
29
00:01:35,230 --> 00:01:36,660
of outlining this form--
30
00:01:36,660 --> 00:01:38,900
and, again, as we will often
do in these types of
31
00:01:38,900 --> 00:01:39,620
techniques--
32
00:01:39,620 --> 00:01:42,290
the place that we'll pick up
the real fine computational
33
00:01:42,290 --> 00:01:44,710
points are in the exercises.
34
00:01:44,710 --> 00:01:48,730
We'll use the lectures to just
outline what the technique is
35
00:01:48,730 --> 00:01:49,760
and how its used.
36
00:01:49,760 --> 00:01:52,160
But suppose, now, that we
have the quotient of two
37
00:01:52,160 --> 00:01:54,560
polynomials and we want
to integrate it.
38
00:01:54,560 --> 00:02:00,920
Now, the idea is this, there are
two key steps that dictate
39
00:02:00,920 --> 00:02:02,510
this particular method.
40
00:02:02,510 --> 00:02:06,470
The first thing is, we saw it
in the last lecture, that we
41
00:02:06,470 --> 00:02:10,419
can handle denominators if they
involve nothing worse
42
00:02:10,419 --> 00:02:13,700
than linear and quadratic
polynomials.
43
00:02:13,700 --> 00:02:16,250
In other words, we know how to
integrate something like
44
00:02:16,250 --> 00:02:18,740
'dx/x' plus a constant.
45
00:02:18,740 --> 00:02:23,480
We know now how to integrate 'dx
/ ''ax squared' plus 'bx'
46
00:02:23,480 --> 00:02:27,020
plus 'c'', so we know how
to handle that type, OK?
47
00:02:27,020 --> 00:02:28,680
And the second important
thing--
48
00:02:28,680 --> 00:02:30,970
and I'll amplify this,
too, as we go along--
49
00:02:30,970 --> 00:02:34,560
the second important thing is
that, theoretically, every
50
00:02:34,560 --> 00:02:37,470
real polynomial can be factored
into linear and
51
00:02:37,470 --> 00:02:38,960
quadratic terms.
52
00:02:38,960 --> 00:02:41,460
Now, this is a little bit
misleading, if you try to read
53
00:02:41,460 --> 00:02:43,560
more into this than what
it really says.
54
00:02:43,560 --> 00:02:48,640
It doesn't mean that we can
always find a factorization
55
00:02:48,640 --> 00:02:49,260
quite simply.
56
00:02:49,260 --> 00:02:51,240
In other words, we may,
at best, be able to
57
00:02:51,240 --> 00:02:52,720
approximate a root.
58
00:02:52,720 --> 00:02:54,290
We'll have to use things like
we've talked about in the
59
00:02:54,290 --> 00:02:57,630
notes so far: Newton's method,
and things like this, linear
60
00:02:57,630 --> 00:03:00,290
approximations, tangential
approximations.
61
00:03:00,290 --> 00:03:02,260
We may have to approximate
the roots.
62
00:03:02,260 --> 00:03:05,810
But that, theoretically, if
a polynomial with real
63
00:03:05,810 --> 00:03:09,280
coefficients has degree greater
than 2, it has, at
64
00:03:09,280 --> 00:03:11,040
least, one real root.
65
00:03:11,040 --> 00:03:13,940
In other words, we can keep
factoring things out this way.
66
00:03:13,940 --> 00:03:16,950
The only thing we can't do is to
guarantee that, once we get
67
00:03:16,950 --> 00:03:20,530
down to a quadratic, that we can
get real roots out of this
68
00:03:20,530 --> 00:03:21,220
particular thing.
69
00:03:21,220 --> 00:03:24,190
In fact, the classic example
is to visualize 'x
70
00:03:24,190 --> 00:03:25,670
squared plus 1'.
71
00:03:25,670 --> 00:03:29,680
In other words, we can't factor
'x squared plus 1'
72
00:03:29,680 --> 00:03:32,580
unless we introduce
non-real numbers.
73
00:03:32,580 --> 00:03:36,330
Remember the technique you can
write, that 'x squared plus 1'
74
00:03:36,330 --> 00:03:40,330
is 'x squared' minus 'i squared'
where 'i' is the
75
00:03:40,330 --> 00:03:41,890
square root of minus 1.
76
00:03:41,890 --> 00:03:46,265
And this factors into 'x plus
i' times 'x minus i', et
77
00:03:46,265 --> 00:03:48,880
cetera, but these are
non-real numbers.
78
00:03:48,880 --> 00:03:52,130
In other words, you can't always
factor a quadratic to
79
00:03:52,130 --> 00:03:52,830
get real numbers.
80
00:03:52,830 --> 00:03:57,750
In fact, if you recall the
quadratic formula of this:
81
00:03:57,750 --> 00:04:01,280
square root of 'b squared' minus
'4 ac', if 'b squared'
82
00:04:01,280 --> 00:04:04,740
is less than '4 ac', what's
inside the square root sign is
83
00:04:04,740 --> 00:04:08,320
negative, and that leads
to non-real roots.
84
00:04:08,320 --> 00:04:09,810
So, you see, we can't
always factor
85
00:04:09,810 --> 00:04:12,160
quadratics into real factors.
86
00:04:12,160 --> 00:04:15,670
By the way, don't identify
things that were difficult to
87
00:04:15,670 --> 00:04:18,290
factor with things that
can't be factored.
88
00:04:18,290 --> 00:04:20,519
You know, a lot of times, we
think of, say-- here's an
89
00:04:20,519 --> 00:04:23,380
example I thought you might
enjoy seeing here--
90
00:04:23,380 --> 00:04:26,070
take, for example, ''x'
to the fourth plus 1'.
91
00:04:26,070 --> 00:04:28,620
That looks something like it
belongs to the family 'x
92
00:04:28,620 --> 00:04:30,270
squared plus 1'.
93
00:04:30,270 --> 00:04:32,570
And 'x squared plus 1'
can't be factored.
94
00:04:32,570 --> 00:04:34,020
You might think that
''x' to the fourth
95
00:04:34,020 --> 00:04:35,660
plus 1' can't be factored.
96
00:04:35,660 --> 00:04:38,940
Now, again, it's not important
that you understand what made
97
00:04:38,940 --> 00:04:40,890
me think of these tricks
or what have you.
98
00:04:40,890 --> 00:04:43,920
What I do want to show is that
even situations where the
99
00:04:43,920 --> 00:04:46,660
polynomials might look like a
known factor, they really do.
100
00:04:46,660 --> 00:04:49,870
For example, with ''x' to the
fourth plus 1', we can write
101
00:04:49,870 --> 00:04:53,980
this in the disguised form ''x
squared plus 1' squared', see?
102
00:04:53,980 --> 00:04:56,150
That's, what? ''x' to
the fourth plus 1'.
103
00:04:56,150 --> 00:04:58,820
But there's a middle term here
of 2 'x squared', so I
104
00:04:58,820 --> 00:05:00,970
subtract off the
2 'x squared'.
105
00:05:00,970 --> 00:05:03,500
Now, this has the form, the
sum and difference of 2
106
00:05:03,500 --> 00:05:07,290
squares, namely, ''x squared
plus 1' squared' and, also,
107
00:05:07,290 --> 00:05:10,400
the square root of 2
times 'x squared'.
108
00:05:10,400 --> 00:05:14,530
In other words, I can write this
as this plus the square
109
00:05:14,530 --> 00:05:18,240
root of 2 times 'x' times
this minus the square
110
00:05:18,240 --> 00:05:19,990
root of '2 times x'.
111
00:05:19,990 --> 00:05:23,710
Observing, that even though
the square root of 2 is an
112
00:05:23,710 --> 00:05:27,480
irrational number, it is,
nonetheless, a real number.
113
00:05:27,480 --> 00:05:29,420
The important point that I want
to point out, though, as
114
00:05:29,420 --> 00:05:31,720
far as setting up this technique
called partial
115
00:05:31,720 --> 00:05:35,020
fractions, is that, whether
it's easy or not easy, the
116
00:05:35,020 --> 00:05:38,070
fact remains that, when we
have a polynomial in our
117
00:05:38,070 --> 00:05:42,760
denominator, it can always be
factored into a combination of
118
00:05:42,760 --> 00:05:47,350
linear and quadratic factors
using real numbers.
119
00:05:47,350 --> 00:05:50,610
Well, it's difficult to factor
some of these things, so, by
120
00:05:50,610 --> 00:05:52,610
way of illustration,
let me pick out
121
00:05:52,610 --> 00:05:54,470
one that comes factored.
122
00:05:54,470 --> 00:05:57,000
Let me start with a particular
problem here.
123
00:05:59,500 --> 00:06:04,670
Let's take the integral 'dx/
''x minus 1' times 'x
124
00:06:04,670 --> 00:06:06,010
squared plus 1''.
125
00:06:06,010 --> 00:06:08,120
So, what is the integrand
in this case?
126
00:06:08,120 --> 00:06:12,620
It's 1/ ''x minus 1' times
'x squared plus 1''.
127
00:06:12,620 --> 00:06:13,790
Now, the idea is this.
128
00:06:13,790 --> 00:06:15,490
See, here's a quadratic factor,
129
00:06:15,490 --> 00:06:17,900
here's a linear factor.
130
00:06:17,900 --> 00:06:21,560
What fractions do I have
to add to wind up with
131
00:06:21,560 --> 00:06:22,380
something like this?
132
00:06:22,380 --> 00:06:25,220
Well, again-- and this is going
to be another example of
133
00:06:25,220 --> 00:06:28,190
our old adage that it's easier
to scramble an egg than to
134
00:06:28,190 --> 00:06:29,540
unscramble one.
135
00:06:29,540 --> 00:06:32,430
You see, given two fractions,
it's one thing
136
00:06:32,430 --> 00:06:34,050
to find their sum.
137
00:06:34,050 --> 00:06:36,130
Given the sum, it's quite
another thing,
138
00:06:36,130 --> 00:06:39,550
computationally, to find what
fractions you had to add to
139
00:06:39,550 --> 00:06:40,520
get that sum.
140
00:06:40,520 --> 00:06:41,560
The idea is this.
141
00:06:41,560 --> 00:06:44,570
If you wind up with a
denominator that has an 'x
142
00:06:44,570 --> 00:06:48,210
minus 1' term and an 'x squared
plus 1' term, it
143
00:06:48,210 --> 00:06:52,860
appears that you must have
started with terms, say, what?
144
00:06:52,860 --> 00:06:55,380
In other words, you must have
had one term which had a
145
00:06:55,380 --> 00:06:58,640
denominator of 'x minus 1' and a
term which had a denominator
146
00:06:58,640 --> 00:06:59,980
of 'x squared plus 1'.
147
00:06:59,980 --> 00:07:01,800
Because, you see, if I start
with these kind of
148
00:07:01,800 --> 00:07:04,860
denominators when I
cross-multiply and put things
149
00:07:04,860 --> 00:07:08,580
over a common denominator, I
will wind up with 'x minus 1'
150
00:07:08,580 --> 00:07:10,500
times 'x squared plus 1'.
151
00:07:10,500 --> 00:07:13,930
The question that comes up is
what shall our numerators be?
152
00:07:13,930 --> 00:07:16,970
And here's the main reason why
we picked the degree of the
153
00:07:16,970 --> 00:07:19,800
numerator to be less than the
degree of the denominator.
154
00:07:19,800 --> 00:07:23,660
Notice, for example, in this
particular case, the numerator
155
00:07:23,660 --> 00:07:26,650
has degree 0, namely, the
highest power of 'x' to the
156
00:07:26,650 --> 00:07:29,260
peers is the 'x'
to the 0 term.
157
00:07:29,260 --> 00:07:32,050
On the other hand, the
denominator has degree 3.
158
00:07:32,050 --> 00:07:35,130
See, there's an 'x cubed'
term in the denominator.
159
00:07:35,130 --> 00:07:36,820
You see, if they're doing
'x' to the fourth in the
160
00:07:36,820 --> 00:07:38,740
numerator, I could have
multiplied out the
161
00:07:38,740 --> 00:07:42,990
denominator, divided it into the
numerator, and just kept
162
00:07:42,990 --> 00:07:45,930
carrying out the division long
enough until I wound up with a
163
00:07:45,930 --> 00:07:50,820
remainder which was less than
a degree, less than a cubic.
164
00:07:50,820 --> 00:07:51,780
In other words, less
than a third
165
00:07:51,780 --> 00:07:53,570
degree polynomial remainder.
166
00:07:53,570 --> 00:07:54,540
That's not the point.
167
00:07:54,540 --> 00:07:57,210
The point is that as long as the
degree of the numerator is
168
00:07:57,210 --> 00:07:59,850
less than the degree of the
denominator, it means that the
169
00:07:59,850 --> 00:08:02,290
terms that we're adding must
have the degree of the
170
00:08:02,290 --> 00:08:04,940
numerator less than the degree
of the denominator.
171
00:08:04,940 --> 00:08:06,710
See, it's like adding
fractions.
172
00:08:06,710 --> 00:08:09,850
If you start with one fraction
whose numerator is greater
173
00:08:09,850 --> 00:08:12,900
than the denominator then,
certainly, any sum that you
174
00:08:12,900 --> 00:08:15,410
get is going to be
bigger than 1.
175
00:08:15,410 --> 00:08:18,960
In other words, if you want to
wind-up, dealing with positive
176
00:08:18,960 --> 00:08:21,640
numbers, and you want to wind-up
with a fraction which
177
00:08:21,640 --> 00:08:24,470
is less than 1, it stands to
reason that all of the
178
00:08:24,470 --> 00:08:27,740
fractions that you're adding
must be less than 1.
179
00:08:27,740 --> 00:08:30,740
So, if I'm going to wind-up
adding quotients of
180
00:08:30,740 --> 00:08:34,570
polynomials to get a sum in
which the degree of the
181
00:08:34,570 --> 00:08:37,850
numerator is less than the
degree of the denominator, it
182
00:08:37,850 --> 00:08:40,679
means that every one of the
terms in my sum must have this
183
00:08:40,679 --> 00:08:42,100
particular property.
184
00:08:42,100 --> 00:08:45,180
In other words, with this as
a hint, I say look at, my
185
00:08:45,180 --> 00:08:48,790
denominator here is 'x minus
1', that's degree one.
186
00:08:48,790 --> 00:08:52,720
That means my numerator can't
be greater than degree 0.
187
00:08:52,720 --> 00:08:55,150
But degree 0 means a constant.
188
00:08:55,150 --> 00:08:59,400
So I say OK, that means that
this has to form some constant
189
00:08:59,400 --> 00:09:01,950
over 'x minus 1'.
190
00:09:01,950 --> 00:09:03,910
Now, I look at this
denominator.
191
00:09:03,910 --> 00:09:05,160
It's quadratic.
192
00:09:05,160 --> 00:09:09,500
And I say to myself I'm starting
out with a quadratic,
193
00:09:09,500 --> 00:09:12,020
the degree of the numerator
can't be greater than the
194
00:09:12,020 --> 00:09:13,470
degree of the denominator.
195
00:09:13,470 --> 00:09:15,980
Since the degree of the
denominator is 2, that means
196
00:09:15,980 --> 00:09:20,040
the degree of the numerator
can't be more than 1.
197
00:09:20,040 --> 00:09:23,300
And the most general first
degree polynomial has the
198
00:09:23,300 --> 00:09:24,040
form, what?
199
00:09:24,040 --> 00:09:27,240
Some constant times 'x'
plus a constant.
200
00:09:27,240 --> 00:09:30,290
So what I'm saying is,
OK, to wind-up with
201
00:09:30,290 --> 00:09:31,690
something of the form--
202
00:09:31,690 --> 00:09:35,670
well, to wind-up with 1/ ''x
minus 1' times 'x squared plus
203
00:09:35,670 --> 00:09:39,030
1'', I had better start with
something of the form 'A/ 'x
204
00:09:39,030 --> 00:09:44,330
minus 1'' plus ''Bx plus C' over
'x squared plus 1'' where
205
00:09:44,330 --> 00:09:46,650
'A', 'B', and 'C'
are constants.
206
00:09:46,650 --> 00:09:49,870
The key point is that if we
weren't sure that the degree
207
00:09:49,870 --> 00:09:51,880
of the numerator were less
than the degree of the
208
00:09:51,880 --> 00:09:54,380
denominator, we would
not know where
209
00:09:54,380 --> 00:09:55,930
to stop in our numerator.
210
00:09:55,930 --> 00:09:59,130
In other words, by this
convention, we're sure that
211
00:09:59,130 --> 00:10:02,090
the degree of the numerator in
any one of these terms can't
212
00:10:02,090 --> 00:10:04,580
be greater than the degree
of the corresponding
213
00:10:04,580 --> 00:10:06,700
denominator, you see?
214
00:10:06,700 --> 00:10:09,280
And by the way, notice, for
example, if it turns out that
215
00:10:09,280 --> 00:10:10,630
we put too much in--
216
00:10:10,630 --> 00:10:13,230
for example, suppose it turns
out that this numerator here
217
00:10:13,230 --> 00:10:14,620
should only be a constant--
218
00:10:14,620 --> 00:10:18,170
there's no law against having
'B' turn out to be 0.
219
00:10:18,170 --> 00:10:20,660
By the way, again, what these
things are called--
220
00:10:20,660 --> 00:10:23,540
just to increase our
vocabulary--
221
00:10:23,540 --> 00:10:25,740
what I'm going to do now
is called the method of
222
00:10:25,740 --> 00:10:27,370
undetermined coefficients.
223
00:10:27,370 --> 00:10:31,060
You see, what I know is that I
must have the form 'A over 'x
224
00:10:31,060 --> 00:10:35,050
minus 1' plus ''Bx plus C'
over 'x squared plus 1''.
225
00:10:35,050 --> 00:10:37,560
What I don't know is,
specifically, how to choose
226
00:10:37,560 --> 00:10:39,650
the values of 'A',
'B', and 'C'.
227
00:10:39,650 --> 00:10:41,620
And the technique works
something like this.
228
00:10:41,620 --> 00:10:45,730
What we do is we put this over
a common denominator.
229
00:10:45,730 --> 00:10:47,540
What will the common
denominator be?
230
00:10:47,540 --> 00:10:50,960
It'll be 'x minus 1' times
'x squared plus 1'.
231
00:10:50,960 --> 00:10:53,200
How will I put this over
a common denominator?
232
00:10:53,200 --> 00:10:58,780
It'll be 'A' times 'x squared
plus 1', plus 'Bx plus C'
233
00:10:58,780 --> 00:11:00,520
times 'x minus 1'.
234
00:11:00,520 --> 00:11:03,130
Now, this is supposed
to be an identity.
235
00:11:03,130 --> 00:11:07,700
Now, if two fractions are
identical and the denominators
236
00:11:07,700 --> 00:11:10,350
are the same, which is what they
will be after I put this
237
00:11:10,350 --> 00:11:14,180
over a common denominator, the
only way they can be identical
238
00:11:14,180 --> 00:11:16,140
is for the numerators
to be identical.
239
00:11:16,140 --> 00:11:19,330
So, you see, what I'm going to
do is to cross-multiply here
240
00:11:19,330 --> 00:11:21,740
to obtain the numerator of
the right-hand side.
241
00:11:21,740 --> 00:11:24,690
And I will equate that to the
numerator on the left-hand
242
00:11:24,690 --> 00:11:26,380
side, which is 1.
243
00:11:26,380 --> 00:11:26,820
All right.
244
00:11:26,820 --> 00:11:29,230
Now, what is the numerator
on the right-hand side?
245
00:11:29,230 --> 00:11:32,520
It's 'A' times 'x squared
plus 1', that's 'A x
246
00:11:32,520 --> 00:11:34,240
squared plus A'.
247
00:11:34,240 --> 00:11:38,770
Then it's going to be 'x minus
1' times 'Bx plus C'.
248
00:11:38,770 --> 00:11:44,610
That's going to be 'B x squared'
minus 'Bx plus C'
249
00:11:44,610 --> 00:11:48,110
times 'x minus C'.
250
00:11:48,110 --> 00:11:51,170
And now the idea something like
this, and I'll come back
251
00:11:51,170 --> 00:11:53,730
to this in a few moments to
hammer this home from a
252
00:11:53,730 --> 00:11:54,780
different point of view.
253
00:11:54,780 --> 00:11:57,720
What is the coefficient of 'x
squared' on the right-hand
254
00:11:57,720 --> 00:11:59,390
side of the equation?
255
00:11:59,390 --> 00:12:01,940
The coefficient of 'x squared'
on the right-hand side of the
256
00:12:01,940 --> 00:12:04,600
equation is 'A plus B'.
257
00:12:04,600 --> 00:12:08,060
What is the coefficient of 'x
squared' on the left-hand side
258
00:12:08,060 --> 00:12:09,160
of the equation?
259
00:12:09,160 --> 00:12:12,080
And at first glance you say
there is no 'x squared' on the
260
00:12:12,080 --> 00:12:13,460
left-hand side of
the equation.
261
00:12:13,460 --> 00:12:16,070
What that means, of course, is
that the coefficient of 'x
262
00:12:16,070 --> 00:12:19,220
squared' on the left-hand side
of the equation is 0.
263
00:12:19,220 --> 00:12:22,080
So what we say is OK, the
coefficients of 'x squared'
264
00:12:22,080 --> 00:12:25,110
must match up, therefore,
'A plus B', which is a
265
00:12:25,110 --> 00:12:27,930
coefficient of 'x squared' on
the right-hand side, must
266
00:12:27,930 --> 00:12:30,780
equal 0, which is the
coefficient of 'x squared' on
267
00:12:30,780 --> 00:12:32,170
the left-hand side.
268
00:12:32,170 --> 00:12:36,590
In a similar way, the
coefficient of 'x' is 'C minus
269
00:12:36,590 --> 00:12:38,870
B' on the right-hand side.
270
00:12:38,870 --> 00:12:42,410
The coefficient of 'x' on
the left-hand side is 0.
271
00:12:42,410 --> 00:12:46,080
Consequently, 'C minus
B' must be 0.
272
00:12:46,080 --> 00:12:49,620
And, finally, the constant term
on the right-hand side is
273
00:12:49,620 --> 00:12:52,140
given by 'A minus C'.
274
00:12:52,140 --> 00:12:55,650
On the left-hand side, the
right-hand term is 1.
275
00:12:55,650 --> 00:12:59,050
Consequently, 'A minus
C' must equal 1.
276
00:12:59,050 --> 00:13:01,660
What do I wind-up with?
277
00:13:01,660 --> 00:13:04,470
Three equations with
three unknowns.
278
00:13:04,470 --> 00:13:06,730
Well, there are a number of ways
of handling these things.
279
00:13:06,730 --> 00:13:09,670
The easiest one I see, off-hand,
is I notice if I add
280
00:13:09,670 --> 00:13:13,750
the first two equations, I
wind-up with 'A plus C' is 0.
281
00:13:13,750 --> 00:13:17,720
Knowing that 'A plus C' is 0
and, also, that 'A minus C' is
282
00:13:17,720 --> 00:13:22,310
1, it's easy for me to conclude
that 'A' must be 1/2
283
00:13:22,310 --> 00:13:24,260
and 'C' must be minus 1/2.
284
00:13:24,260 --> 00:13:27,180
And, by the way, now knowing
what 'A' and 'C' are, I can
285
00:13:27,180 --> 00:13:30,510
use either of these two
equations to determine 'B',
286
00:13:30,510 --> 00:13:33,430
and 'B' turns out
to be minus 1/2.
287
00:13:33,430 --> 00:13:35,190
In other words, what
does this tell me?
288
00:13:35,190 --> 00:13:39,490
It tells me that if I replace
'A' here by 1/2, and 'B' and
289
00:13:39,490 --> 00:13:44,220
'C' each by minus 1/2, the
right-hand side here will be
290
00:13:44,220 --> 00:13:47,310
an identity for the left-hand
side, that there'll be two
291
00:13:47,310 --> 00:13:49,850
different ways of naming
the same number for
292
00:13:49,850 --> 00:13:51,470
each value of 'x'.
293
00:13:51,470 --> 00:13:54,980
In fact, doing this now out in
more detail, what we've really
294
00:13:54,980 --> 00:13:59,110
shown here, in other words, if
I replace 'A' by 1/2 and 'B'
295
00:13:59,110 --> 00:14:03,320
and 'C' each by minus 1/2, then
I factor the 1/2 out.
296
00:14:03,320 --> 00:14:08,830
What I've shown is that one over
the quantity 'x minus 1'
297
00:14:08,830 --> 00:14:13,040
times 'x squared plus 1' is
equal to 1/2 times the
298
00:14:13,040 --> 00:14:18,006
quantity '1 over 'x minus 1''
minus the quantity ''x plus 1'
299
00:14:18,006 --> 00:14:20,870
over 'x squared plus 1''.
300
00:14:20,870 --> 00:14:23,670
By the way, I can separate this
into two terms, each of
301
00:14:23,670 --> 00:14:25,840
which has a denominator
of 'x squared plus
302
00:14:25,840 --> 00:14:27,420
1', and I get, what?
303
00:14:27,420 --> 00:14:34,220
1/2 times 1/ 'x minus 1' minus
1/ ''2 times x'/ 'x squared
304
00:14:34,220 --> 00:14:41,230
plus 1'' minus 1/2 times '1/
'x squared plus 1''.
305
00:14:41,230 --> 00:14:43,110
Now, the key is this.
306
00:14:43,110 --> 00:14:45,460
I didn't really want this,
what I wanted was, what?
307
00:14:45,460 --> 00:14:47,340
This was to be my integrand.
308
00:14:47,340 --> 00:14:51,080
What I wanted was to integrate
this with respect to 'x'.
309
00:14:51,080 --> 00:14:55,880
Well, if given this identity
over here, the integral of
310
00:14:55,880 --> 00:14:59,850
'dx'/ ''x minus 1' times 'x
squared plus 1'', recalling
311
00:14:59,850 --> 00:15:02,410
that the integral of a sum is
a sum of the integrals, et
312
00:15:02,410 --> 00:15:04,940
cetera, can now be
written as, what?
313
00:15:04,940 --> 00:15:11,600
It's 1/2 integral 'dx'/ 'x minus
1' minus 1/2 integral
314
00:15:11,600 --> 00:15:18,680
'x'/ 'x squared plus 1' times
'dx' minus 1/2 integral 1/ 'x
315
00:15:18,680 --> 00:15:21,050
squared plus 1' times 'dx'.
316
00:15:21,050 --> 00:15:22,200
Now, here's the point.
317
00:15:22,200 --> 00:15:25,010
Notice that every one of my
denominators, now, is either
318
00:15:25,010 --> 00:15:27,030
linear or quadratic.
319
00:15:27,030 --> 00:15:30,470
In fact, without going through
the details here again, if I
320
00:15:30,470 --> 00:15:34,830
let 'u' equal 'x minus 1' in
this example, this reduces
321
00:15:34,830 --> 00:15:38,510
this to the form 'du/u',
in other words, a ''u
322
00:15:38,510 --> 00:15:40,710
to the n' du' form.
323
00:15:40,710 --> 00:15:46,330
If I let 'u' equal 'x squared
plus 1' over here, if 'u' is
324
00:15:46,330 --> 00:15:50,990
'x squared plus 1', notice
that 'du' is '2 xdx'.
325
00:15:50,990 --> 00:15:55,480
So my numerator becomes a
constant multiple of 'du' and,
326
00:15:55,480 --> 00:15:58,280
again, I have a ''u of
the n' du' form.
327
00:15:58,280 --> 00:16:01,910
And finally, if I look at my
last integral here, notice
328
00:16:01,910 --> 00:16:05,230
that this is the sum of two
squares, which suggests the
329
00:16:05,230 --> 00:16:07,280
circular trigonometric functions
that we were talking
330
00:16:07,280 --> 00:16:08,840
about last time.
331
00:16:08,840 --> 00:16:11,320
Or, if you wish, you can
go to tables and look
332
00:16:11,320 --> 00:16:11,940
these things up.
333
00:16:11,940 --> 00:16:12,780
They're all in there.
334
00:16:12,780 --> 00:16:15,870
But, again, without going
through the details because
335
00:16:15,870 --> 00:16:18,650
this is the easy part, again,
it turns out that once you
336
00:16:18,650 --> 00:16:22,480
have this relationship here we
can integrate this to obtain
337
00:16:22,480 --> 00:16:25,400
log absolute value
'x minus 1'.
338
00:16:25,400 --> 00:16:30,080
The integral here is log,
natural log, absolute value of
339
00:16:30,080 --> 00:16:31,490
'x squared plus 1'.
340
00:16:31,490 --> 00:16:34,170
I can leave the absolute value
signs out because 'x squared
341
00:16:34,170 --> 00:16:36,380
plus 1' has to be, at
least, as big as 1.
342
00:16:36,380 --> 00:16:37,430
It can't be negative.
343
00:16:37,430 --> 00:16:40,330
And finally, either by
trigonometric substitution, or
344
00:16:40,330 --> 00:16:43,610
by memorization, or what have
you, integral of 'dx'/ 'x
345
00:16:43,610 --> 00:16:48,200
squared plus 1' is just the
inverse tangent of 'x'.
346
00:16:48,200 --> 00:16:51,810
In other words, by using partial
fractions and reducing
347
00:16:51,810 --> 00:16:56,510
a complicated polynomial
denominator into a sum of
348
00:16:56,510 --> 00:17:01,840
linear and quadratic terms,
I was able, by knowing my
349
00:17:01,840 --> 00:17:04,670
techniques of last time,
to integrate the given
350
00:17:04,670 --> 00:17:05,839
expression.
351
00:17:05,839 --> 00:17:08,460
And, by the way, I would be a
little remiss at the stage of
352
00:17:08,460 --> 00:17:12,780
the game if I did not take the
time to, once again, reinforce
353
00:17:12,780 --> 00:17:16,960
a very important concept, and
that is it may be difficult,
354
00:17:16,960 --> 00:17:19,819
starting with this,
to get this.
355
00:17:19,819 --> 00:17:23,980
What should not be difficult is,
starting with this, to be
356
00:17:23,980 --> 00:17:27,087
able to differentiate it and
show that you wind-up with 1/
357
00:17:27,087 --> 00:17:31,200
''x minus 1' times 'x
squared plus 1''.
358
00:17:31,200 --> 00:17:34,200
In other words, as usual, with
the inverse derivative, once
359
00:17:34,200 --> 00:17:36,460
you find an answer and you
want to see whether your
360
00:17:36,460 --> 00:17:39,060
answer is correct or not,
all you have to do is
361
00:17:39,060 --> 00:17:43,070
differentiate the answer and see
if you get the integrand.
362
00:17:43,070 --> 00:17:46,130
But, at any rate, this is how
the technique called partial
363
00:17:46,130 --> 00:17:47,330
fractions works.
364
00:17:47,330 --> 00:17:50,220
It works for the quotient
of two polynomials.
365
00:17:50,220 --> 00:17:52,820
And to make the undetermined
coefficients technique work
366
00:17:52,820 --> 00:17:56,170
right, you must assume that the
degree of the numerator is
367
00:17:56,170 --> 00:17:57,900
less than the degree
of the denominator.
368
00:17:57,900 --> 00:18:00,420
And, obviously, in the
exercises, I'll give you some
369
00:18:00,420 --> 00:18:03,010
where the degree of the
numerator is greater than the
370
00:18:03,010 --> 00:18:04,570
degree of the denominator.
371
00:18:04,570 --> 00:18:07,110
And, if you don't perform long
division first, you're going
372
00:18:07,110 --> 00:18:08,690
to get into trouble
trying to find the
373
00:18:08,690 --> 00:18:09,690
answer to the problem.
374
00:18:09,690 --> 00:18:12,800
But all I want to emphasize
here is the technique.
375
00:18:12,800 --> 00:18:15,450
And, by the way, what I want to
do before I go any further,
376
00:18:15,450 --> 00:18:20,440
also, is to emphasize a rather
special property of polynomial
377
00:18:20,440 --> 00:18:21,650
identities.
378
00:18:21,650 --> 00:18:24,070
You recall that undetermined
coefficients
379
00:18:24,070 --> 00:18:25,260
hinged on the following.
380
00:18:25,260 --> 00:18:27,600
And I'll pick a quadratic
to illustrate it with.
381
00:18:27,600 --> 00:18:31,720
Suppose you have two quadratic
expressions in 'x' identically
382
00:18:31,720 --> 00:18:34,920
equal, in other words, 'a sub-2
'x squared'' plus 'a
383
00:18:34,920 --> 00:18:39,380
sub-1 x' plus 'a sub-0' was
identically equal to 'b sub-2
384
00:18:39,380 --> 00:18:43,190
'x squared'' plus 'b
sub-1 x' plus 'b0'.
385
00:18:43,190 --> 00:18:45,900
Notice the technique that we
used was is we said look at,
386
00:18:45,900 --> 00:18:48,470
let's compare the coefficients
of 'x squared'.
387
00:18:48,470 --> 00:18:51,480
Let's equate the coefficients
of 'x', and let's equate the
388
00:18:51,480 --> 00:18:53,750
constant terms.
389
00:18:53,750 --> 00:18:57,350
How do we know that you're
allowed to do this?
390
00:18:57,350 --> 00:19:01,780
Well, let's see if we can show
that this must be the case.
391
00:19:01,780 --> 00:19:04,460
For example, let's do this
without any calculus at all.
392
00:19:04,460 --> 00:19:06,430
Suppose this is an identity.
393
00:19:06,430 --> 00:19:08,830
If this is an identity,
it must be true for
394
00:19:08,830 --> 00:19:10,240
all values of 'x'.
395
00:19:10,240 --> 00:19:13,980
In particular, it must be
true when 'x' is 0.
396
00:19:13,980 --> 00:19:16,750
Notice that, when 'x' is 0,
the left-hand side is 'a
397
00:19:16,750 --> 00:19:21,170
sub-0', the right-hand side is
'B sub-0', and we wind-up with
398
00:19:21,170 --> 00:19:23,860
'a sub-0' equals 'b sub-0'.
399
00:19:23,860 --> 00:19:25,870
See, the constant
terms are equal.
400
00:19:25,870 --> 00:19:29,800
Well, if 'a sub-0' equals 'b
sub-0', we can cancel 'a
401
00:19:29,800 --> 00:19:32,330
sub-0' and 'b sub-0'
from this equation.
402
00:19:32,330 --> 00:19:35,460
That leaves us with this
equaling this.
403
00:19:35,460 --> 00:19:38,930
From this, we can factor out an
'x', and get to 'x' times
404
00:19:38,930 --> 00:19:43,800
'a sub-2 x' plus 'a1' is
identical with 'x' times 'b
405
00:19:43,800 --> 00:19:46,630
sub-2 x' plus 'b1'.
406
00:19:46,630 --> 00:19:50,070
If 'x' is not 0, we can cancel
'x' from both sides of the
407
00:19:50,070 --> 00:19:51,900
equation, and that
shows, what?
408
00:19:51,900 --> 00:19:56,390
That, for any non-zero value of
'x', 'a 2x' plus 'a1' must
409
00:19:56,390 --> 00:20:00,840
be identically equal to
'b 2x' plus 'b1'.
410
00:20:00,840 --> 00:20:04,460
Once we have this formula
established, let's let 'x'
411
00:20:04,460 --> 00:20:06,850
equal 0 in here, and
we see, what?
412
00:20:06,850 --> 00:20:10,670
With 'x' equal to 0, then 'a1'
equals 'b1', in other words,
413
00:20:10,670 --> 00:20:13,240
that the coefficient of 'x' on
the left-hand side equals the
414
00:20:13,240 --> 00:20:15,500
coefficient of 'x' on
the right-hand side.
415
00:20:15,500 --> 00:20:18,410
You can keep on this way, but
a very nice technique to use
416
00:20:18,410 --> 00:20:20,760
here is a reinforcement
of something that we
417
00:20:20,760 --> 00:20:22,120
talked about before.
418
00:20:22,120 --> 00:20:23,710
I think it was when we
were doing implicit
419
00:20:23,710 --> 00:20:25,660
differentiation, and
we talked about
420
00:20:25,660 --> 00:20:27,690
identities verses equations.
421
00:20:27,690 --> 00:20:31,790
You see, if this is an identity,
and what I mean by
422
00:20:31,790 --> 00:20:33,780
an identity, that this--
423
00:20:33,780 --> 00:20:35,800
we're not saying find
what values of 'x'
424
00:20:35,800 --> 00:20:36,860
this is true for.
425
00:20:36,860 --> 00:20:39,840
By an identity, we're saying
look at, these two expressions
426
00:20:39,840 --> 00:20:41,880
are the same for all
values of 'x'.
427
00:20:41,880 --> 00:20:43,150
They're synonyms.
428
00:20:43,150 --> 00:20:45,340
And what we're saying is, if
these two things are synonyms,
429
00:20:45,340 --> 00:20:47,620
their derivatives must
be synonyms.
430
00:20:47,620 --> 00:20:49,400
And all we're saying is look
at, if you want to use
431
00:20:49,400 --> 00:20:53,320
calculus here, differentiate
both sides of this expression.
432
00:20:53,320 --> 00:20:57,740
And you wind-up with '2a sub-2
x' plus 'a1' is identically
433
00:20:57,740 --> 00:21:01,330
equal to '2b sub-2
x' plus 'b1'.
434
00:21:01,330 --> 00:21:04,680
Since these two things are
identical, let's equate their
435
00:21:04,680 --> 00:21:05,950
derivatives again.
436
00:21:05,950 --> 00:21:09,240
See, the derivative of
identities are identical, and
437
00:21:09,240 --> 00:21:13,580
we wind-up with '2 a2' is
identically equal to '2 b2'
438
00:21:13,580 --> 00:21:16,590
and, therefore, 'a2'
must equal 'b2'.
439
00:21:16,590 --> 00:21:20,420
Knowing that 'a2' equals 'b2',
we can come back to this step
440
00:21:20,420 --> 00:21:22,700
and show that 'a1'
equals 'b1'.
441
00:21:22,700 --> 00:21:26,270
And now, knowing that 'a2'
equals 'b2', and 'a1' equals
442
00:21:26,270 --> 00:21:29,415
'b1', we can come back to the
original equation and show
443
00:21:29,415 --> 00:21:32,010
that 'a0' must equal 'b0'.
444
00:21:32,010 --> 00:21:37,410
Now, you may wonder why are we
making all of this ado over
445
00:21:37,410 --> 00:21:39,860
what appears to be a
very obvious thing?
446
00:21:39,860 --> 00:21:42,310
And I would like to give
you a caution here.
447
00:21:42,310 --> 00:21:44,200
I'd like you to beware
of something.
448
00:21:44,200 --> 00:21:47,180
It's something which works very
nicely for polynomials,
449
00:21:47,180 --> 00:21:49,530
but doesn't always have
to work all the time.
450
00:21:49,530 --> 00:21:52,840
In fact, later, when one
studies differential
451
00:21:52,840 --> 00:21:56,330
equations, this becomes a very
important concept, which later
452
00:21:56,330 --> 00:21:58,910
gets the name linear dependence
and linear
453
00:21:58,910 --> 00:22:00,020
independence.
454
00:22:00,020 --> 00:22:01,820
We're not going to go
into that now, but
455
00:22:01,820 --> 00:22:03,180
the key idea is this.
456
00:22:03,180 --> 00:22:08,740
In general, knowing that 'a1 u1'
plus 'a2 u2' is equal to
457
00:22:08,740 --> 00:22:15,750
'b1 u1' plus 'b2 u2', you cannot
say ah, therefore, the
458
00:22:15,750 --> 00:22:18,950
coefficients of 'u1' must be
equal, and the coefficients of
459
00:22:18,950 --> 00:22:20,890
'u2' must be equal.
460
00:22:20,890 --> 00:22:25,140
Don't get me wrong, if 'a1'
equals 'b1', and 'a2' equals
461
00:22:25,140 --> 00:22:30,700
'b2', then, certainly, 'a1 u1'
plus 'a2 u2' is equal to 'b1
462
00:22:30,700 --> 00:22:32,670
u1' plus 'b2 u2'.
463
00:22:32,670 --> 00:22:33,750
I'm not saying that.
464
00:22:33,750 --> 00:22:36,190
What I'm saying is, conversely,
if you start,
465
00:22:36,190 --> 00:22:40,010
knowing that this is true, it
does not follow that they must
466
00:22:40,010 --> 00:22:43,470
match up coefficient
by coefficient, OK?
467
00:22:43,470 --> 00:22:45,630
And you say well, why doesn't
it have to follow?
468
00:22:45,630 --> 00:22:47,180
And I think the best
way to do that is
469
00:22:47,180 --> 00:22:48,870
by means of an example.
470
00:22:48,870 --> 00:22:51,910
For example, in this general
expression, let 'u1' equal
471
00:22:51,910 --> 00:22:55,160
'x', and let 'u2' to be 'x/2'.
472
00:22:55,160 --> 00:22:58,360
Look at the expression
'5x' plus '6 times
473
00:22:58,360 --> 00:23:00,160
x/2', that's, what?
474
00:23:00,160 --> 00:23:03,790
It's '5x' plus '3x' is '8x'.
475
00:23:03,790 --> 00:23:07,280
Look at '3x' plus
'10 times x/2'.
476
00:23:07,280 --> 00:23:11,090
That's '3x' plus '5x',
which is also '8x'.
477
00:23:11,090 --> 00:23:15,890
In other words, 5 times 'x'
plus 6 times 'x/2' is
478
00:23:15,890 --> 00:23:20,860
identically equal to 3 times
'x' plus 10 times 'x/2'.
479
00:23:20,860 --> 00:23:24,270
Yet, you can't say, therefore,
the coefficients of 'x' must
480
00:23:24,270 --> 00:23:28,980
be equal, and the coefficients
of 'x/2' must be equal.
481
00:23:28,980 --> 00:23:31,570
In fact, if you said that,
you'd be saying that 5 is
482
00:23:31,570 --> 00:23:35,670
equal to 3 and 6 is equal
to 10, which, of course,
483
00:23:35,670 --> 00:23:38,620
is not true, OK?
484
00:23:38,620 --> 00:23:42,030
So, at any rate, I just wanted
to show you here the kind of
485
00:23:42,030 --> 00:23:45,800
mathematical rigor and cautions
that have to be taken
486
00:23:45,800 --> 00:23:48,090
if one is going to use a
technique called partial
487
00:23:48,090 --> 00:23:51,420
fractions, what the key
ingredients are.
488
00:23:51,420 --> 00:23:54,590
Now, it turns out that not only
are partial fractions
489
00:23:54,590 --> 00:23:58,190
important in their own right, it
also turns out that partial
490
00:23:58,190 --> 00:24:03,750
fractions handles a rather
difficult type of integral:
491
00:24:03,750 --> 00:24:10,260
one that uses polynomials in
'sine x' and 'cosine x'.
492
00:24:10,260 --> 00:24:12,890
And the reason I wanted to
mention this was not so much
493
00:24:12,890 --> 00:24:14,770
because the technique is nice.
494
00:24:14,770 --> 00:24:17,130
The technique, by the
way, is in the text.
495
00:24:17,130 --> 00:24:19,620
But there's something very
interesting in the text, the
496
00:24:19,620 --> 00:24:21,520
way the author introduces
a topic.
497
00:24:21,520 --> 00:24:23,250
And I thought that
that was worth an
498
00:24:23,250 --> 00:24:24,680
aside in its own right.
499
00:24:29,610 --> 00:24:33,550
In the section where this thing
appears, it says, "It
500
00:24:33,550 --> 00:24:38,210
has been discovered that." The
author makes no attempt to
501
00:24:38,210 --> 00:24:41,330
show logically why one would
expect that a certain thing is
502
00:24:41,330 --> 00:24:42,590
going to work.
503
00:24:42,590 --> 00:24:46,240
All the author says is, "It has
been discovered that." And
504
00:24:46,240 --> 00:24:50,850
this tells a long story that, in
many cases, we wind up with
505
00:24:50,850 --> 00:24:53,090
an integrand that we don't
know how to handle.
506
00:24:53,090 --> 00:24:56,440
We make all sorts of
substitutions in the hope that
507
00:24:56,440 --> 00:24:59,720
we can reduce the given
integrand to a form that we
508
00:24:59,720 --> 00:25:01,200
know how to handle.
509
00:25:01,200 --> 00:25:04,300
Sometimes we're successful,
sometimes we're not.
510
00:25:04,300 --> 00:25:08,310
In the cases where we're not
successful, somebody, either
511
00:25:08,310 --> 00:25:11,490
by clever intuition or what
have you, maybe it's just
512
00:25:11,490 --> 00:25:14,200
blind luck, stumbles
across a technique
513
00:25:14,200 --> 00:25:16,040
that happens to work.
514
00:25:16,040 --> 00:25:18,980
And I, sort of, liked this
particular example in the text
515
00:25:18,980 --> 00:25:21,360
where the author says, "It has
been discovered that,"
516
00:25:21,360 --> 00:25:24,320
because, to me, it's not at
all self-evident, and yet,
517
00:25:24,320 --> 00:25:26,550
it's a rather pretty result.
518
00:25:26,550 --> 00:25:30,520
The result says this: suppose
you make the substitution 'z'
519
00:25:30,520 --> 00:25:33,690
equals tangent 'x/2'.
520
00:25:33,690 --> 00:25:35,920
Now, where do you pull this
out of the hat from?
521
00:25:35,920 --> 00:25:39,650
'z' equals tangent 'x/2', that's
the ingenuity, the
522
00:25:39,650 --> 00:25:41,330
experience, the luck.
523
00:25:41,330 --> 00:25:44,850
But the idea is, let's suppose
that we stumbled across this
524
00:25:44,850 --> 00:25:46,120
one way or the other.
525
00:25:46,120 --> 00:25:52,420
If we translate this equation
into a reference triangle, we
526
00:25:52,420 --> 00:25:53,560
have, what?
527
00:25:53,560 --> 00:25:56,070
We'll call the angle 'x/2'.
528
00:25:56,070 --> 00:25:58,870
And the tangent is 'z', so we'll
make the side opposite
529
00:25:58,870 --> 00:26:01,070
'z', the side-adjacent 1.
530
00:26:01,070 --> 00:26:03,280
That makes the hypotenuse
the square root
531
00:26:03,280 --> 00:26:05,410
of '1 plus z squared'.
532
00:26:05,410 --> 00:26:07,470
Now, watch what happens
when you do this.
533
00:26:07,470 --> 00:26:12,500
See, 'z' is equal to '10x/2',
therefore, 'dz' is the
534
00:26:12,500 --> 00:26:15,180
differential of '10x/2'.
535
00:26:15,180 --> 00:26:19,950
Remember, the differential of
'10x', with respect to 'x', is
536
00:26:19,950 --> 00:26:21,500
secant squared.
537
00:26:21,500 --> 00:26:24,710
But we also, by the chain rule,
have to multiply by a
538
00:26:24,710 --> 00:26:26,980
derivative of 'x/2' with
respect to 'x'.
539
00:26:26,980 --> 00:26:31,960
In other words, if 'z' is
'10x/2', 'dz' is not secant
540
00:26:31,960 --> 00:26:37,750
squared 'x/2', it's 1/2 'secant
squared 'x/2' dx'.
541
00:26:37,750 --> 00:26:40,460
But what's 'secant
squared 'x/2''?
542
00:26:40,460 --> 00:26:43,540
Let's go back and look
at our diagram.
543
00:26:43,540 --> 00:26:47,790
The secant of 'x/2' is the
hypotenuse over side-adjacent.
544
00:26:47,790 --> 00:26:52,010
That's the square root of '1
plus 'z squared'' over 1 and,
545
00:26:52,010 --> 00:26:56,990
therefore, the square of the
secant is just '1 plus 'z
546
00:26:56,990 --> 00:27:00,280
squared'' over 1, see,
'1 plus 'z squared''.
547
00:27:00,280 --> 00:27:03,510
So, with the 1/2 in here, this
becomes '1 plus 'z squared''
548
00:27:03,510 --> 00:27:05,710
over 2 times 'dx'.
549
00:27:05,710 --> 00:27:08,680
And, consequently, if I compare
these two now, notice
550
00:27:08,680 --> 00:27:13,390
that 'dx' is just twice 'dz'/
'1 plus 'z squared''.
551
00:27:13,390 --> 00:27:15,300
In other words, what's
happened to 'dx'?
552
00:27:15,300 --> 00:27:18,350
It's been replaced by a
differential in 'z', which
553
00:27:18,350 --> 00:27:21,100
involves the quotient
of two polynomials.
554
00:27:21,100 --> 00:27:24,140
So, 'dx' comes out very nicely
this way, in terms of, what?
555
00:27:24,140 --> 00:27:26,680
The quotient of two
polynomials.
556
00:27:26,680 --> 00:27:28,840
How about 'sine x'
and 'cosine x'?
557
00:27:28,840 --> 00:27:32,690
And, again, notice the
dependency on identities.
558
00:27:32,690 --> 00:27:37,490
'Sine x' is twice 'sine
'x/2' cosine 'x/2''.
559
00:27:37,490 --> 00:27:41,020
But, from my reference triangle,
I can pick off the
560
00:27:41,020 --> 00:27:44,150
trigonometric functions
of 'x/2' very easily.
561
00:27:44,150 --> 00:27:48,400
Namely, the sine of 'x/2' is
just 'z' over the square root
562
00:27:48,400 --> 00:27:53,300
of '1 plus 'z squared'', and the
cosine of 'x/2' is just 1
563
00:27:53,300 --> 00:27:56,280
over the square root of
'1 plus 'z squared''.
564
00:27:56,280 --> 00:28:00,390
Plugging that in here, I find
that 'sine x' is twice 'z'
565
00:28:00,390 --> 00:28:03,960
over the square root of '1 plus
'z squared'' times 1 over
566
00:28:03,960 --> 00:28:06,190
the square root of '1
plus 'z squared''.
567
00:28:06,190 --> 00:28:08,500
Multiplying this out,
I find, what?
568
00:28:08,500 --> 00:28:12,610
That 'sine x' is twice 'z', and
the square root of '1 plus
569
00:28:12,610 --> 00:28:16,020
'z squared'' times itself is
just '1 plus 'z squared''.
570
00:28:16,020 --> 00:28:20,420
In other words, 'sine x' is
'2z'/ '1 plus 'z squared''.
571
00:28:20,420 --> 00:28:22,700
In other words, with the
substitution, what
572
00:28:22,700 --> 00:28:25,320
happens to 'sine x'?
573
00:28:25,320 --> 00:28:29,870
It becomes '2z'/ '1 plus 'z
squared'', which is also the
574
00:28:29,870 --> 00:28:33,530
quotient of two polynomials
in 'z', OK?
575
00:28:33,530 --> 00:28:36,610
Finally, how about 'cosine x'?
576
00:28:36,610 --> 00:28:40,650
What identity can we use
to reduce 'cosine
577
00:28:40,650 --> 00:28:43,360
x' in terms of 'x/2'?
578
00:28:43,360 --> 00:28:45,270
Why do we want the 'x/2'?
579
00:28:45,270 --> 00:28:48,870
Again, notice that even though
we may not have invented this
580
00:28:48,870 --> 00:28:52,740
substitution by ourself,
once it's invented, the
581
00:28:52,740 --> 00:28:56,260
relationship to the angle 'x/2'
becomes very apparent.
582
00:28:56,260 --> 00:28:59,200
At any rate, notice the identity
that says that
583
00:28:59,200 --> 00:29:03,500
'cosine 2x' is 'cosine squared
x' minus 'sine squared x'
584
00:29:03,500 --> 00:29:08,010
translates into 'cosine x' is
cosine squared of half the
585
00:29:08,010 --> 00:29:11,330
angle minus sine squared
of half the angle.
586
00:29:11,330 --> 00:29:17,400
Well, 'cosine squared 'x/2'',
well, 'cosine 'x/2'' is just 1
587
00:29:17,400 --> 00:29:20,570
over the square root of '1 plus
'c squared'', so 'cosine
588
00:29:20,570 --> 00:29:24,600
squared 'x/2'' is just 1/
'1 plus 'z squared''.
589
00:29:24,600 --> 00:29:28,620
Similarly, 'sine squared 'x/2''
is just 'z squared'
590
00:29:28,620 --> 00:29:30,630
over '1 plus 'z squared''.
591
00:29:30,630 --> 00:29:34,450
And what we find is that 'cosine
x' is '1 minus 'z
592
00:29:34,450 --> 00:29:38,500
squared'' over '1 plus
'z squared''.
593
00:29:38,500 --> 00:29:40,020
And, again, what's happened?
594
00:29:40,020 --> 00:29:43,620
'Cosine x' is now expressable
as the quotient of two
595
00:29:43,620 --> 00:29:45,520
polynomials in 'z'.
596
00:29:45,520 --> 00:29:48,840
In fact, as an application of
this, let's come back to an
597
00:29:48,840 --> 00:29:51,340
integrand that's been giving
us some trouble for
598
00:29:51,340 --> 00:29:52,800
quite some time now.
599
00:29:52,800 --> 00:29:57,350
Let's look at the integral
'secant x dx' and see if we
600
00:29:57,350 --> 00:30:00,750
can't use this technique that
we've just learned to solve
601
00:30:00,750 --> 00:30:02,780
this particular problem.
602
00:30:02,780 --> 00:30:06,750
Notice that secant is 1 over
cosine, therefore, integral
603
00:30:06,750 --> 00:30:11,020
'secant x dx' is integral
'dx' over 'cosine x'.
604
00:30:11,020 --> 00:30:13,010
Now, let's come back here
for a moment just
605
00:30:13,010 --> 00:30:14,280
to refresh our memories.
606
00:30:14,280 --> 00:30:20,290
We saw that 'dx' was two 'dz'/
'1 plus 'z squared''.
607
00:30:20,290 --> 00:30:26,320
We saw that 'cosine x' was '1
minus 'z squared'' over '1
608
00:30:26,320 --> 00:30:30,515
plus 'z squared'', therefore, '1
over cosine x' will just be
609
00:30:30,515 --> 00:30:31,840
the reciprocal of this.
610
00:30:31,840 --> 00:30:35,740
In other words, coming back here
now, we can replace 'dx'
611
00:30:35,740 --> 00:30:40,920
by '2 dz' over '1 plus 'z
squared'' and '1 over cosine
612
00:30:40,920 --> 00:30:43,950
x' by '1 plus 'z squared''
over '1 minus
613
00:30:43,950 --> 00:30:46,840
'z squared'', OK?
614
00:30:46,840 --> 00:30:50,710
Therefore, in terms of this
substitution, integral 'secant
615
00:30:50,710 --> 00:30:57,290
x dx' is just twice integral
'dz'/ '1 minus 'z squared''.
616
00:30:57,290 --> 00:30:59,010
But, what is this integrand?
617
00:30:59,010 --> 00:31:04,030
This integrand is the quotient
of two polynomials in 'z',
618
00:31:04,030 --> 00:31:07,590
therefore, I could use partial
fractions here, write this as
619
00:31:07,590 --> 00:31:13,600
a term: something over '1 plus
z' plus something over '1
620
00:31:13,600 --> 00:31:17,270
minus z', et cetera, solve the
problem in terms of 'z', and
621
00:31:17,270 --> 00:31:22,460
then, remembering that 'z' is
'tangent 'x/2'', I can then
622
00:31:22,460 --> 00:31:25,596
replace 'z' by what its equal
to in terms of 'x', and in
623
00:31:25,596 --> 00:31:27,460
that way solve the problem.
624
00:31:27,460 --> 00:31:30,020
Now, you see, what I want you to
see again, that comes true
625
00:31:30,020 --> 00:31:33,590
here all the time, is how we
are continually looking for
626
00:31:33,590 --> 00:31:39,650
ways of reducing integrals to
equivalent integrals, but,
627
00:31:39,650 --> 00:31:43,290
hopefully, ones that are more
familiar to us, meaning, what?
628
00:31:43,290 --> 00:31:45,670
Ones that we are
able to handle.
629
00:31:45,670 --> 00:31:49,000
In the next lesson, we're going
to find a very powerful
630
00:31:49,000 --> 00:31:51,950
technique, which is far more
general than partial
631
00:31:51,950 --> 00:31:55,560
fractions, a technique which is
used over, and over, again,
632
00:31:55,560 --> 00:31:58,920
which is probably the single
most important technique.
633
00:31:58,920 --> 00:32:00,970
But, I won't say any more
about that right now.
634
00:32:00,970 --> 00:32:03,560
We'll continue with this
discussion next time.
635
00:32:03,560 --> 00:32:07,400
And, until next time,
good bye.
636
00:32:07,400 --> 00:32:10,390
MALE VOICE: Funding for the
publication of this video was
637
00:32:10,390 --> 00:32:15,110
provided by the Gabriella and
Paul Rosenbaum Foundation.
638
00:32:15,110 --> 00:32:19,280
Help OCW continue to provide
free and open access to MIT
639
00:32:19,280 --> 00:32:23,740
courses by making a donation
at ocw.mit.edu/donate.