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PROFESSOR: So we're
going to continue
9
00:00:24,560 --> 00:00:29,420
to talk about trig integrals
and trig substitutions.
10
00:00:29,420 --> 00:00:31,520
This is maybe the
most technical part
11
00:00:31,520 --> 00:00:34,560
of this course, which maybe is
why professor Jerison decided
12
00:00:34,560 --> 00:00:37,520
to just take a leave, go AWOL
just now and let me take over
13
00:00:37,520 --> 00:00:38,610
for him.
14
00:00:38,610 --> 00:00:41,960
But I'll do my best to help
you learn this technique
15
00:00:41,960 --> 00:00:45,680
and it'll be useful for you.
16
00:00:45,680 --> 00:00:48,880
So we've talked about
trig integrals involving
17
00:00:48,880 --> 00:00:51,460
sines and cosines yesterday.
18
00:00:51,460 --> 00:00:53,060
There's another
whole world out there
19
00:00:53,060 --> 00:00:58,850
that involves these other trig
polynomials-- trig functions,
20
00:00:58,850 --> 00:01:00,520
secant and tangent.
21
00:01:00,520 --> 00:01:03,460
Let me just make a little table
to remind you what they are.
22
00:01:03,460 --> 00:01:05,210
Because I have trouble
remembering myself,
23
00:01:05,210 --> 00:01:07,630
so I enjoy the
opportunity to go back
24
00:01:07,630 --> 00:01:09,321
to remind myself of this stuff.
25
00:01:09,321 --> 00:01:09,820
Let's see.
26
00:01:09,820 --> 00:01:15,860
The secant is one over one of
those things, which one is it?
27
00:01:15,860 --> 00:01:18,420
It's weird, it's
1 over the cosine.
28
00:01:18,420 --> 00:01:24,710
And the cosecant
is 1 over the sin.
29
00:01:24,710 --> 00:01:26,800
Of course the tangent, we know.
30
00:01:26,800 --> 00:01:31,840
It's the sine over the
cosine and the cotangent
31
00:01:31,840 --> 00:01:36,000
is the other way around.
32
00:01:36,000 --> 00:01:37,700
So when you put a
"co" in front of it,
33
00:01:37,700 --> 00:01:43,590
it exchanges sine and cosine.
34
00:01:43,590 --> 00:01:46,840
Well, I have a few identities
involving tangent and secant
35
00:01:46,840 --> 00:01:50,270
up there, in that little
prepared blackboard up above.
36
00:01:50,270 --> 00:01:53,680
Maybe I'll just go
through and check that out
37
00:01:53,680 --> 00:01:57,604
to make sure that we're all
on the same page with them.
38
00:01:57,604 --> 00:01:59,020
So I'm going to
claim that there's
39
00:01:59,020 --> 00:02:00,560
this trig identity at the top.
40
00:02:00,560 --> 00:02:07,310
Secant squared is
1 plus the tangent.
41
00:02:07,310 --> 00:02:10,590
So let's just check that out.
42
00:02:10,590 --> 00:02:12,670
So the secant is
1 over the cosine,
43
00:02:12,670 --> 00:02:14,890
so secant squared is
1 over cosine squared.
44
00:02:14,890 --> 00:02:17,100
And then whenever you
see a 1 in trigonometry,
45
00:02:17,100 --> 00:02:23,010
you'll always have the option
of writing as cos^2 + sin^2.
46
00:02:27,890 --> 00:02:32,210
And if I do that, then I can
divide the cos^2 into that
47
00:02:32,210 --> 00:02:34,300
first term.
48
00:02:34,300 --> 00:02:36,380
And I get 1 + sin^2 / cos^2.
49
00:02:36,380 --> 00:02:37,940
/ ^2.
50
00:02:37,940 --> 00:02:42,430
Which is the tangent squared.
51
00:02:42,430 --> 00:02:44,180
So there you go.
52
00:02:44,180 --> 00:02:47,820
That checks the first one.
53
00:02:47,820 --> 00:02:49,320
That's the main
trig identity that's
54
00:02:49,320 --> 00:02:51,330
going to be behind what
I talk about today.
55
00:02:51,330 --> 00:02:53,460
That's the trigonometry
identity part.
56
00:02:53,460 --> 00:02:57,040
How about this
piece of calculus.
57
00:02:57,040 --> 00:03:05,490
Can we calculate what the
derivative of tan x is?
58
00:03:05,490 --> 00:03:12,350
Actually, I'm going to
do that on this board.
59
00:03:12,350 --> 00:03:19,110
So tan x = sin x / cos x.
60
00:03:19,110 --> 00:03:21,250
So I think I was with
you when we learned
61
00:03:21,250 --> 00:03:23,500
about the quotient rule.
62
00:03:23,500 --> 00:03:26,030
Computing the derivative
of a quotient.
63
00:03:26,030 --> 00:03:30,340
And the rule is, you
take the numerator
64
00:03:30,340 --> 00:03:34,330
and you-- sorry, you take the
derivative of the numerator,
65
00:03:34,330 --> 00:03:36,510
which is cosine.
66
00:03:36,510 --> 00:03:38,570
And you multiply it
by the denominator,
67
00:03:38,570 --> 00:03:41,590
so that gives you cos^2.
68
00:03:41,590 --> 00:03:46,480
And then you take the numerator,
take minus the numerator,
69
00:03:46,480 --> 00:03:49,400
and multiply that by the
derivative of the denominator,
70
00:03:49,400 --> 00:03:53,240
which is -sin x.
71
00:03:53,240 --> 00:03:57,870
And you put all that over the
square of the denominator.
72
00:03:57,870 --> 00:04:02,490
And now I look at that and
before my eyes I see the same
73
00:04:02,490 --> 00:04:07,420
trig identity, cos^2 +
sin^2 = 1, appearing there.
74
00:04:07,420 --> 00:04:13,850
This is 1 / cos^2(x)
which is secant squared.
75
00:04:13,850 --> 00:04:15,022
And, good.
76
00:04:15,022 --> 00:04:16,230
So that's what the claim was.
77
00:04:16,230 --> 00:04:20,420
The derivative of the tangent
is the secant squared.
78
00:04:20,420 --> 00:04:22,260
That immediately
gives you an integral.
79
00:04:22,260 --> 00:04:25,240
Namely, the integral of
secant squared is the tangent.
80
00:04:25,240 --> 00:04:28,510
That's the fundamental
theorem of calculus.
81
00:04:28,510 --> 00:04:32,360
So we verified the
first integral there.
82
00:04:32,360 --> 00:04:34,490
Well, let's just do
the second one as well.
83
00:04:34,490 --> 00:04:40,280
So if I want to
differentiate the secant,
84
00:04:40,280 --> 00:04:41,360
derivative of the secant.
85
00:04:41,360 --> 00:04:46,500
So that's d/dx of
1 over the cosine.
86
00:04:46,500 --> 00:04:47,817
And again, I have a quotient.
87
00:04:47,817 --> 00:04:49,900
This one's a little easier
because the numerator's
88
00:04:49,900 --> 00:04:51,010
so simple.
89
00:04:51,010 --> 00:04:54,310
So I take the derivative of
the numerator, which is 0.
90
00:04:54,310 --> 00:04:56,580
And then I take the
numerator, I take
91
00:04:56,580 --> 00:05:00,830
minus the numerator times the
derivative of the denominator.
92
00:05:00,830 --> 00:05:05,610
Which is -sin x,
and put all that
93
00:05:05,610 --> 00:05:09,900
over the square of
the same denominator.
94
00:05:09,900 --> 00:05:12,809
So one minus sign came
from the quotient rule,
95
00:05:12,809 --> 00:05:14,350
and the other one
came because that's
96
00:05:14,350 --> 00:05:16,580
the derivative of the cosine.
97
00:05:16,580 --> 00:05:21,010
But they cancel, and
so I get sin / cos^2,
98
00:05:21,010 --> 00:05:25,640
which is sin / cos times 1 /
cos and so that's the secant,
99
00:05:25,640 --> 00:05:32,160
that's 1 / cos, times tan x.
100
00:05:32,160 --> 00:05:34,190
So, not hard.
101
00:05:34,190 --> 00:05:36,300
That verifies that the
derivative of secant
102
00:05:36,300 --> 00:05:37,990
is secant tangent.
103
00:05:37,990 --> 00:05:41,800
And it tells you that the
integral of that weird thing
104
00:05:41,800 --> 00:05:43,830
in case you ever want
to know, the integral
105
00:05:43,830 --> 00:05:47,910
of the secant tangent
is the secant.
106
00:05:47,910 --> 00:05:50,470
Well, there are a
couple more integrals
107
00:05:50,470 --> 00:05:54,000
that I want to do for you.
108
00:05:54,000 --> 00:05:57,050
Where I can't sort of
work backwards like that.
109
00:05:57,050 --> 00:06:07,060
Let's calculate the
integral of the tangent.
110
00:06:07,060 --> 00:06:09,380
Just do this straight out.
111
00:06:09,380 --> 00:06:22,610
So the tangent is the sine
divided by the cosine.
112
00:06:22,610 --> 00:06:28,520
And now there's a habit of
mind, that I hope you get into.
113
00:06:28,520 --> 00:06:33,310
When you see the cosine and
you're calculating an integral
114
00:06:33,310 --> 00:06:35,950
like this, it's
useful to remember
115
00:06:35,950 --> 00:06:37,700
what the derivative
of the cosine is.
116
00:06:37,700 --> 00:06:41,420
Because maybe it shows up
somewhere else in the integral.
117
00:06:41,420 --> 00:06:43,540
And that happens here.
118
00:06:43,540 --> 00:06:50,300
So that suggests we make
a substitution. u = cos x.
119
00:06:50,300 --> 00:06:55,620
Which means du = -sin x dx.
120
00:06:55,620 --> 00:06:59,020
That's the numerator,
except for the minus sign.
121
00:06:59,020 --> 00:07:08,570
And so I can rewrite this
as, under the substitution,
122
00:07:08,570 --> 00:07:13,120
I can rewrite this as
-du, that's the numerator,
123
00:07:13,120 --> 00:07:19,600
sin x dx is -du, divided by u.
124
00:07:19,600 --> 00:07:22,150
Well, I know how to
do that integral too.
125
00:07:22,150 --> 00:07:24,210
That gives me the
natural log, doesn't it.
126
00:07:24,210 --> 00:07:31,150
So this is -ln(u)
plus a constant.
127
00:07:31,150 --> 00:07:32,360
I'm not quite done.
128
00:07:32,360 --> 00:07:34,750
I have to back-substitute
and replace
129
00:07:34,750 --> 00:07:39,080
this new variable that I've made
up, called u, with what it is.
130
00:07:39,080 --> 00:07:46,760
And what you get is -ln(cos x).
131
00:07:46,760 --> 00:07:50,410
So the integral of the
tangent is minus log cosine.
132
00:07:50,410 --> 00:07:55,590
Now, you find these
tables of integrals
133
00:07:55,590 --> 00:07:56,910
in the back of the book.
134
00:07:56,910 --> 00:07:58,320
Things like that.
135
00:07:58,320 --> 00:08:01,060
I'm not sure how much
memorization Professor Jerison
136
00:08:01,060 --> 00:08:04,210
is going to ask
of you, but there
137
00:08:04,210 --> 00:08:05,770
is a certain amount
of memorization
138
00:08:05,770 --> 00:08:06,937
that goes on in calculus.
139
00:08:06,937 --> 00:08:08,520
And this is one of
the kinds of things
140
00:08:08,520 --> 00:08:12,380
that you probably want to know.
141
00:08:12,380 --> 00:08:15,450
Let me do one more integral.
142
00:08:15,450 --> 00:08:18,100
I think I'm making my way
through a prepared board here,
143
00:08:18,100 --> 00:08:20,350
let's see.
144
00:08:20,350 --> 00:08:23,060
Good.
145
00:08:23,060 --> 00:08:28,940
So the integral of the
tangent is minus log cosine.
146
00:08:28,940 --> 00:08:36,090
I'd also like to know what the
integral of the secant of x is.
147
00:08:36,090 --> 00:08:41,280
And I don't know a way to
kind of go straight at this,
148
00:08:41,280 --> 00:08:45,580
but let me show you a way to
think your way through to it.
149
00:08:45,580 --> 00:08:52,720
If I take these two facts,
tangent prime is what it is,
150
00:08:52,720 --> 00:08:56,280
and secant prime is what it
is, and add them together,
151
00:08:56,280 --> 00:08:57,420
I get this fact.
152
00:08:57,420 --> 00:09:05,400
That the derivative of
the sec x + tan x is,
153
00:09:05,400 --> 00:09:08,350
well, it's the sum
of these two things.
154
00:09:08,350 --> 00:09:11,060
Secant squared plus
secant tangent.
155
00:09:11,060 --> 00:09:15,180
And there's a secant that
occurs in both of those terms.
156
00:09:15,180 --> 00:09:17,140
So I'll factor it out.
157
00:09:17,140 --> 00:09:21,440
And that gives me,
I'll put it over here.
158
00:09:21,440 --> 00:09:24,516
There's the secant of x
that occurs in both terms.
159
00:09:24,516 --> 00:09:26,390
And then in one term,
there's another secant.
160
00:09:26,390 --> 00:09:32,740
And in the other term,
there's a tangent.
161
00:09:32,740 --> 00:09:36,970
So that's interesting somehow,
because this same term appears
162
00:09:36,970 --> 00:09:40,200
on both sides of this equation.
163
00:09:40,200 --> 00:09:49,160
Let's write u, for
that sec x + tan x.
164
00:09:49,160 --> 00:09:55,120
And so the equation that
I get is u' = u sec x.
165
00:10:00,930 --> 00:10:03,860
I've just made a
direct substitution.
166
00:10:03,860 --> 00:10:05,620
Just decide that
I'm going to write u
167
00:10:05,620 --> 00:10:07,730
for that single
thing that occurs
168
00:10:07,730 --> 00:10:09,600
on both sides of the equation.
169
00:10:09,600 --> 00:10:16,440
So u' is on the left, and
u * sec x is on the right.
170
00:10:16,440 --> 00:10:18,560
Well, there's my secant.
171
00:10:18,560 --> 00:10:20,210
That I was trying to integrate.
172
00:10:20,210 --> 00:10:28,560
And what it tells
you is that = u' / u.
173
00:10:28,560 --> 00:10:35,550
Just divide both sides by u,
and I get this equation. u' / u,
174
00:10:35,550 --> 00:10:36,910
that has a name.
175
00:10:36,910 --> 00:10:38,880
Not sure that
professor Jerison's
176
00:10:38,880 --> 00:10:41,604
used this in this
class, but u' / u,
177
00:10:41,604 --> 00:10:43,270
we've actually used
something like that.
178
00:10:43,270 --> 00:10:45,560
It's on the board right now.
179
00:10:45,560 --> 00:10:47,650
It's a logarithmic derivative.
180
00:10:47,650 --> 00:10:57,750
It is the derivative of the
national logarithm of u.
181
00:10:57,750 --> 00:10:59,940
Maybe it's easier to read
this from right to left,
182
00:10:59,940 --> 00:11:02,780
if I want to calculate the
derivative of the logarithm,
183
00:11:02,780 --> 00:11:06,380
well, the chain rule says I
get the derivative of u times
184
00:11:06,380 --> 00:11:11,100
the derivative of the log
function, which is 1 / u.
185
00:11:11,100 --> 00:11:25,964
So often u' / u is called
the logarithmic derivative.
186
00:11:25,964 --> 00:11:27,130
But it's done what I wanted.
187
00:11:27,130 --> 00:11:31,740
Because it's expressed the
secant as a derivative.
188
00:11:31,740 --> 00:11:38,630
And I guess I should
put in what u is.
189
00:11:38,630 --> 00:11:46,090
It's the secant
plus the tangent.
190
00:11:46,090 --> 00:11:48,500
And so that implies
that the integral--
191
00:11:48,500 --> 00:11:49,510
Integrate both sides.
192
00:11:49,510 --> 00:11:56,840
That says that the integral of
sec x dx, is ln(sec x + tan x).
193
00:12:02,280 --> 00:12:09,500
So that's the last line in this
little memo that I created.
194
00:12:09,500 --> 00:12:14,500
That we can use now for
the rest of the class.
195
00:12:14,500 --> 00:12:19,320
Any questions about that trick?
196
00:12:19,320 --> 00:12:23,740
It's a trick, I have nothing
more to say about it.
197
00:12:23,740 --> 00:12:24,540
OK.
198
00:12:24,540 --> 00:12:30,210
So, the next thing
I-- oh yes, so now
199
00:12:30,210 --> 00:12:33,060
I want to make the point
that using these rules
200
00:12:33,060 --> 00:12:38,350
and some thought,
you can now integrate
201
00:12:38,350 --> 00:12:41,650
most trigonometric polynomials.
202
00:12:41,650 --> 00:12:44,850
Most things that involve
powers of sines and cosines
203
00:12:44,850 --> 00:12:48,080
and tangents and secants
and everything else.
204
00:12:48,080 --> 00:12:56,790
For example, let's try to
integrate the integral of sec^4
205
00:12:56,790 --> 00:12:59,180
x.
206
00:12:59,180 --> 00:13:01,914
Big power of the
secant function.
207
00:13:01,914 --> 00:13:03,830
Well, there are too many
secants there for me.
208
00:13:03,830 --> 00:13:05,680
So let's take some away.
209
00:13:05,680 --> 00:13:09,730
And I can take them away by
using that trig identity,
210
00:13:09,730 --> 00:13:12,300
sec^2 = 1 + tan^2.
211
00:13:12,300 --> 00:13:16,680
So I'm going to replace two
of those secants by 1 + tan^2.
212
00:13:21,790 --> 00:13:25,190
That leaves me
with two left over.
213
00:13:25,190 --> 00:13:27,920
Now there was method
to my madness.
214
00:13:27,920 --> 00:13:31,020
Because I've got a secant
squared left over there.
215
00:13:31,020 --> 00:13:35,280
And secant squared is the
derivative of tangent.
216
00:13:35,280 --> 00:13:40,970
So that suggests a substitution.
217
00:13:40,970 --> 00:13:48,440
Namely, let's say, let's let u =
tan x, so that du = sec^2 x dx.
218
00:13:50,980 --> 00:13:55,700
And I have both terms
that occur in my integral
219
00:13:55,700 --> 00:14:01,800
sitting there very nicely.
220
00:14:01,800 --> 00:14:05,020
So this is the possibility
of making this substitution
221
00:14:05,020 --> 00:14:08,240
and seeing a secant
squared up here as part
222
00:14:08,240 --> 00:14:10,330
of the differential here.
223
00:14:10,330 --> 00:14:13,910
That's why it was a good
idea for me to take two
224
00:14:13,910 --> 00:14:16,350
of the secants and
write them as 1 + tan^2.
225
00:14:19,170 --> 00:14:22,250
So now I can continue this.
226
00:14:22,250 --> 00:14:25,780
Under that
substitution, I get 1.
227
00:14:25,780 --> 00:14:32,810
Oh yeah, and I should add
the other fact, that--
228
00:14:32,810 --> 00:14:38,460
Well I guess it's obvious
that tangent squared is u^2.
229
00:14:38,460 --> 00:14:41,070
So I get 1 + u^2.
230
00:14:41,070 --> 00:14:50,110
And then du-- sec^2
sec^2 x dx, that is du.
231
00:14:50,110 --> 00:14:52,430
Well that's pretty
easy to integrate.
232
00:14:52,430 --> 00:14:56,690
So I get u + u^3 / 3.
233
00:14:56,690 --> 00:14:57,950
Plus a constant.
234
00:14:57,950 --> 00:15:00,020
And then I just have
to back-substitute.
235
00:15:00,020 --> 00:15:03,600
Put things back in terms
of the original variables.
236
00:15:03,600 --> 00:15:12,220
And that gives me tan x
plus tangent cubed over 3.
237
00:15:12,220 --> 00:15:15,610
And there's the answer.
238
00:15:15,610 --> 00:15:17,070
So we could spend
a lot more time
239
00:15:17,070 --> 00:15:21,810
doing more examples of this
kind of polynomial trig thing.
240
00:15:21,810 --> 00:15:27,440
It's probably best for you to
do some practice on your own.
241
00:15:27,440 --> 00:15:30,020
Because I want to talk
about other things, also.
242
00:15:30,020 --> 00:15:32,800
And what I want to
talk about is the use
243
00:15:32,800 --> 00:15:39,380
of these trig identities in
making really trig substitution
244
00:15:39,380 --> 00:15:51,590
integration.
245
00:15:51,590 --> 00:15:54,320
So we did a little
bit of this yesterday,
246
00:15:54,320 --> 00:15:56,990
and I'll show you some
more examples today.
247
00:15:56,990 --> 00:15:59,520
Let's start with a pretty hard
example right off the bat.
248
00:15:59,520 --> 00:16:07,700
So this is going to be the
integral of dx over x^2 times
249
00:16:07,700 --> 00:16:10,200
the square root of 1+x^2.
250
00:16:14,690 --> 00:16:18,120
It's a pretty bad
looking integral.
251
00:16:18,120 --> 00:16:20,190
So how can we approach this?
252
00:16:20,190 --> 00:16:25,390
Well, the square root is the
ugliest part of the integral,
253
00:16:25,390 --> 00:16:27,240
I think.
254
00:16:27,240 --> 00:16:30,160
What we should try to do
is write this square root
255
00:16:30,160 --> 00:16:31,600
in some nicer way.
256
00:16:31,600 --> 00:16:37,180
That is, figure out a way
to write 1+x^2 as a square.
257
00:16:37,180 --> 00:16:40,140
That'll get rid of
the square root.
258
00:16:40,140 --> 00:16:44,540
So there is an example of a
way to write 1 plus something
259
00:16:44,540 --> 00:16:46,140
squared in a different way.
260
00:16:46,140 --> 00:16:47,760
And it's right up there.
261
00:16:47,760 --> 00:16:50,790
sec^2 = 1 + tan^2.
262
00:16:50,790 --> 00:16:53,450
So I want to use that idea.
263
00:16:53,450 --> 00:16:56,650
And when I see this
form, that suggests
264
00:16:56,650 --> 00:16:59,520
that we make a trig
substitution and write
265
00:16:59,520 --> 00:17:04,682
x as the tangent of
some new variable.
266
00:17:04,682 --> 00:17:06,140
Which you might as
well call theta,
267
00:17:06,140 --> 00:17:09,060
to because it's like an angle.
268
00:17:09,060 --> 00:17:15,960
Then 1+x^2 is the
secant squared.
269
00:17:15,960 --> 00:17:18,620
According to that trig identity.
270
00:17:18,620 --> 00:17:23,970
And so the square root
of 1+x^2 is sec(theta).
271
00:17:29,780 --> 00:17:31,070
Right?
272
00:17:31,070 --> 00:17:36,220
So this identity is the
reason that the substitution
273
00:17:36,220 --> 00:17:37,120
is going to help us.
274
00:17:37,120 --> 00:17:39,430
Because it gets rid
of the square root
275
00:17:39,430 --> 00:17:43,550
and replaces it by some
other trig function.
276
00:17:43,550 --> 00:17:46,080
I'd better be able to
get rid of the dx, too.
277
00:17:46,080 --> 00:17:48,520
That's part of the
substitution process.
278
00:17:48,520 --> 00:17:50,520
But we can do that,
because I know
279
00:17:50,520 --> 00:17:52,440
what the derivative
of the tangent is.
280
00:17:52,440 --> 00:17:56,070
It's secant squared.
281
00:17:56,070 --> 00:17:58,880
So dx / d theta is sec^2(theta).
282
00:17:58,880 --> 00:18:01,150
So dx is sec^2(theta) d theta.
283
00:18:04,460 --> 00:18:07,350
So let's just substitute
all of that stuff
284
00:18:07,350 --> 00:18:10,280
in, and rewrite the
entire integral in terms
285
00:18:10,280 --> 00:18:11,750
of our new variable, theta.
286
00:18:11,750 --> 00:18:14,570
So dx is in the numerator.
287
00:18:14,570 --> 00:18:18,210
That's sec^2(theta) d theta.
288
00:18:18,210 --> 00:18:21,197
And then the denominator,
well, it has an x^2.
289
00:18:21,197 --> 00:18:22,030
That's tan^2(theta).
290
00:18:24,740 --> 00:18:28,660
And then there's
this square root.
291
00:18:28,660 --> 00:18:31,690
And we know what that
is in terms of theta.
292
00:18:31,690 --> 00:18:32,730
It's sec(theta).
293
00:18:36,470 --> 00:18:41,920
OK, now. we've done
the trig substitution.
294
00:18:41,920 --> 00:18:43,940
I've gotten rid of
the square root,
295
00:18:43,940 --> 00:18:46,050
I've got everything in
terms of trig functions
296
00:18:46,050 --> 00:18:48,020
of the new variable.
297
00:18:48,020 --> 00:18:49,860
Pretty complicated
trig function.
298
00:18:49,860 --> 00:18:52,550
This often happens, you wind
up with a complete scattering
299
00:18:52,550 --> 00:18:55,050
of different trig functions in
the numerator and denominator
300
00:18:55,050 --> 00:18:56,250
and everything.
301
00:18:56,250 --> 00:18:58,950
A systematic thing to do
here is to put everything
302
00:18:58,950 --> 00:19:09,110
in terms of sines and cosines.
303
00:19:09,110 --> 00:19:12,600
Unless you can see right away,
how it's going to simplify,
304
00:19:12,600 --> 00:19:15,010
the systematic thing
to do is to rewrite
305
00:19:15,010 --> 00:19:17,110
in terms of sines and cosines.
306
00:19:17,110 --> 00:19:19,300
So let's do that.
307
00:19:19,300 --> 00:19:20,260
So let's see.
308
00:19:20,260 --> 00:19:23,520
The secant squared,
secant is 1 over cosine.
309
00:19:23,520 --> 00:19:28,920
So I'm going to put a cosine
squared in the denominator.
310
00:19:28,920 --> 00:19:32,070
Oh, I guess the first
thing I can do is cancel.
311
00:19:32,070 --> 00:19:33,270
Let's do that.
312
00:19:33,270 --> 00:19:34,200
That's clever.
313
00:19:34,200 --> 00:19:35,700
You were all thinking that too.
314
00:19:35,700 --> 00:19:37,220
Cancel those.
315
00:19:37,220 --> 00:19:39,860
So now I just get one
cosine denominator
316
00:19:39,860 --> 00:19:44,382
from the secant there
in the numerator.
317
00:19:44,382 --> 00:19:46,840
It's still pretty complicated,
secant over tangent squared,
318
00:19:46,840 --> 00:19:48,110
who knows.
319
00:19:48,110 --> 00:19:49,410
Well, we'll find out.
320
00:19:49,410 --> 00:19:52,450
Because the tangent
is sine over cosine.
321
00:19:52,450 --> 00:19:55,940
So I should put a sine
squared where the tangent was,
322
00:19:55,940 --> 00:19:59,330
and a cosine squared up there.
323
00:19:59,330 --> 00:20:02,180
And I still have d theta.
324
00:20:02,180 --> 00:20:04,300
And now you see some
more cancellation occurs.
325
00:20:04,300 --> 00:20:07,950
That's the virtue of writing
things out in this way.
326
00:20:07,950 --> 00:20:14,940
So now, the square here
cancels with this cosine.
327
00:20:14,940 --> 00:20:20,580
And I'm left with cos(theta)
d theta / sin^2(theta).
328
00:20:25,340 --> 00:20:26,860
That's a little simpler.
329
00:20:26,860 --> 00:20:33,300
And it puts me in a position to
use the same idea I just used.
330
00:20:33,300 --> 00:20:35,100
I see the sine here.
331
00:20:35,100 --> 00:20:37,090
I might look around
in this integral
332
00:20:37,090 --> 00:20:39,940
to see if its derivative
occurs anywhere.
333
00:20:39,940 --> 00:20:43,890
The differential of
the sine is the cosine.
334
00:20:43,890 --> 00:20:49,720
And so I'm very much inclined
to make another substitution.
335
00:20:49,720 --> 00:20:54,250
Say, u, direct
substitution this time.
336
00:20:54,250 --> 00:20:56,150
And say u is the
cosine of theta.
337
00:20:56,150 --> 00:20:59,110
Because then du-- Oh, I'm sorry.
338
00:20:59,110 --> 00:21:01,850
Say, u is the sine of theta.
339
00:21:01,850 --> 00:21:05,980
Because then du is
cos(theta) d theta.
340
00:21:13,540 --> 00:21:20,140
And then this integral becomes,
well, the numerator just is du.
341
00:21:20,140 --> 00:21:22,660
The denominator is u^2.
342
00:21:22,660 --> 00:21:25,550
And I think we can
break out the champagne,
343
00:21:25,550 --> 00:21:28,169
because we can
integrate that one.
344
00:21:28,169 --> 00:21:29,710
Finally get rid of
the integral sign.
345
00:21:29,710 --> 00:21:30,210
Yes sir.
346
00:21:30,210 --> 00:21:37,842
STUDENT: [INAUDIBLE]
347
00:21:37,842 --> 00:21:40,300
PROFESSOR: OK, how do I know
to make u equal to sine rather
348
00:21:40,300 --> 00:21:41,640
than cosine.
349
00:21:41,640 --> 00:21:45,540
Because I want to see
du appear up here.
350
00:21:45,540 --> 00:21:49,360
If I'd had a sine up here,
that would be a signal to me
351
00:21:49,360 --> 00:21:53,510
that maybe I should say
let u be the cosine.
352
00:21:53,510 --> 00:21:54,280
OK?
353
00:21:54,280 --> 00:21:56,980
Also, because this thing in
the denominator is something
354
00:21:56,980 --> 00:21:57,870
I want to get rid of.
355
00:21:57,870 --> 00:21:59,050
It's in the denominator.
356
00:21:59,050 --> 00:22:01,030
So I'll get rid of it
by wishful thinking
357
00:22:01,030 --> 00:22:04,940
and just call it something else.
358
00:22:04,940 --> 00:22:07,010
It works pretty
well in this case.
359
00:22:07,010 --> 00:22:10,400
Wishful thinking doesn't
always work so well.
360
00:22:10,400 --> 00:22:18,510
So I integrate u^(-2) du, and
I get - -1/u plus a constant,
361
00:22:18,510 --> 00:22:21,340
and I'm done with the
calculus part of this problem.
362
00:22:21,340 --> 00:22:22,580
I've done the integral now.
363
00:22:22,580 --> 00:22:25,736
Gotten rid of the integral sign.
364
00:22:25,736 --> 00:22:27,360
But I'm not quite
done with the problem
365
00:22:27,360 --> 00:22:29,640
yet, because I have
to work my way back
366
00:22:29,640 --> 00:22:31,320
through two substitutions.
367
00:22:31,320 --> 00:22:33,120
First, this one.
368
00:22:33,120 --> 00:22:34,620
And then this one.
369
00:22:34,620 --> 00:22:38,800
So this first substitution isn't
so bad to get rid of, to undo,
370
00:22:38,800 --> 00:22:40,200
to back-substitute.
371
00:22:40,200 --> 00:22:43,630
Because u is just sin(theta).
372
00:22:43,630 --> 00:22:46,740
And so 1/u is, I
guess a fancy way
373
00:22:46,740 --> 00:22:52,540
to write it is the
cosecant of theta.
374
00:22:52,540 --> 00:22:54,890
1 over the sine is the cosecant.
375
00:22:54,890 --> 00:23:00,257
So I get -csc(theta)
plus a constant.
376
00:23:00,257 --> 00:23:01,590
Is there a question in the back?
377
00:23:01,590 --> 00:23:02,090
Yes sir?
378
00:23:02,090 --> 00:23:07,010
STUDENT: [INAUDIBLE]
379
00:23:07,010 --> 00:23:09,000
PROFESSOR: I'm sorry,
my hearing is so bad.
380
00:23:09,000 --> 00:23:12,500
STUDENT: [INAUDIBLE]
381
00:23:12,500 --> 00:23:16,070
PROFESSOR: How did I
know this substitution
382
00:23:16,070 --> 00:23:19,950
in the first place.
383
00:23:19,950 --> 00:23:22,510
It's because of the 1 + x^2.
384
00:23:22,510 --> 00:23:24,510
And I want to make use
of the trig identity
385
00:23:24,510 --> 00:23:26,380
in the upper left-hand corner.
386
00:23:26,380 --> 00:23:29,430
I'll make you a table in a
few minutes that will put
387
00:23:29,430 --> 00:23:30,870
all this in a bigger context.
388
00:23:30,870 --> 00:23:32,350
And I think it'll help you then.
389
00:23:32,350 --> 00:23:35,880
OK, I'll promise.
390
00:23:35,880 --> 00:23:39,570
So, what I want to try
to talk about right
391
00:23:39,570 --> 00:23:43,710
now is how to rewrite
a term like this.
392
00:23:43,710 --> 00:23:47,240
A trig term like this,
back in terms of x.
393
00:23:47,240 --> 00:23:51,020
So I want to undo this
trick substitution.
394
00:23:51,020 --> 00:23:55,190
This is a trig sub.
395
00:23:55,190 --> 00:23:59,350
And what I want to do now is
try to undo that trig sub.
396
00:23:59,350 --> 00:24:00,970
And I'll show you
a general method
397
00:24:00,970 --> 00:24:03,210
for undoing trig substitutions.
398
00:24:03,210 --> 00:24:05,100
This happens quite often.
399
00:24:05,100 --> 00:24:07,150
I don't know what the
cosecant of theta is.
400
00:24:07,150 --> 00:24:11,290
But I do know what the
tangent of theta is.
401
00:24:11,290 --> 00:24:14,060
So I want to make a
relation between them.
402
00:24:14,060 --> 00:24:20,920
OK, so undoing.
403
00:24:20,920 --> 00:24:24,420
Trig subs.
404
00:24:24,420 --> 00:24:29,420
So let's go back to where
trigonometry always comes from,
405
00:24:29,420 --> 00:24:32,130
this right angled triangle.
406
00:24:32,130 --> 00:24:35,540
The theta in the corner,
and then these three sides.
407
00:24:35,540 --> 00:24:37,590
This one's called
the hypotenuse.
408
00:24:37,590 --> 00:24:41,430
This one is called
the adjacent side,
409
00:24:41,430 --> 00:24:45,690
and that one's called
the opposite side.
410
00:24:45,690 --> 00:24:52,040
And now, let's find out where
x lies in this triangle.
411
00:24:52,040 --> 00:24:55,500
Let's try to write the sides
of this triangle in terms of x.
412
00:24:55,500 --> 00:24:58,670
And what I know is, x
is the tangent of theta.
413
00:24:58,670 --> 00:25:01,310
So the tangent of theta,
tangent of this angle,
414
00:25:01,310 --> 00:25:03,830
is opposite divided by adjacent.
415
00:25:03,830 --> 00:25:05,890
Did you learn SOH CAH TOA?
416
00:25:05,890 --> 00:25:09,520
OK, so it's opposite
divided by adjacent.
417
00:25:09,520 --> 00:25:10,580
Is the tangent.
418
00:25:10,580 --> 00:25:13,650
So there are different
ways to do that,
419
00:25:13,650 --> 00:25:15,530
but why not just do
it in the simplest way
420
00:25:15,530 --> 00:25:21,950
and suppose that the adjacent
is 1, and the opposite is x.
421
00:25:21,950 --> 00:25:24,530
This is correct now, isn't it?
422
00:25:24,530 --> 00:25:27,890
I get the correct value
for the tangent of theta
423
00:25:27,890 --> 00:25:32,910
by saying that the lengths
of those are 1 and x.
424
00:25:32,910 --> 00:25:39,160
And that means that the
hypotenuse has length 1 + x^2.
425
00:25:39,160 --> 00:25:40,840
Well, here's a triangle.
426
00:25:40,840 --> 00:25:44,060
I'm interested in computing
the cosecant of theta.
427
00:25:44,060 --> 00:25:50,840
Where's that appear
in the triangle?
428
00:25:50,840 --> 00:25:51,510
Well, let's see.
429
00:25:51,510 --> 00:25:55,570
The cosecant of theta
is 1 over the sine.
430
00:25:55,570 --> 00:26:02,830
And the sine is opposite
over hypotenuse.
431
00:26:02,830 --> 00:26:15,240
So the cosecant is
hypotenuse over opposite.
432
00:26:15,240 --> 00:26:20,680
And the hypotenuse is the
square root of 1 + x^2,
433
00:26:20,680 --> 00:26:24,680
and the opposite is x.
434
00:26:24,680 --> 00:26:26,200
And so I've done it.
435
00:26:26,200 --> 00:26:29,650
I've undone the
trig substitution.
436
00:26:29,650 --> 00:26:32,100
I've figured out what
this cosecant of theta
437
00:26:32,100 --> 00:26:35,360
is, in terms of x.
438
00:26:35,360 --> 00:26:39,490
And so the final answer is
minus the square root of 1+x^2,
439
00:26:39,490 --> 00:26:43,470
over x, plus a constant,
and there's an answer
440
00:26:43,470 --> 00:26:53,030
to the original problem.
441
00:26:53,030 --> 00:26:57,160
This took two boards
to go through this.
442
00:26:57,160 --> 00:27:00,800
I illustrated several things.
443
00:27:00,800 --> 00:27:02,870
Actually, this
three half boards.
444
00:27:02,870 --> 00:27:05,190
I illustrated this use
of trig substitution,
445
00:27:05,190 --> 00:27:07,910
and I'll come back
to that in a second.
446
00:27:07,910 --> 00:27:10,870
I illustrated patience.
447
00:27:10,870 --> 00:27:15,320
I illustrated rewriting things
in terms of sines and cosines,
448
00:27:15,320 --> 00:27:17,080
and then making a
direct substitution
449
00:27:17,080 --> 00:27:19,550
to evaluate an
integral like this.
450
00:27:19,550 --> 00:27:23,520
And then there's this undoing
all of those substitutions.
451
00:27:23,520 --> 00:27:25,670
And it culminated with
undoing the trig sub.
452
00:27:25,670 --> 00:27:31,100
So let's play a game here.
453
00:27:31,100 --> 00:27:35,070
Why don't we play
the game where you
454
00:27:35,070 --> 00:27:46,970
give me-- So, there's a step in
here that I should have done.
455
00:27:46,970 --> 00:27:56,390
I should've said this is
-cos(arctan(theta)) plus
456
00:27:56,390 --> 00:27:58,812
a constant.
457
00:27:58,812 --> 00:28:00,520
The most straightforward
thing you can do
458
00:28:00,520 --> 00:28:02,760
is to say since x is
the tangent of theta,
459
00:28:02,760 --> 00:28:06,950
that means that-- sorry,
if x, that means that theta
460
00:28:06,950 --> 00:28:08,970
is the arctangent of x.
461
00:28:08,970 --> 00:28:11,761
And so let's just put in
theta as the arctangent of x,
462
00:28:11,761 --> 00:28:12,760
and that's what you get.
463
00:28:12,760 --> 00:28:15,020
So really, what I
just did for you
464
00:28:15,020 --> 00:28:19,720
was to show you a way to compute
some trig function applied
465
00:28:19,720 --> 00:28:23,900
to the inverse of
another trig function.
466
00:28:23,900 --> 00:28:29,860
I computed cosecant of the
arctangent by this trick.
467
00:28:29,860 --> 00:28:32,260
So now, let's play the
game where you give me
468
00:28:32,260 --> 00:28:35,150
a trig function and an
inverse trig function,
469
00:28:35,150 --> 00:28:46,950
and I try to compute
what the composite is.
470
00:28:46,950 --> 00:28:47,510
OK.
471
00:28:47,510 --> 00:28:58,530
So who can give me
a trig function.
472
00:28:58,530 --> 00:29:03,310
Has to be one of
these standard ones.
473
00:29:03,310 --> 00:29:03,900
STUDENT: Tan.
474
00:29:03,900 --> 00:29:04,870
PROFESSOR: Tangent.
475
00:29:04,870 --> 00:29:07,034
Alright.
476
00:29:07,034 --> 00:29:07,950
How about another one?
477
00:29:07,950 --> 00:29:11,120
STUDENT: Sine.
478
00:29:11,120 --> 00:29:12,350
PROFESSOR: Sine.
479
00:29:12,350 --> 00:29:15,180
Do we have agreement on sine.
480
00:29:15,180 --> 00:29:16,910
STUDENT: [INAUDIBLE]
481
00:29:16,910 --> 00:29:17,760
PROFESSOR: Secant?
482
00:29:17,760 --> 00:29:27,802
STUDENT: [INAUDIBLE]
483
00:29:27,802 --> 00:29:29,510
PROFESSOR: Right, csc
has the best cheer.
484
00:29:29,510 --> 00:29:30,301
So that's the game.
485
00:29:30,301 --> 00:29:34,867
We have to compute, try to
compute, that composite.
486
00:29:34,867 --> 00:29:35,950
Something wrong with this?
487
00:29:35,950 --> 00:29:45,459
STUDENT: [INAUDIBLE]
488
00:29:45,459 --> 00:29:47,000
PROFESSOR: What does
acceptable mean?
489
00:29:47,000 --> 00:29:48,810
Don't you think--
so the question is,
490
00:29:48,810 --> 00:29:51,900
isn't this a perfectly
acceptable final answer.
491
00:29:51,900 --> 00:29:53,760
It's a correct final answer.
492
00:29:53,760 --> 00:29:56,850
But this is much
more insightful.
493
00:29:56,850 --> 00:30:00,919
And after all the original
thing was involving square roots
494
00:30:00,919 --> 00:30:02,460
and things, this is
the kind of thing
495
00:30:02,460 --> 00:30:03,920
you might hope for is an answer.
496
00:30:03,920 --> 00:30:07,720
This is just a nicer
answer for sure.
497
00:30:07,720 --> 00:30:10,210
And likely to be more
useful to you when you go on
498
00:30:10,210 --> 00:30:13,830
and use that answer
for something else.
499
00:30:13,830 --> 00:30:18,610
OK, so let's try
to do this this.
500
00:30:18,610 --> 00:30:22,260
Undo a trig substitution
that involved a cosecant.
501
00:30:22,260 --> 00:30:23,910
And I manipulate
around, and I find
502
00:30:23,910 --> 00:30:27,540
myself trying to find out
what's the tangent of theta.
503
00:30:27,540 --> 00:30:29,240
So here's how we go about it.
504
00:30:29,240 --> 00:30:33,970
I draw this triangle.
505
00:30:33,970 --> 00:30:37,590
Theta is the angle here.
506
00:30:37,590 --> 00:30:43,570
This is the adjacent,
opposite, hypotenuse.
507
00:30:43,570 --> 00:30:49,030
So, the first thing is how can
I make the cosecant appear here,
508
00:30:49,030 --> 00:30:49,950
csc x.
509
00:30:49,950 --> 00:30:52,640
What dimensions should I
give to the sides in order
510
00:30:52,640 --> 00:31:01,400
for the cosecant of x,
sorry, in order for theta
511
00:31:01,400 --> 00:31:02,880
to be the cosecant of x.
512
00:31:02,880 --> 00:31:05,280
This thing is theta.
513
00:31:05,280 --> 00:31:17,810
So, that means that
the cosecant of x--
514
00:31:17,810 --> 00:31:24,830
that means the cosecant
of theta should be x.
515
00:31:24,830 --> 00:31:28,970
Theta is the arccosecant, so
x is the cosecant of theta.
516
00:31:28,970 --> 00:31:31,980
So, what'll I take the sides
to be, to get the cosecant?
517
00:31:31,980 --> 00:31:37,050
The cosecant is 1 over the sine.
518
00:31:37,050 --> 00:31:47,960
And the sine is the opposite
over the hypotenuse.
519
00:31:47,960 --> 00:31:49,910
So I get hypotenuse
over opposite.
520
00:31:49,910 --> 00:31:53,750
And that's supposed
to be what x is.
521
00:31:53,750 --> 00:31:56,540
So I could make the
opposite anything
522
00:31:56,540 --> 00:31:59,190
I want, but the simplest
thing is to make it 1.
523
00:31:59,190 --> 00:32:00,080
Let's do that.
524
00:32:00,080 --> 00:32:04,300
And then what does that mean
about the rest of the sides?
525
00:32:04,300 --> 00:32:06,142
Hypotenuse had better be x.
526
00:32:06,142 --> 00:32:07,350
And then I've recovered this.
527
00:32:07,350 --> 00:32:12,350
So here's a triangle that
exhibits the correct angle.
528
00:32:12,350 --> 00:32:14,680
This remaining side is
going to be useful to us.
529
00:32:14,680 --> 00:32:20,550
And it is the square
root of x^2 - 1.
530
00:32:20,550 --> 00:32:24,060
So I've got a triangle of
the correct angle theta,
531
00:32:24,060 --> 00:32:27,400
and now I want to compute
the tangent of that angle.
532
00:32:27,400 --> 00:32:28,350
Well, that's easy.
533
00:32:28,350 --> 00:32:31,020
That's opposite
divided by adjacent.
534
00:32:31,020 --> 00:32:34,690
So I get 1 over the
square root of x^2 - 1.
535
00:32:38,270 --> 00:32:40,590
Very flexible tool
that'll be useful to you
536
00:32:40,590 --> 00:32:42,200
in many different times.
537
00:32:42,200 --> 00:32:44,170
Whenever you have to
undo a trig substitution,
538
00:32:44,170 --> 00:32:49,100
this is likely to be useful.
539
00:32:49,100 --> 00:32:51,740
OK, that was a good game.
540
00:32:51,740 --> 00:32:52,740
No winners in this game.
541
00:32:52,740 --> 00:32:53,700
We're all winners.
542
00:32:53,700 --> 00:33:01,970
No losers, we're all winners.
543
00:33:01,970 --> 00:33:02,740
OK.
544
00:33:02,740 --> 00:33:06,340
So, good.
545
00:33:06,340 --> 00:33:09,810
So let me make this table of the
different trig substitutions,
546
00:33:09,810 --> 00:33:11,010
and how they can be useful.
547
00:33:11,010 --> 00:33:20,520
Summary of trig substitutions.
548
00:33:20,520 --> 00:33:28,540
So over here, we
have, if you see,
549
00:33:28,540 --> 00:33:58,090
so if your integrand contains,
make a substitution to get.
550
00:33:58,090 --> 00:34:01,086
So if your integrand
contains, I'll
551
00:34:01,086 --> 00:34:02,710
write these things
out as square roots.
552
00:34:02,710 --> 00:34:07,670
If it contains the
square root a^2 - x^2,
553
00:34:07,670 --> 00:34:10,560
this is what we talked
about on Thursday.
554
00:34:10,560 --> 00:34:14,470
When I was trying to find the
area of that piece of a circle.
555
00:34:14,470 --> 00:34:18,130
There, I suggested that we
should make the substitution
556
00:34:18,130 --> 00:34:20,260
x = a cos(theta).
557
00:34:22,890 --> 00:34:28,920
Or, x = a sin(theta).
558
00:34:28,920 --> 00:34:31,400
Either one works just as well.
559
00:34:31,400 --> 00:34:36,430
And there's no way to
prefer one over the other.
560
00:34:36,430 --> 00:34:41,110
And when you make the
substitution, x = a cos(theta),
561
00:34:41,110 --> 00:34:44,030
you get a^2 - a^2 cos^2(theta).
562
00:34:44,030 --> 00:34:45,340
theta.
563
00:34:45,340 --> 00:34:48,280
1 - cos^2 is sin^2.
564
00:34:48,280 --> 00:34:50,620
So you get a sin(theta).
565
00:34:55,270 --> 00:34:59,880
So this expression becomes
equal to this expression
566
00:34:59,880 --> 00:35:03,950
under that substitution.
567
00:35:03,950 --> 00:35:04,800
And then you go on.
568
00:35:04,800 --> 00:35:06,626
Then you've gotten rid
of the square root,
569
00:35:06,626 --> 00:35:08,250
and you've got a
trigonometric integral
570
00:35:08,250 --> 00:35:10,610
that you have to try to do.
571
00:35:10,610 --> 00:35:14,680
If you made the substitution a
sin(theta), you'd get a a^2 -
572
00:35:14,680 --> 00:35:21,580
a^2 sin^2, which
is a cos(theta).
573
00:35:21,580 --> 00:35:24,530
And then you can
go ahead as well.
574
00:35:24,530 --> 00:35:26,430
We just saw another example.
575
00:35:26,430 --> 00:35:29,510
Namely, if you have a^2 + x^2.
576
00:35:29,510 --> 00:35:35,530
That's like the example we had
up here. a = 1 in this example.
577
00:35:35,530 --> 00:35:36,510
What did we do?
578
00:35:36,510 --> 00:35:42,150
We tried the substitution
x = a tan(theta).
579
00:35:42,150 --> 00:35:44,910
And the reason is that I can
plug into the trig identity
580
00:35:44,910 --> 00:35:46,950
up here in the upper left.
581
00:35:46,950 --> 00:35:51,210
And replace a^2 +
x^2 by a sec(theta).
582
00:35:55,020 --> 00:35:59,790
Square root of the
secant squared.
583
00:35:59,790 --> 00:36:03,410
There's one more
thing in this table.
584
00:36:03,410 --> 00:36:06,520
Sort of, the only
remaining sum or difference
585
00:36:06,520 --> 00:36:07,544
of terms like this.
586
00:36:07,544 --> 00:36:09,460
And that's what happens
if you have x^2 - a^2.
587
00:36:14,050 --> 00:36:17,790
So there, I think we can make
a substitution a sec(theta).
588
00:36:21,300 --> 00:36:27,031
Because, after all,
sec^2(theta)-- so x^2 - a^2--
589
00:36:27,031 --> 00:36:27,530
Sorry.
590
00:36:32,310 --> 00:36:34,790
Let's see what happens when
I make that substitution.
591
00:36:34,790 --> 00:36:42,850
x^2 - a^2 = a^2
sec^2(theta) - a^2.
592
00:36:42,850 --> 00:36:45,460
Under this substitution.
593
00:36:45,460 --> 00:36:48,690
That's sec^2 - 1.
594
00:36:48,690 --> 00:36:51,390
Well, put the 1
on the other side.
595
00:36:51,390 --> 00:36:53,720
And you find tan^2, coming out.
596
00:36:53,720 --> 00:36:56,710
So this is a a^2 tan^2(theta).
597
00:36:59,440 --> 00:37:02,680
And so that's what you
get, a times tan(theta).
598
00:37:02,680 --> 00:37:07,850
After I take the square
root, I get a tan(theta).
599
00:37:07,850 --> 00:37:13,080
So these are the three basic
trig substitution forms.
600
00:37:13,080 --> 00:37:15,580
Where trig substitutions
are useful to get
601
00:37:15,580 --> 00:37:17,920
rid of expressions like
this, and replace them
602
00:37:17,920 --> 00:37:22,670
by trigonometric expressions.
603
00:37:22,670 --> 00:37:25,140
And then you use this trick,
you do the integral if you can
604
00:37:25,140 --> 00:37:27,265
and then you use this trick
to get rid of the theta
605
00:37:27,265 --> 00:37:37,460
at the end.
606
00:37:37,460 --> 00:37:40,630
So now, the last thing I
want to talk about today
607
00:37:40,630 --> 00:37:59,180
is called completing the square.
608
00:37:59,180 --> 00:38:03,440
And that comes in
because unfortunately,
609
00:38:03,440 --> 00:38:08,350
not every square
root of a quadratic
610
00:38:08,350 --> 00:38:10,390
has such a simple form.
611
00:38:10,390 --> 00:38:16,200
You will often encounter
things that are not just
612
00:38:16,200 --> 00:38:18,080
the square root of
something simple.
613
00:38:18,080 --> 00:38:19,140
Like one of these forms.
614
00:38:19,140 --> 00:38:27,537
Like there might be a
middle term in there.
615
00:38:27,537 --> 00:38:29,120
I don't actually
have time to show you
616
00:38:29,120 --> 00:38:33,200
an example of how this comes out
in a sort of practical example.
617
00:38:33,200 --> 00:38:35,760
But it does happen
quite frequently.
618
00:38:35,760 --> 00:38:38,580
And so I want to show you
how to deal with things
619
00:38:38,580 --> 00:38:40,550
like the following example.
620
00:38:40,550 --> 00:38:47,380
Let's try to integrate
dx over x^2 + 4x,
621
00:38:47,380 --> 00:38:55,090
the square root of x^2 + 4x.
622
00:38:55,090 --> 00:38:59,010
So there's a square root of
some square, some quadratic.
623
00:38:59,010 --> 00:39:02,010
It's very much
like this business.
624
00:39:02,010 --> 00:39:04,830
But it isn't of
any of these forms.
625
00:39:04,830 --> 00:39:06,360
And so what I want
to do is show you
626
00:39:06,360 --> 00:39:10,540
how to rewrite it in one of
those forms using substitution,
627
00:39:10,540 --> 00:39:11,120
again.
628
00:39:11,120 --> 00:39:14,520
All this is about substitution.
629
00:39:14,520 --> 00:39:26,100
So the game is to rewrite
quadratic as something
630
00:39:26,100 --> 00:39:30,950
like x plus something or other.
631
00:39:30,950 --> 00:39:32,470
Plus some other constant.
632
00:39:32,470 --> 00:39:35,690
So write it, try to write
it, in the form of a square
633
00:39:35,690 --> 00:39:42,340
plus or minus another constant.
634
00:39:42,340 --> 00:39:45,170
And then we'll go on from there.
635
00:39:45,170 --> 00:39:51,110
So let's do that in this
case. x^2, x^2 + 4x.
636
00:39:51,110 --> 00:39:53,390
Well, if you square
this form out,
637
00:39:53,390 --> 00:39:57,280
then the middle term
is going to be 2ax.
638
00:39:57,280 --> 00:40:00,920
So that, since I have
a middle term here,
639
00:40:00,920 --> 00:40:03,640
I pretty much know
what a has to be.
640
00:40:03,640 --> 00:40:08,220
The only choice in order to
get something like x^2 + 4x out
641
00:40:08,220 --> 00:40:11,440
of this, is to take a to be 2.
642
00:40:11,440 --> 00:40:16,950
Because then, this
is what you get.
643
00:40:16,950 --> 00:40:19,790
This isn't quite right yet, but
let's compute what I have here.
644
00:40:19,790 --> 00:40:24,490
x^2 + 4x, so far
so good, plus 4,
645
00:40:24,490 --> 00:40:26,190
and I don't have a plus 4 here.
646
00:40:26,190 --> 00:40:31,394
So I have to fix that
by subtracting 4.
647
00:40:31,394 --> 00:40:32,310
So that's what I mean.
648
00:40:32,310 --> 00:40:33,393
I've completed the square.
649
00:40:33,393 --> 00:40:38,370
The word for this process of
eliminating the middle term
650
00:40:38,370 --> 00:40:41,850
by using the square of
an expression like that.
651
00:40:41,850 --> 00:40:44,830
That's called
completing the square.
652
00:40:44,830 --> 00:40:48,010
And we can use that process
to compute this integral.
653
00:40:48,010 --> 00:40:51,040
So let's do that.
654
00:40:51,040 --> 00:40:54,170
So I can rewrite this integral,
rewrite this denominator
655
00:40:54,170 --> 00:41:00,100
like this.
656
00:41:00,100 --> 00:41:03,860
And then I'm going to try to use
one of these forms over here.
657
00:41:03,860 --> 00:41:07,900
So in order to get a
single variable there,
658
00:41:07,900 --> 00:41:12,500
instead of something
complicated like x + 2,
659
00:41:12,500 --> 00:41:15,090
I'm inclined to come up
with another variable name
660
00:41:15,090 --> 00:41:20,330
and write it equal, write x +
2 as that other variable name.
661
00:41:20,330 --> 00:41:29,950
So here's another little
direct substitution. u = x + 2.
662
00:41:29,950 --> 00:41:31,260
Figure out what du is.
663
00:41:31,260 --> 00:41:36,610
That's pretty easy.
664
00:41:36,610 --> 00:41:41,580
And then rewrite the
integral in those terms.
665
00:41:41,580 --> 00:41:44,880
So dx = du.
666
00:41:44,880 --> 00:41:47,270
And then in the
denominator I have,
667
00:41:47,270 --> 00:41:49,100
well, I have the
square root of that.
668
00:41:49,100 --> 00:41:55,100
Oh yeah, so I think as part of
this I'll write out what x^2 +
669
00:41:55,100 --> 00:41:56,550
4x is.
670
00:41:56,550 --> 00:41:59,520
The point is, it's
equal to u^2 - 4.
671
00:42:04,170 --> 00:42:06,540
x^2 + 4x = u^2 - 4.
672
00:42:12,950 --> 00:42:20,260
There's the data box containing
the substitution data.
673
00:42:20,260 --> 00:42:22,640
And so now I can put that in.
674
00:42:22,640 --> 00:42:24,960
I have x^2 + 4x there.
675
00:42:24,960 --> 00:42:31,330
In terms of u, that's u^2 - 4.
676
00:42:31,330 --> 00:42:33,690
Well, now I'm in a happier
position because I can look
677
00:42:33,690 --> 00:42:37,060
for u^2 - 4 for something
like that in my table here.
678
00:42:37,060 --> 00:42:40,430
And it actually sits down here.
679
00:42:40,430 --> 00:42:43,365
So except for the
use of the letter x
680
00:42:43,365 --> 00:42:45,470
here instead of u over there.
681
00:42:45,470 --> 00:42:47,510
That tells me what I want.
682
00:42:47,510 --> 00:42:55,100
So to handle this, what I should
use is a trig substitution.
683
00:42:55,100 --> 00:42:58,020
And the trig substitution
that's suggested is,
684
00:42:58,020 --> 00:43:03,870
according to the bottom
line with a = 2, so a^2 = 4.
685
00:43:03,870 --> 00:43:07,600
The suggestion is,
I should take x--
686
00:43:07,600 --> 00:43:13,909
But I'd better not use
the letter x any more.
687
00:43:13,909 --> 00:43:15,950
But I don't have a letter
x, I have the letter u.
688
00:43:15,950 --> 00:43:21,310
I should take u
equal to 2 secant.
689
00:43:21,310 --> 00:43:23,110
And then some letter
I haven't used before.
690
00:43:23,110 --> 00:43:28,820
And theta is available.
691
00:43:28,820 --> 00:43:30,560
This is a look-up table process.
692
00:43:30,560 --> 00:43:33,240
I see the square
root of u^2 - 4,
693
00:43:33,240 --> 00:43:35,250
I see that that's of this form.
694
00:43:35,250 --> 00:43:38,230
I'm instructed to make
this substitution.
695
00:43:38,230 --> 00:43:40,540
And that's what I just did.
696
00:43:40,540 --> 00:43:43,600
Let's see how it works out.
697
00:43:43,600 --> 00:43:47,690
So that means the du is 2, OK.
698
00:43:47,690 --> 00:43:50,480
What's the derivative
of the secant?
699
00:43:50,480 --> 00:43:52,660
Secant tangent.
700
00:43:52,660 --> 00:43:59,050
So du = 2 sec(theta) tan(theta).
701
00:43:59,050 --> 00:44:05,690
And u^2 - 4 is,
here's the payoff.
702
00:44:05,690 --> 00:44:07,170
I'm supposed to
be able to rewrite
703
00:44:07,170 --> 00:44:09,350
that in terms of the tangent.
704
00:44:09,350 --> 00:44:19,650
According to this. u u^2 - 4
is 4 secant squared minus 4.
705
00:44:19,650 --> 00:44:22,450
And secant squared minus
1 is tangent squared.
706
00:44:22,450 --> 00:44:24,400
So this is 4 tan^2(theta).
707
00:44:30,890 --> 00:44:36,807
Right, yeah?
708
00:44:36,807 --> 00:44:37,640
STUDENT: [INAUDIBLE]
709
00:44:37,640 --> 00:44:42,200
PROFESSOR: But I squared it.
710
00:44:42,200 --> 00:44:44,180
And now I'll square root it.
711
00:44:44,180 --> 00:44:50,650
And I'll get a 2 and this
tangent will go away.
712
00:44:50,650 --> 00:44:56,130
So there's my data box
for this substitution.
713
00:44:56,130 --> 00:45:15,560
And let's go on
to another board.
714
00:45:15,560 --> 00:45:20,980
So where I'm at is the integral
of du over the square root
715
00:45:20,980 --> 00:45:22,010
of u^2 - 4.
716
00:45:26,850 --> 00:45:30,060
And I have all the
data I need here
717
00:45:30,060 --> 00:45:33,090
to rewrite that
in terms of theta.
718
00:45:33,090 --> 00:45:38,170
So du = 2 sec(theta)
tan(theta) d theta.
719
00:45:41,720 --> 00:45:44,140
And the denominator
is 2 tan(theta).
720
00:45:47,760 --> 00:45:48,590
Ha.
721
00:45:48,590 --> 00:45:53,040
Well, so some very nice
simplification happens here.
722
00:45:53,040 --> 00:45:55,940
The 2's cancel.
723
00:45:55,940 --> 00:45:58,260
And the tangents cancel.
724
00:45:58,260 --> 00:46:01,900
And I'm left with trying to work
with the integral, sec(theta) d
725
00:46:01,900 --> 00:46:03,920
theta.
726
00:46:03,920 --> 00:46:06,610
And luckily enough at the
very beginning of the hour,
727
00:46:06,610 --> 00:46:08,480
I worked out how to
compute the integral
728
00:46:08,480 --> 00:46:09,885
of the secant of theta.
729
00:46:09,885 --> 00:46:11,670
And there it is.
730
00:46:11,670 --> 00:46:25,000
So this is ln(sec(theta) +
tan(theta)) plus a constant.
731
00:46:25,000 --> 00:46:27,660
And we're done with
the calculus part.
732
00:46:27,660 --> 00:46:29,390
There's no more integral there.
733
00:46:29,390 --> 00:46:31,490
But I still am not quite
done with the problem,
734
00:46:31,490 --> 00:46:37,180
because again I have these two
substitutions to try to undo.
735
00:46:37,180 --> 00:46:40,220
So let's undo them one by one.
736
00:46:40,220 --> 00:46:42,240
Let's see.
737
00:46:42,240 --> 00:46:44,130
I have this trig
substitution here.
738
00:46:44,130 --> 00:46:47,210
And I could use my triangle
trick, if I need to.
739
00:46:47,210 --> 00:46:49,280
But maybe I don't need to.
740
00:46:49,280 --> 00:46:51,220
Let's see, do I know
what the secant of theta
741
00:46:51,220 --> 00:46:53,090
is in terms of u?
742
00:46:53,090 --> 00:46:54,620
Well, I do.
743
00:46:54,620 --> 00:46:55,730
So I get ln(u/2).
744
00:46:58,980 --> 00:47:01,740
Do I know what the
tangent is in terms of u?
745
00:47:01,740 --> 00:47:02,730
Well, I do.
746
00:47:02,730 --> 00:47:04,180
It's here.
747
00:47:04,180 --> 00:47:06,500
So I lucked out, in this case.
748
00:47:06,500 --> 00:47:09,930
And I don't have to go through
and use that triangle trick.
749
00:47:09,930 --> 00:47:15,700
So the tangent of theta is the
square root of u^2 - 4, over 2.
750
00:47:19,000 --> 00:47:20,080
Good.
751
00:47:20,080 --> 00:47:24,220
So I've undone this
trig substitution.
752
00:47:24,220 --> 00:47:28,440
I'm not quite done yet because
my answer is involved with u.
753
00:47:28,440 --> 00:47:30,780
And what I wanted
originally was x.
754
00:47:30,780 --> 00:47:33,780
But this direct substitution
that I started with
755
00:47:33,780 --> 00:47:35,250
is really easy to deal with.
756
00:47:35,250 --> 00:47:39,780
I can just put x + 2
every time I see a u.
757
00:47:39,780 --> 00:47:45,590
So this is the natural logarithm
of (x+2)/2 plus the square
758
00:47:45,590 --> 00:47:46,090
root...
759
00:47:46,090 --> 00:47:51,330
What's going to happen when I
put x + 2 in place for u here?
760
00:47:51,330 --> 00:47:54,490
You know what you get.
761
00:47:54,490 --> 00:47:59,230
You get exactly what
we started with.
762
00:47:59,230 --> 00:48:00,260
Right?
763
00:48:00,260 --> 00:48:04,850
I put x + 2 in
place of the u here.
764
00:48:04,850 --> 00:48:13,210
I get x^2 + 4x.
765
00:48:13,210 --> 00:48:15,330
So I've gotten back
to a function purely
766
00:48:15,330 --> 00:48:21,200
in terms of x OK, that's
a good place to quit.
767
00:48:21,200 --> 00:48:24,840
Have a great little
one-day break.
768
00:48:24,840 --> 00:48:27,320
I guess this class doesn't
meet on Monday anyway.
769
00:48:27,320 --> 00:48:28,477
Bye.