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