1
00:00:06,910 --> 00:00:08,380
Welcome back to recitation.

2
00:00:08,380 --> 00:00:10,460
In this video, I'd
like us to find

3
00:00:10,460 --> 00:00:12,270
an antiderivative
of the function 1

4
00:00:12,270 --> 00:00:15,884
over x squared minus 8x plus 1.

5
00:00:15,884 --> 00:00:18,550
So I'll give you a while to work
on it, and then I'll come back,

6
00:00:18,550 --> 00:00:19,883
and I'll show you how I started.

7
00:00:27,710 --> 00:00:28,890
So welcome back.

8
00:00:28,890 --> 00:00:32,570
Well, what we'd like to do is,
find an antiderivative to 1

9
00:00:32,570 --> 00:00:34,814
over x squared minus 8x plus 1.

10
00:00:34,814 --> 00:00:36,230
And how we're going
to do that, is

11
00:00:36,230 --> 00:00:38,700
we're going to use the
technique completing the square.

12
00:00:38,700 --> 00:00:40,200
And I'm going to
set up the problem,

13
00:00:40,200 --> 00:00:41,350
I'm going to get it
to a certain place,

14
00:00:41,350 --> 00:00:43,017
and then I'm going
to let you finish it.

15
00:00:43,017 --> 00:00:45,016
And how do you know if
you got the right answer?

16
00:00:45,016 --> 00:00:47,200
Well, you actually take a
derivative of your answer,

17
00:00:47,200 --> 00:00:51,180
and see if it gives you back 1
over x squared minus 8x plus 1.

18
00:00:51,180 --> 00:00:52,430
That's how you can check.

19
00:00:52,430 --> 00:00:53,420
So let's start off.

20
00:00:53,420 --> 00:00:55,080
If I want to
complete the square,

21
00:00:55,080 --> 00:01:01,230
let's just remind
ourselves how to complete

22
00:01:01,230 --> 00:01:04,660
the square on this quadratic.

23
00:01:08,480 --> 00:01:13,490
So I'd like something
right here that

24
00:01:13,490 --> 00:01:15,251
makes this a perfect square.

25
00:01:15,251 --> 00:01:15,750
Right?

26
00:01:15,750 --> 00:01:18,570
Well, the 8 here in the middle,
if I want a perfect square,

27
00:01:18,570 --> 00:01:20,020
if you think about
this, I'm going

28
00:01:20,020 --> 00:01:22,020
to have an x minus--
I need a number here

29
00:01:22,020 --> 00:01:24,845
that when I multiply it by 2,
that's where this 8 comes from,

30
00:01:24,845 --> 00:01:26,210
it gives me 8.

31
00:01:26,210 --> 00:01:28,871
So obviously I need
this number to be a 4.

32
00:01:28,871 --> 00:01:29,370
Right?

33
00:01:29,370 --> 00:01:31,460
Which puts what here?

34
00:01:31,460 --> 00:01:34,450
Puts a 16 here, right?

35
00:01:34,450 --> 00:01:37,750
So just to double check, what
I'm looking for is a number

36
00:01:37,750 --> 00:01:41,110
here-- I need a number right
here that when I multiply by 2,

37
00:01:41,110 --> 00:01:43,740
gives me negative 8, and
then I need to figure out

38
00:01:43,740 --> 00:01:45,090
what it squares to.

39
00:01:45,090 --> 00:01:47,630
So that number is negative
4, and it squares to 16.

40
00:01:47,630 --> 00:01:49,660
But obviously this isn't
what I have, right?

41
00:01:49,660 --> 00:01:51,020
I have plus 1.

42
00:01:51,020 --> 00:01:54,090
So what have I had to do
to get from here to here?

43
00:01:54,090 --> 00:01:56,710
Well, I had plus 1 and
now I have plus 16,

44
00:01:56,710 --> 00:01:58,460
so obviously I've added 15.

45
00:01:58,460 --> 00:02:01,874
So I have to subtract
15 to keep this,

46
00:02:01,874 --> 00:02:05,560
to keep these three lines
all equal to each other.

47
00:02:05,560 --> 00:02:07,840
So to understand
where that comes from,

48
00:02:07,840 --> 00:02:10,090
let me just remind
you, my denominator

49
00:02:10,090 --> 00:02:11,040
was looking like this.

50
00:02:11,040 --> 00:02:13,540
I'd like it to have
a perfect square,

51
00:02:13,540 --> 00:02:18,170
and then subtract a
constant, or add a constant.

52
00:02:18,170 --> 00:02:20,131
Right now I have
something that-- I don't

53
00:02:20,131 --> 00:02:21,380
have a perfect square in here.

54
00:02:21,380 --> 00:02:23,330
I can't make this
into a perfect square

55
00:02:23,330 --> 00:02:26,480
unless I add a certain amount
to the constant right here.

56
00:02:26,480 --> 00:02:28,840
So I had to add a 15
to the constant here.

57
00:02:28,840 --> 00:02:30,560
Notice 16 minus 15 is 1.

58
00:02:30,560 --> 00:02:33,500
That's my check, also,
that the lines are equal.

59
00:02:33,500 --> 00:02:37,430
And so what I've done, is I've
added 15 and subtracted 15,

60
00:02:37,430 --> 00:02:40,870
and then I put this
plus 16 into here. x

61
00:02:40,870 --> 00:02:43,750
minus 4 quantity squared is
exactly these first three

62
00:02:43,750 --> 00:02:45,050
terms.

63
00:02:45,050 --> 00:02:46,950
And then I keep the minus 15.

64
00:02:46,950 --> 00:02:50,450
Now, you might say,
why did you do this?

65
00:02:50,450 --> 00:02:52,870
So let's make sure we
understand why we're

66
00:02:52,870 --> 00:02:54,345
completing the square on this.

67
00:02:54,345 --> 00:02:58,442
If we come back, I'm going
to put this line in place

68
00:02:58,442 --> 00:02:59,900
of what's in the
denominator there,

69
00:02:59,900 --> 00:03:02,090
because these three
things are all equal.

70
00:03:02,090 --> 00:03:05,780
So this is actually
the integral of dx

71
00:03:05,780 --> 00:03:12,970
over x minus 4 quantity
squared minus 15.

72
00:03:12,970 --> 00:03:15,230
Now you might say, Christine,
this looks no easier.

73
00:03:15,230 --> 00:03:17,250
I don't know why you did this.

74
00:03:17,250 --> 00:03:21,100
But it actually is one of our
favorite, or least favorite,

75
00:03:21,100 --> 00:03:25,230
depending on how you feel about
it, types of tricks we use now,

76
00:03:25,230 --> 00:03:26,980
which is the trig substitution.

77
00:03:26,980 --> 00:03:28,730
So some people love
this because they just

78
00:03:28,730 --> 00:03:30,090
have to memorize
a little formula,

79
00:03:30,090 --> 00:03:32,423
and some people love it because
they can draw a triangle

80
00:03:32,423 --> 00:03:34,780
and understand what they choose.

81
00:03:34,780 --> 00:03:37,140
I'm going to show you, remind
you what the formula was

82
00:03:37,140 --> 00:03:39,310
you saw in class.

83
00:03:39,310 --> 00:03:42,930
I believe Professor Jerison
said something like this.

84
00:03:42,930 --> 00:03:47,900
If the denominator is in the
form u squared minus a squared,

85
00:03:47,900 --> 00:03:54,570
this implies that you make
u equal to a secant theta.

86
00:03:54,570 --> 00:03:58,080
Now, he probably wrote it
as x, but I wrote it as u

87
00:03:58,080 --> 00:04:00,300
for a very specific reason.

88
00:04:00,300 --> 00:04:03,690
Because here I have x minus 4.

89
00:04:03,690 --> 00:04:06,150
I have x minus 4
quantity squared.

90
00:04:06,150 --> 00:04:08,400
Now, this is where it gets
a little rough, right?

91
00:04:08,400 --> 00:04:11,560
This is not a perfect
square, but it

92
00:04:11,560 --> 00:04:14,475
is the perfect square--
it is the square

93
00:04:14,475 --> 00:04:16,800
of the square root of 15.

94
00:04:16,800 --> 00:04:20,810
So I can write the denominator
in the form, something squared

95
00:04:20,810 --> 00:04:23,069
minus something else squared.

96
00:04:23,069 --> 00:04:24,860
And again, you might
say, why is this good?

97
00:04:24,860 --> 00:04:26,600
Well, what we're going
to be able to do,

98
00:04:26,600 --> 00:04:28,880
is we're going to be able
to rewrite this in terms

99
00:04:28,880 --> 00:04:30,540
of trigonometric
functions, which will

100
00:04:30,540 --> 00:04:32,660
make it much simpler to solve.

101
00:04:32,660 --> 00:04:35,870
So let's use what
Professor Jerison gave us.

102
00:04:35,870 --> 00:04:41,030
And so what we see, is that
this is u and this is a.

103
00:04:41,030 --> 00:04:41,530
Right?

104
00:04:41,530 --> 00:04:50,782
So I get x minus 4 is equal to
square root of 15 secant theta.

105
00:04:50,782 --> 00:04:52,740
Now, you might not like
this square root of 15,

106
00:04:52,740 --> 00:04:53,750
but it's just hanging out.

107
00:04:53,750 --> 00:04:54,970
It's not causing any problems.

108
00:04:54,970 --> 00:04:56,090
It's just a number
there, so we'll

109
00:04:56,090 --> 00:04:57,390
keep it a square root of 15.

110
00:04:57,390 --> 00:04:59,426
So you don't have
to worry about it.

111
00:04:59,426 --> 00:05:00,550
Now what's the point again?

112
00:05:00,550 --> 00:05:01,966
Let me just remind
you, the object

113
00:05:01,966 --> 00:05:05,250
is to get this in terms
of the trig functions.

114
00:05:05,250 --> 00:05:07,850
So we should anticipate
that probably we'll

115
00:05:07,850 --> 00:05:11,090
have some tangent
functions to go with this.

116
00:05:11,090 --> 00:05:13,520
And there are two
reasons to think that.

117
00:05:13,520 --> 00:05:15,590
The first reason to think
that is at some point,

118
00:05:15,590 --> 00:05:17,550
I have to find dx.

119
00:05:17,550 --> 00:05:20,891
Well, the derivative of secant
involves secant and tangent,

120
00:05:20,891 --> 00:05:21,390
right?

121
00:05:21,390 --> 00:05:23,910
So that's going to pull in a
tangent function somewhere.

122
00:05:23,910 --> 00:05:25,951
I'm also going to have a
tangent function show up

123
00:05:25,951 --> 00:05:26,725
somewhere else.

124
00:05:26,725 --> 00:05:28,350
And where that's
going to be, is coming

125
00:05:28,350 --> 00:05:31,680
from this denominator, this
expression in the denominator.

126
00:05:31,680 --> 00:05:34,240
Because there's a certain
trig identity that we should

127
00:05:34,240 --> 00:05:36,980
have memorized, but
I'll just remind you.

128
00:05:36,980 --> 00:05:39,490
I'll write it here and
put a star next to it.

129
00:05:39,490 --> 00:05:43,045
It's 1 plus tangent
squared theta is

130
00:05:43,045 --> 00:05:46,096
equal to secant squared theta.

131
00:05:46,096 --> 00:05:48,345
So this is a-- I'll even put
a star on the other side.

132
00:05:48,345 --> 00:05:49,870
So we should really
remember this.

133
00:05:49,870 --> 00:05:51,115
Now, where does it come from?

134
00:05:51,115 --> 00:05:54,120
It comes from the cosine squared
theta plus sine squared theta

135
00:05:54,120 --> 00:05:56,010
equals 1 identity.

136
00:05:56,010 --> 00:05:58,850
You can divide everything
by cosine squared theta

137
00:05:58,850 --> 00:06:00,640
and get this one.

138
00:06:00,640 --> 00:06:03,630
So we have this identity,
and so if you notice,

139
00:06:03,630 --> 00:06:08,200
we're going to be able to
manipulate the expression right

140
00:06:08,200 --> 00:06:10,706
here, and get the
denominator to look

141
00:06:10,706 --> 00:06:11,830
like tangent squared theta.

142
00:06:11,830 --> 00:06:16,494
So let's do some of that
work off to the right here.

143
00:06:16,494 --> 00:06:17,660
So what did I say we needed?

144
00:06:17,660 --> 00:06:18,660
We have this expression.

145
00:06:18,660 --> 00:06:21,310
We need dx, so let's
find-- actually, no.

146
00:06:21,310 --> 00:06:22,770
Let's find the
denominator first,

147
00:06:22,770 --> 00:06:24,940
because I was just
talking about it.

148
00:06:24,940 --> 00:06:27,640
So if I look at what
x minus 4 squared is,

149
00:06:27,640 --> 00:06:30,190
I'm going to substitute
in this expression.

150
00:06:30,190 --> 00:06:35,850
So x minus 4 squared
minus 15 is the same

151
00:06:35,850 --> 00:06:42,670
as, based on this substitution,
square root 15 secant theta

152
00:06:42,670 --> 00:06:48,020
squared minus 15, which
is 15 secant squared

153
00:06:48,020 --> 00:06:53,970
theta minus 15, which, just
to hammer home the point,

154
00:06:53,970 --> 00:06:59,530
is 15 times the quantity
secant squared theta minus 1.

155
00:06:59,530 --> 00:07:00,030
OK?

156
00:07:00,030 --> 00:07:01,480
Everybody follows, hopefully.

157
00:07:01,480 --> 00:07:05,110
All I've done is the
substitution I made,

158
00:07:05,110 --> 00:07:08,880
and then I started expanding,
or I squared this term,

159
00:07:08,880 --> 00:07:11,050
and I factored out the 15.

160
00:07:11,050 --> 00:07:13,460
And now let's go back
to my start expression.

161
00:07:13,460 --> 00:07:17,290
What is secant
squared theta minus 1?

162
00:07:17,290 --> 00:07:19,170
It's tangent squared theta.

163
00:07:19,170 --> 00:07:24,200
So we get 15 tangent
squared theta.

164
00:07:24,200 --> 00:07:28,940
So that is actually what the
denominator of our integral

165
00:07:28,940 --> 00:07:30,480
is going to be over there.

166
00:07:30,480 --> 00:07:34,610
So I'm going to come in and
put that part-- actually,

167
00:07:34,610 --> 00:07:36,990
let me even put this here, too.

168
00:07:36,990 --> 00:07:46,170
So right now, our denominator
is 15 tangent squared theta.

169
00:07:46,170 --> 00:07:47,920
So far, so good.

170
00:07:47,920 --> 00:07:50,610
But of course, if I put a
dx up here, I'm in trouble.

171
00:07:50,610 --> 00:07:53,270
Because I have, it's a
function of theta now.

172
00:07:53,270 --> 00:07:56,470
So I need to write this-- I
shouldn't write in terms of x.

173
00:07:56,470 --> 00:07:58,870
I need to figure out what
it is in terms of theta.

174
00:07:58,870 --> 00:08:03,310
And to do that, we again use
the substitution that we made.

175
00:08:03,310 --> 00:08:05,620
Which is just above
the starred expression.

176
00:08:05,620 --> 00:08:10,650
It was that x minus 4 equals
square root 15 secant theta.

177
00:08:10,650 --> 00:08:12,670
This is going to
allow us to find

178
00:08:12,670 --> 00:08:15,950
what d theta is in terms of dx.

179
00:08:15,950 --> 00:08:16,750
OK?

180
00:08:16,750 --> 00:08:19,506
So let's do that.

181
00:08:19,506 --> 00:08:21,130
So I'm not done, by
the way, over here.

182
00:08:21,130 --> 00:08:23,160
I'm not done I've
got a little gap I've

183
00:08:23,160 --> 00:08:25,270
got to fill in the numerator.

184
00:08:25,270 --> 00:08:27,480
So let's come back over here.

185
00:08:27,480 --> 00:08:30,549
So now we have x
minus 4-- let me just

186
00:08:30,549 --> 00:08:31,590
write that one more time.

187
00:08:36,500 --> 00:08:39,979
So we get dx is equal to
the square root of 15.

188
00:08:39,979 --> 00:08:42,630
Well, what's the
derivative of secant theta?

189
00:08:42,630 --> 00:08:49,120
It's secant theta
tangent theta d theta.

190
00:08:49,120 --> 00:08:51,480
So now I have all
the pieces I need.

191
00:08:51,480 --> 00:08:52,870
And I'm actually
going to rewrite

192
00:08:52,870 --> 00:08:55,640
the whole thing over
here underneath,

193
00:08:55,640 --> 00:08:57,990
so that I can work with
it a little bit more.

194
00:08:57,990 --> 00:09:01,000
So the dx is in the numerator.

195
00:09:01,000 --> 00:09:11,720
Square root 15 secant theta tan
theta d theta, all over 15 tan

196
00:09:11,720 --> 00:09:13,550
squared theta.

197
00:09:13,550 --> 00:09:15,300
Now, this might still
look a little messy,

198
00:09:15,300 --> 00:09:18,590
but we can simplify
it some more.

199
00:09:18,590 --> 00:09:22,920
We divide out by one tangent,
we'll pull this out in front.

200
00:09:22,920 --> 00:09:24,620
And notice, what's secant?

201
00:09:24,620 --> 00:09:28,060
Secant theta is 1
over cosine theta,

202
00:09:28,060 --> 00:09:30,620
and tangent theta is sine
theta over cosine theta.

203
00:09:30,620 --> 00:09:32,460
So let's write that down.

204
00:09:32,460 --> 00:09:36,510
So this becomes square
root 15 over 15.

205
00:09:36,510 --> 00:09:38,790
We'll just leave it out there.

206
00:09:38,790 --> 00:09:41,130
It's not hurting anyone.

207
00:09:41,130 --> 00:09:45,630
So we get a 1 over cosine theta
times-- well, tangent theta,

208
00:09:45,630 --> 00:09:48,090
1 over tangent theta is
cotangent theta also.

209
00:09:48,090 --> 00:09:49,680
There's another way
to think about it.

210
00:09:49,680 --> 00:09:56,820
So it's cosine theta
over sine theta d theta.

211
00:09:56,820 --> 00:09:58,200
So these divide out.

212
00:09:58,200 --> 00:10:01,170
And I'm left with, I'm taking
now an antiderivative of 1

213
00:10:01,170 --> 00:10:05,000
over sine theta, which
is cosecant theta.

214
00:10:05,000 --> 00:10:06,550
So I have to find
an antiderivative

215
00:10:06,550 --> 00:10:07,980
of cosecant theta.

216
00:10:07,980 --> 00:10:10,620
Well, you can find that
with the exact same strategy

217
00:10:10,620 --> 00:10:14,140
you found, or I should say,
that Professor Jerison used

218
00:10:14,140 --> 00:10:16,080
in class-- or maybe
it was actually

219
00:10:16,080 --> 00:10:18,130
Professor Miller
in that lecture--

220
00:10:18,130 --> 00:10:21,840
to find an antiderivative
of secant theta.

221
00:10:21,840 --> 00:10:24,250
So you can do the same kind
of thing with cosecant theta,

222
00:10:24,250 --> 00:10:26,208
because they have the
same kind of derivatives.

223
00:10:26,208 --> 00:10:29,610
Cosecant and cotangent have
very similar-looking derivatives

224
00:10:29,610 --> 00:10:31,090
to tangent and secant.

225
00:10:31,090 --> 00:10:33,090
Same kinds of relationships.

226
00:10:33,090 --> 00:10:35,190
So you can actually find
that antiderivative.

227
00:10:35,190 --> 00:10:37,190
So this is some constant
we don't care about.

228
00:10:43,150 --> 00:10:46,810
And once you find that, this
will be in terms of theta.

229
00:10:46,810 --> 00:10:49,400
Your final answer needs
to be in terms of x,

230
00:10:49,400 --> 00:10:51,760
but you saw how to
do that, actually.

231
00:10:51,760 --> 00:10:54,040
You just need to
make a triangle that

232
00:10:54,040 --> 00:10:58,690
represents the relationship
between x and theta.

233
00:10:58,690 --> 00:11:00,832
So I'll draw a picture
of that triangle,

234
00:11:00,832 --> 00:11:02,790
then I'll give a little
summary of what we did,

235
00:11:02,790 --> 00:11:03,975
and then we'll stop.

236
00:11:03,975 --> 00:11:06,087
So let me draw a picture
of that triangle.

237
00:11:06,087 --> 00:11:08,045
So from here, all you
would do is actually find

238
00:11:08,045 --> 00:11:09,959
this antiderivative,
and then you

239
00:11:09,959 --> 00:11:12,000
would have to make the
right kind of substitution

240
00:11:12,000 --> 00:11:12,850
in terms of theta.

241
00:11:12,850 --> 00:11:16,330
We want to know, how do we find
that, do that substitution.

242
00:11:16,330 --> 00:11:19,690
So the triangle is going to
come from the following thing.

243
00:11:19,690 --> 00:11:26,630
We know x minus 4, again, is
square root of 15 secant theta.

244
00:11:26,630 --> 00:11:28,840
So I'm going to make this theta.

245
00:11:28,840 --> 00:11:31,700
Secant theta, well, it's
1 over cosine theta,

246
00:11:31,700 --> 00:11:33,930
cosine is adjacent
over hypotenuse,

247
00:11:33,930 --> 00:11:37,430
so secant is hypotenuse
over adjacent.

248
00:11:37,430 --> 00:11:38,010
Right?

249
00:11:38,010 --> 00:11:39,700
That's the relationship.

250
00:11:39,700 --> 00:11:43,740
So x minus 4 over 15 is
equal to the hypotenuse

251
00:11:43,740 --> 00:11:46,030
over the adjacent.

252
00:11:46,030 --> 00:11:47,130
Did I square root?

253
00:11:47,130 --> 00:11:48,230
Sorry.

254
00:11:48,230 --> 00:11:49,450
Square root.

255
00:11:49,450 --> 00:11:53,510
So the hypotenuse is x
minus 4, the adjacent

256
00:11:53,510 --> 00:11:57,050
is square root of 15, and then
now I can fill in the opposite

257
00:11:57,050 --> 00:11:58,270
by Pythagorean theorem.

258
00:11:58,270 --> 00:11:58,770
Right?

259
00:11:58,770 --> 00:12:02,830
I just take this squared,
and I subtract this squared,

260
00:12:02,830 --> 00:12:04,440
and then I take the square root.

261
00:12:04,440 --> 00:12:11,870
So I get the square root of
x minus 4 squared minus 15.

262
00:12:11,870 --> 00:12:13,660
So whatever I have
in terms of theta,

263
00:12:13,660 --> 00:12:15,460
I just look at this triangle.

264
00:12:15,460 --> 00:12:18,910
If I had in my
answer sine theta,

265
00:12:18,910 --> 00:12:23,250
I would replace sine theta
by this square root divided

266
00:12:23,250 --> 00:12:24,300
by x minus 4.

267
00:12:24,300 --> 00:12:25,799
Because that's
what sine theta is.

268
00:12:25,799 --> 00:12:28,090
And so from-- that's how I
finish this type of problem,

269
00:12:28,090 --> 00:12:28,589
always.

270
00:12:28,589 --> 00:12:31,590
I want to have a picture of
this triangle, label a theta,

271
00:12:31,590 --> 00:12:34,800
use my substitution to give
me what two of the sides are,

272
00:12:34,800 --> 00:12:37,200
use the Pythagorean theorem
to get the third side.

273
00:12:37,200 --> 00:12:38,237
So that's the strategy.

274
00:12:38,237 --> 00:12:40,820
So let's go back and just remind
ourselves where we came from.

275
00:12:40,820 --> 00:12:42,819
We're going to go all the
way to the other side.

276
00:12:42,819 --> 00:12:45,480
This was a long, long problem.

277
00:12:45,480 --> 00:12:47,260
So what did we do
in this problem?

278
00:12:47,260 --> 00:12:50,420
I wanted us to find an
antiderivative of something.

279
00:12:50,420 --> 00:12:54,720
And right away, we can't
use partial fractions,

280
00:12:54,720 --> 00:12:57,110
because we can't
factor out an x here.

281
00:12:57,110 --> 00:13:00,540
So I'm forced to use
completing the square.

282
00:13:00,540 --> 00:13:02,560
So I completed the square first.

283
00:13:02,560 --> 00:13:05,660
That was the little algebra
that I had to do first.

284
00:13:05,660 --> 00:13:07,500
Then once I have that
little bit of algebra,

285
00:13:07,500 --> 00:13:11,910
I get into a situation where I'm
set up for a trig substitution.

286
00:13:11,910 --> 00:13:15,070
So then I had to start off
and do some trig substituting.

287
00:13:15,070 --> 00:13:18,200
And the things you have to do
to make a trig substitution work

288
00:13:18,200 --> 00:13:21,442
are, pick the right substitution
that makes sense, which

289
00:13:21,442 --> 00:13:22,650
you were given that in class.

290
00:13:22,650 --> 00:13:25,210
You can also figure it out
from a triangle picture,

291
00:13:25,210 --> 00:13:26,530
if you wanted to.

292
00:13:26,530 --> 00:13:28,780
And then you have to make
sure you substitute not just

293
00:13:28,780 --> 00:13:32,130
for the expression, the
function of x, but also the dx.

294
00:13:32,130 --> 00:13:35,690
So we did all that, and then
we came over, further, further,

295
00:13:35,690 --> 00:13:37,350
further, further, further, here.

296
00:13:37,350 --> 00:13:39,920
And we had everything
in terms of theta.

297
00:13:39,920 --> 00:13:42,080
So then we had to
look at-- we had

298
00:13:42,080 --> 00:13:44,320
all these trigonometric
functions of theta.

299
00:13:44,320 --> 00:13:46,730
We simplified that
as far as we could.

300
00:13:46,730 --> 00:13:48,520
We got one we could find.

301
00:13:48,520 --> 00:13:50,864
Then we finally, we take
the antiderivative there,

302
00:13:50,864 --> 00:13:52,405
and then in the very
end, we're going

303
00:13:52,405 --> 00:13:55,500
to substitute in for theta,
using the triangle we've

304
00:13:55,500 --> 00:13:57,810
drawn up here.

305
00:13:57,810 --> 00:13:58,800
So!

306
00:13:58,800 --> 00:14:00,950
I think that's where I'm
going to stop this one.

307
00:14:00,950 --> 00:14:03,811
Also, I ran out of board
space, so I have to stop.