Introduction to trig substitution

WEBVTT
Kind: captions
Language: en

00:00:00.790 --> 00:00:04.490
Let's try to evaluate
the indefinite integral 1

00:00:04.490 --> 00:00:10.980
over the square root of
1 minus x squared dx.

00:00:10.980 --> 00:00:13.149
And you might try something
like u substitution,

00:00:13.149 --> 00:00:14.940
but you'll find out
that it doesn't get you

00:00:14.940 --> 00:00:16.590
very far with this integral.

00:00:16.590 --> 00:00:18.180
And so instead, we
will do something

00:00:18.180 --> 00:00:20.060
called trig substitution.

00:00:20.060 --> 00:00:23.400
And the whole idea behind
trigonometric substitution

00:00:23.400 --> 00:00:27.740
is to substitute for x
with a trig function,

00:00:27.740 --> 00:00:30.670
so that this might simplify
using some trig identities.

00:00:30.670 --> 00:00:33.240
And maybe it'll turn this
entire integral into something

00:00:33.240 --> 00:00:36.050
that we can actually integrate.

00:00:36.050 --> 00:00:38.380
And so the trig
substitution that we'll do--

00:00:38.380 --> 00:00:41.590
and the big clue here is when
we see this 1 minus x squared.

00:00:41.590 --> 00:00:44.060
And in fact, if we ever
see an a squared minus an x

00:00:44.060 --> 00:00:46.420
squared-- so 1 is
1 squared, minus x

00:00:46.420 --> 00:00:48.160
squared-- it's a
pretty good clue

00:00:48.160 --> 00:00:51.050
that we would want to
substitute with something times

00:00:51.050 --> 00:00:51.690
sine of theta.

00:00:51.690 --> 00:00:54.340
In this case, if we just
make the substitution,

00:00:54.340 --> 00:00:58.710
if we say that x is equal
to-- let me do it over here.

00:00:58.710 --> 00:01:01.166
If we say that x is
equal to sine of theta,

00:01:01.166 --> 00:01:03.290
then what we have under
the radical right over here

00:01:03.290 --> 00:01:07.480
is going to become 1 minus sine
squared theta, which is cosine

00:01:07.480 --> 00:01:08.340
squared theta.

00:01:08.340 --> 00:01:10.298
And then we could take
the square root of that.

00:01:10.298 --> 00:01:13.040
And that might simplify all
of this business a good bit.

00:01:13.040 --> 00:01:15.415
And we just want to make sure
that we're not artificially

00:01:15.415 --> 00:01:18.030
constraining our x more
than what the domain here

00:01:18.030 --> 00:01:19.970
would otherwise constrain our x.

00:01:19.970 --> 00:01:25.570
So if you look over here,
the x right over here

00:01:25.570 --> 00:01:28.490
has to have an absolute
value less than 1.

00:01:28.490 --> 00:01:32.200
Or another way of saying it,
the domain right over here, x

00:01:32.200 --> 00:01:38.550
has to be greater than negative
1 and less than positive 1.

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If it was 1, then
the denominator right

00:01:40.950 --> 00:01:42.270
here is going to be 0.

00:01:42.270 --> 00:01:45.210
If it was greater than 1
or less than negative 1,

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then all of a
sudden, you're going

00:01:46.690 --> 00:01:48.460
to have a negative
under the radical sign.

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So already, our
domain is constrained

00:01:50.320 --> 00:01:52.850
for these values of x.

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So if we say x is sine
theta, this actually

00:01:55.340 --> 00:01:57.840
allows x to be anywhere
between negative 1 and 1,

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including negative 1 and 1.

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So we're not constraining x
more than the domain has already

00:02:02.180 --> 00:02:03.440
constrained it.

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Since we're going to have to
unwind the substitution later

00:02:06.200 --> 00:02:08.600
on, let's also
define theta in terms

00:02:08.600 --> 00:02:11.420
of x and in a way
that's not consistent

00:02:11.420 --> 00:02:13.090
with this top statement.

00:02:13.090 --> 00:02:17.220
So let's also say that
theta is equal to arcsin,

00:02:17.220 --> 00:02:20.500
or inverse sine, of x.

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And what does this do?

00:02:21.660 --> 00:02:24.640
Well, this actually constrains
what our theta is going to be.

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If I draw a unit circle--
my best to draw one.

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So this is my unit circle.

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And if I come up with
a particular x that's

00:02:40.550 --> 00:02:43.180
going to be essentially
the potential sine value,

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arcsin is going to
give an angle that

00:02:45.190 --> 00:02:48.960
is in the first or
fourth quadrant.

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So if we say that this is
our x, then the arcsin of x

00:02:53.470 --> 00:02:56.140
is going to be this
angle, not this one

00:02:56.140 --> 00:02:58.130
out here in the second quadrant.

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If we say that this is
our x, then our arcsin

00:03:00.430 --> 00:03:03.530
is going to be this angle.

00:03:03.530 --> 00:03:06.360
So this right over
here constrains

00:03:06.360 --> 00:03:11.250
theta to be less than
or equal to pi/2,

00:03:11.250 --> 00:03:13.970
so less than or equal to
that, and greater than

00:03:13.970 --> 00:03:15.650
or equal to negative pi/2.

00:03:15.650 --> 00:03:18.730
This is the way the arcsin
function is typically defined.

00:03:18.730 --> 00:03:23.070
It will return a value between
negative pi/2 and pi/2.

00:03:23.070 --> 00:03:26.350
Or another way of saying that,
in this range right over here

00:03:26.350 --> 00:03:28.010
on the unit circle.

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And this will be
important later on when

00:03:29.760 --> 00:03:32.295
we try to simplify
this expression.

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But let's actually try
to simplify it now.

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Before we do that, we
could substitute x,

00:03:38.740 --> 00:03:41.020
but we're also going to
have to substitute for dx.

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So let's figure out
what dx is equal to.

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dx is equal to--
well, the derivative

00:03:45.550 --> 00:03:51.030
of sine theta with respect to
theta is cosine theta d theta.

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So now we're ready to actually
evaluate, or at least simplify.

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So our original integral becomes
the integral of 1 times dx.

00:03:59.950 --> 00:04:02.550
Well, dx is cosine
theta d theta.

00:04:02.550 --> 00:04:07.380
Cosine theta-- I'll put
the d theta out here--

00:04:07.380 --> 00:04:13.920
over the square root
of 1 minus x squared,

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which is now going to be 1
minus sine squared theta.

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Well, what is 1 minus
sine squared theta?

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Well, that's going to
be cosine squared theta.

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Let me write this down.

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1 minus sine squared theta is
equal to cosine squared theta.

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This comes straight out of
the unit circle definition

00:04:33.860 --> 00:04:34.640
of trig functions.

00:04:34.640 --> 00:04:37.620
Cosine squared plus sine
squared is equal to 1.

00:04:37.620 --> 00:04:40.670
So we can rewrite
this as being equal

00:04:40.670 --> 00:04:47.050
to the indefinite integral
of cosine theta d theta

00:04:47.050 --> 00:04:53.585
over the square root of
cosine squared theta.

00:04:53.585 --> 00:04:55.460
Now, we have to be a
little bit careful here.

00:04:55.460 --> 00:04:57.626
We have to think about
whether cosine theta is going

00:04:57.626 --> 00:04:59.320
to be either
positive or negative

00:04:59.320 --> 00:05:00.720
or maybe we don't know.

00:05:00.720 --> 00:05:03.377
If we square a number and
then take its principal root,

00:05:03.377 --> 00:05:05.460
we're essentially taking
the absolute value of it.

00:05:05.460 --> 00:05:07.087
If cosine of theta
was, say, negative 1

00:05:07.087 --> 00:05:09.170
and we square it and then
take the principal root,

00:05:09.170 --> 00:05:10.130
it'll get 1.

00:05:10.130 --> 00:05:13.080
If cosine theta was positive
1, then we still get 1.

00:05:13.080 --> 00:05:14.940
So right now, all we
can do, without knowing

00:05:14.940 --> 00:05:17.340
more information about
cosine of theta, is

00:05:17.340 --> 00:05:21.080
say that this can be
rewritten as cosine theta

00:05:21.080 --> 00:05:26.425
over the absolute
value of cosine theta.

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If you square something and then
take the positive square root,

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the principal
square root, you're

00:05:30.425 --> 00:05:32.680
essentially just taking
its absolute value.

00:05:32.680 --> 00:05:34.810
And so it's really tempting
to simplify this more,

00:05:34.810 --> 00:05:36.740
but we can't quite do
it without understanding

00:05:36.740 --> 00:05:39.600
what cosine of theta is
going to be equal to.

00:05:39.600 --> 00:05:42.320
Well, lucky for us, we've
already constrained theta

00:05:42.320 --> 00:05:44.211
to be equal to arcsin of x.

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And so that says
that theta is going

00:05:45.710 --> 00:05:48.440
to be between negative
pi/2 and pi/2.

00:05:48.440 --> 00:05:50.230
And in this range
of thetas, we're

00:05:50.230 --> 00:05:52.840
always in kind of
the positive x-axis.

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So cosine of theta, which is
the x-coordinate for wherever

00:05:58.290 --> 00:06:02.210
we sit on the unit circle, is
always going to be positive.

00:06:02.210 --> 00:06:04.460
And since cosine of theta--
so let me make this clear.

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Arcsin of theta
is equal to arcsin

00:06:06.620 --> 00:06:09.120
of x tells us that
cosine of theta

00:06:09.120 --> 00:06:12.597
is going to be greater
than or equal to 0.

00:06:12.597 --> 00:06:14.180
It's going to be
non-negative, I guess

00:06:14.180 --> 00:06:16.346
is the best way to say it,
not necessarily positive.

00:06:16.346 --> 00:06:21.606
It could still be equal to 0.

00:06:21.606 --> 00:06:22.980
So it's going to
be non-negative.

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And since it's non-negative,
the absolute value

00:06:25.060 --> 00:06:27.240
of a non-negative number
is just the number itself.

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So this is just going to
be equal to the integral--

00:06:30.490 --> 00:06:33.440
and I'll just do it in one
color-- of cosine theta

00:06:33.440 --> 00:06:36.590
over cosine theta
d theta, which is

00:06:36.590 --> 00:06:41.810
just equal to the integral of d
theta, which is equal to theta

00:06:41.810 --> 00:06:45.940
plus C. And now all we have to
do is unwind this substitution.

00:06:45.940 --> 00:06:48.800
We already know that theta
is equal to arcsin of x.

00:06:48.800 --> 00:06:55.420
So we could say this is
equal to arcsin x plus C.

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And we're done.

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Using trig substitution,
we were able to evaluate

00:06:59.290 --> 00:07:00.380
this indefinite integral.

00:07:00.380 --> 00:07:02.421
And another way you could
think about it is we've

00:07:02.421 --> 00:07:05.320
just figured out what the
derivative of arcsin of x is.

00:07:05.320 --> 00:07:06.940
By taking the
antiderivative of this,

00:07:06.940 --> 00:07:10.350
we got arcsin of x plus C. If
we take the derivative of arcsin

00:07:10.350 --> 00:07:13.874
of x, we get this
business right over here.


Another substitution with x=sin (theta)

WEBVTT
Kind: captions
Language: en

00:00:00.520 --> 00:00:03.370
Let's see if we can evaluate
this indefinite integral.

00:00:03.370 --> 00:00:05.230
And the clue that
trig substitution

00:00:05.230 --> 00:00:07.400
might be appropriate is
what we see right over here

00:00:07.400 --> 00:00:09.280
in the denominator
under the radical.

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In general, if you see
something of the form

00:00:11.200 --> 00:00:14.929
a squared minus x squared, it
tends to be a pretty good idea,

00:00:14.929 --> 00:00:17.470
not always, but it's a good clue
that it might be a good idea

00:00:17.470 --> 00:00:22.156
to make the substitution x
is equal to a sine theta.

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Because if you do
that, then this

00:00:23.530 --> 00:00:27.540
will become a squared
minus a squared sine theta.

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And if you factor
out the a squared,

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you can start leveraging-- this
is a squared-- you can leverage

00:00:33.550 --> 00:00:35.170
one of the most
basic trig identities

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that this right over here
is cosine squared theta

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and maybe simplify
the expression.

00:00:40.134 --> 00:00:41.550
Now, you're probably
saying, well,

00:00:41.550 --> 00:00:44.530
this 8 minus 2x squared,
it's not as obvious

00:00:44.530 --> 00:00:46.610
that it's a squared
minus x squared.

00:00:46.610 --> 00:00:47.800
But we could simplify this.

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Or I guess we can
write it in a way

00:00:49.258 --> 00:00:50.780
that it starts to
have this pattern.

00:00:50.780 --> 00:00:55.350
You can rewrite 8 minus 2--
let me write it right under it.

00:00:55.350 --> 00:00:58.570
You could write 8
minus 2x squared

00:00:58.570 --> 00:01:04.629
as, if we factor out a 2, as
2 times 4 minus x squared.

00:01:04.629 --> 00:01:07.340
And now this very clearly
has a pattern a squared

00:01:07.340 --> 00:01:08.040
minus x squared.

00:01:08.040 --> 00:01:12.600
You could write this as 2 times
2 squared minus x squared.

00:01:12.600 --> 00:01:16.520
So in this case, a
would be equal to 2.

00:01:16.520 --> 00:01:18.350
So let's make that substitution.

00:01:18.350 --> 00:01:21.040
Let's make the
substitution that x

00:01:21.040 --> 00:01:27.010
is going to be equal
to 2 sine theta and dx

00:01:27.010 --> 00:01:31.910
is going to be equal to
2 cosine theta d theta.

00:01:31.910 --> 00:01:34.350
So what's this part under
the expression going to be?

00:01:34.350 --> 00:01:36.683
Well, we already started
simplifying it right over here.

00:01:36.683 --> 00:01:41.790
It's going to become 2 times
2 squared minus x squared.

00:01:41.790 --> 00:01:44.310
x squared is 2 sine
theta, so x squared

00:01:44.310 --> 00:01:49.440
is going to be 2 squared
sine theta squared.

00:01:49.440 --> 00:01:51.850
And now we can factor
out the 2 squared.

00:01:51.850 --> 00:01:56.610
So this is going to
be 2 times 2 squared

00:01:56.610 --> 00:02:00.800
times 1 minus sine
squared theta.

00:02:00.800 --> 00:02:03.250
2 times 2 squared, well
that's just going to be 8,

00:02:03.250 --> 00:02:08.500
times cosine squared theta.

00:02:08.500 --> 00:02:10.270
That's what we have
under the radical.

00:02:10.270 --> 00:02:11.140
So let's do that.

00:02:11.140 --> 00:02:13.480
Let's rewrite this
thing up here.

00:02:13.480 --> 00:02:16.840
So we're going to
have-- and I'll

00:02:16.840 --> 00:02:19.700
take the pi outside
of the integration.

00:02:19.700 --> 00:02:21.940
So we're going to
have pi times dx.

00:02:21.940 --> 00:02:24.800
dx is 2 cosine theta d theta.

00:02:24.800 --> 00:02:26.810
So it's going to be--
let me make it clear.

00:02:26.810 --> 00:02:30.350
So dx-- I want to
do that in blue.

00:02:30.350 --> 00:02:35.090
dx right over there is
2 cosine theta d theta.

00:02:35.090 --> 00:02:37.680
So let me write
that, 2 cosine theta.

00:02:37.680 --> 00:02:39.180
And I'll write the
d theta out here.

00:02:39.180 --> 00:02:41.346
I could have written
it in the numerator.

00:02:41.346 --> 00:02:42.720
And then here in
the denominator,

00:02:42.720 --> 00:02:44.890
I'm going to the square
root of this business,

00:02:44.890 --> 00:02:48.990
the square root of 8
cosine squared theta.

00:02:48.990 --> 00:02:50.450
So the square root
of that is going

00:02:50.450 --> 00:02:51.890
to be 2 square roots of 2.

00:02:51.890 --> 00:02:54.250
The square root of 8
is 2 square roots of 2.

00:02:54.250 --> 00:02:55.240
So let me write this.

00:02:55.240 --> 00:02:56.865
So let me make it
clear what I'm doing.

00:02:56.865 --> 00:02:59.670
So this right over here is going
to be the square root of this,

00:02:59.670 --> 00:03:01.730
which is 2 square roots of 2.

00:03:01.730 --> 00:03:02.900
That's the square root of 8.

00:03:02.900 --> 00:03:04.730
And the square root
of cosine squared

00:03:04.730 --> 00:03:06.540
theta is going to
be cosine theta.

00:03:09.660 --> 00:03:11.160
Now, you might be
saying, hey, wait,

00:03:11.160 --> 00:03:13.440
if I take the square root
of something squared,

00:03:13.440 --> 00:03:16.190
then wouldn't that just be the
absolute value of cosine theta?

00:03:16.190 --> 00:03:18.290
In order to take away
the absolute value,

00:03:18.290 --> 00:03:21.470
I'd have to assume that
cosine theta is positive.

00:03:21.470 --> 00:03:25.600
But we can make that assumption
that cosine theta is positive,

00:03:25.600 --> 00:03:30.700
because if we look right here at
this part of our substitution,

00:03:30.700 --> 00:03:33.950
if we wanted to solve for theta,
you'd divide both sides by 2,

00:03:33.950 --> 00:03:37.900
and you'd get x of 2 is
equal to sine of theta.

00:03:37.900 --> 00:03:44.456
Or we could say that theta is
equal to arcsin of x over 2.

00:03:44.456 --> 00:03:48.380
Now the arcsin function, as
it is traditionally defined,

00:03:48.380 --> 00:03:52.390
will return theta that is
between negative pi over 2

00:03:52.390 --> 00:03:53.510
and pi over 2.

00:03:53.510 --> 00:03:55.870
And in that range,
cosine of theta

00:03:55.870 --> 00:03:57.895
is always going to be positive.

00:03:57.895 --> 00:03:59.770
So we don't have to
write the absolute value.

00:03:59.770 --> 00:04:02.380
We know cosine
theta is positive.

00:04:02.380 --> 00:04:04.910
So now we can start to simplify.

00:04:04.910 --> 00:04:07.550
Cosine theta cancels
out with cosine theta.

00:04:07.550 --> 00:04:08.990
This 2 cancels out with this 2.

00:04:08.990 --> 00:04:11.310
We can bring this square
root of 2 outside.

00:04:11.310 --> 00:04:15.560
And so we are left with
pi over this square root

00:04:15.560 --> 00:04:21.709
of 2 times the indefinite
integral of just d theta.

00:04:21.709 --> 00:04:24.700
And this is just going
to be equal to pi over

00:04:24.700 --> 00:04:29.660
the square root of 2
times theta plus c.

00:04:29.660 --> 00:04:30.810
And we're almost done.

00:04:30.810 --> 00:04:33.420
We just have to rewrite
this in terms of x.

00:04:33.420 --> 00:04:36.726
And we already know that theta
is equal to arcsin of x over 2.

00:04:36.726 --> 00:04:39.962
So we can say that this
indefinite integral,

00:04:39.962 --> 00:04:41.670
or the antiderivative
of this expression,

00:04:41.670 --> 00:04:44.700
is going to be pi over
the square root of 2

00:04:44.700 --> 00:04:50.790
times arcsin of x over 2 plus c.

00:04:50.790 --> 00:04:51.415
And we're done.

00:04:51.415 --> 00:04:53.706
Some people like a square
root of 2 in the denominator.

00:04:53.706 --> 00:04:55.700
If you want to remove
it, you can multiply this

00:04:55.700 --> 00:04:59.640
by square root of 2
over square root of 2,

00:04:59.640 --> 00:05:00.950
and that will simplify it.

00:05:00.950 --> 00:05:02.866
But right now, I'll just
leave the denominator

00:05:02.866 --> 00:05:04.160
in irrational form.

00:05:04.160 --> 00:05:07.071
And this right over here
is our antiderivative.


Trig substitution with tangent

WEBVTT
Kind: captions
Language: en

00:00:00.690 --> 00:00:04.410
Let's see if we can evaluate
the indefinite integral 1

00:00:04.410 --> 00:00:09.430
over plus 9 plus x squared dx.

00:00:09.430 --> 00:00:12.040
And we know that if you
have the pattern a squared

00:00:12.040 --> 00:00:14.230
minus x squared, it
could be a good idea

00:00:14.230 --> 00:00:18.180
to make the substitution,
x is equal to a sine theta.

00:00:18.180 --> 00:00:20.470
But we don't see that
pattern over here.

00:00:20.470 --> 00:00:24.030
Instead, what we see is
a squared plus x squared.

00:00:24.030 --> 00:00:26.794
And in this context, it tends
to be a good idea-- it's not

00:00:26.794 --> 00:00:28.960
always going to work, but
it never hurts to try out.

00:00:28.960 --> 00:00:31.160
This is a little bit of
an art here-- to try out,

00:00:31.160 --> 00:00:34.100
x is equal to a tangent theta.

00:00:34.100 --> 00:00:36.224
Now you might say,
Sal, why is that?

00:00:36.224 --> 00:00:37.640
Well, let's make
that substitution

00:00:37.640 --> 00:00:40.010
and see how this
thing would simplify.

00:00:40.010 --> 00:00:44.490
This thing would become a
squared plus a squared tangent

00:00:44.490 --> 00:00:48.360
squared theta, which
is a squared times 1

00:00:48.360 --> 00:00:51.630
plus tangent squared theta.

00:00:51.630 --> 00:00:54.742
And this right over here,
we could reprove it.

00:00:54.742 --> 00:00:56.450
Actually, let me just
reprove it for you.

00:00:56.450 --> 00:00:59.450
This is going to become
a squared times-- this

00:00:59.450 --> 00:01:02.500
is 1 could be written
as cosine squared theta

00:01:02.500 --> 00:01:04.750
over cosine squared theta.

00:01:04.750 --> 00:01:07.720
Tangent squared is
sine squared theta

00:01:07.720 --> 00:01:10.542
over cosine squared theta.

00:01:10.542 --> 00:01:13.000
And this is why I picked cosine
squared as the denominator,

00:01:13.000 --> 00:01:14.900
so that I can add the two.

00:01:14.900 --> 00:01:18.160
And this is going to become
a squared times cosine

00:01:18.160 --> 00:01:21.900
squared theta plus
sine squared theta,

00:01:21.900 --> 00:01:25.270
all of that over
cosine squared theta.

00:01:25.270 --> 00:01:27.240
This numerator from the
unit circle definition

00:01:27.240 --> 00:01:29.330
of trig functions becomes 1.

00:01:29.330 --> 00:01:31.380
So this is 1 over
cosine squared theta.

00:01:31.380 --> 00:01:34.900
And everything simplifies
to a squared secant

00:01:34.900 --> 00:01:38.850
squared theta, which
might simplify things.

00:01:38.850 --> 00:01:41.520
So let's see if we have
what's going on over here.

00:01:41.520 --> 00:01:42.700
We can rewrite this.

00:01:42.700 --> 00:01:47.580
So 9 plus x squared
you could rewrite

00:01:47.580 --> 00:01:50.770
as 3 squared plus x squared.

00:01:50.770 --> 00:01:54.800
In this case, a
would be equal to 3.

00:01:54.800 --> 00:01:56.630
So we want to make
the substitution,

00:01:56.630 --> 00:02:01.990
x is equal to 3
tangent of theta.

00:02:01.990 --> 00:02:03.900
And if we wanted
to solve for x, you

00:02:03.900 --> 00:02:06.235
can divide both sides by
3, because we're later

00:02:06.235 --> 00:02:07.860
going to have to undo
the substitution.

00:02:07.860 --> 00:02:11.230
x over 3 is equal
to tangent theta,

00:02:11.230 --> 00:02:18.190
or theta is equal to arctangent
or inverse tangent of x over 3.

00:02:18.190 --> 00:02:20.474
Now we're also going to have
to figure out what dx is.

00:02:20.474 --> 00:02:22.640
We're also going to have
to figure out what that is.

00:02:22.640 --> 00:02:25.395
So let's take the
derivative or we'll write it

00:02:25.395 --> 00:02:26.750
in differential form.

00:02:26.750 --> 00:02:30.400
dx is equal to 3 derivative
of tangent theta with respect

00:02:30.400 --> 00:02:35.282
to theta is secant
squared theta d theta.

00:02:35.282 --> 00:02:36.990
So now it looks like
we're armed with all

00:02:36.990 --> 00:02:40.850
of the things necessary to
rewrite this entire integral.

00:02:40.850 --> 00:02:43.820
It's going to be equal to
the indefinite integral.

00:02:43.820 --> 00:02:45.650
You're going to
have dx here, which

00:02:45.650 --> 00:02:51.450
is equal to 3 secant
squared theta d theta.

00:02:51.450 --> 00:02:52.880
That's the dx.

00:02:52.880 --> 00:02:59.080
And all of that's going
to be over this business

00:02:59.080 --> 00:03:02.302
right over here, our a
squared plus our x squared.

00:03:02.302 --> 00:03:04.510
Now we already know what
that's going to simplify to.

00:03:04.510 --> 00:03:06.380
Our a squared plus
our x squared is

00:03:06.380 --> 00:03:08.260
going to-- since we
made this substitution,

00:03:08.260 --> 00:03:12.070
x is equal to 3
tangent theta is going

00:03:12.070 --> 00:03:16.610
to simplify to a squared
secant squared theta.

00:03:16.610 --> 00:03:24.530
So this is going to simplify
to 9 secant squared theta.

00:03:24.530 --> 00:03:27.610
And you could essentially
go through this logic.

00:03:27.610 --> 00:03:30.720
You're going to get 9 plus
9 tangent squared theta.

00:03:30.720 --> 00:03:35.740
Factor out a 9, you get 9
times 1 plus tan squared theta.

00:03:35.740 --> 00:03:38.130
And so that's going
to be 9 times secant

00:03:38.130 --> 00:03:40.650
squared theta, exactly
what we have here.

00:03:40.650 --> 00:03:43.200
Lucky for us, we have
the secant squared.

00:03:43.200 --> 00:03:44.520
It's canceling out.

00:03:44.520 --> 00:03:46.020
The secant squared
is canceling out.

00:03:46.020 --> 00:03:47.390
You have 3 over 9.

00:03:47.390 --> 00:03:48.850
This whole thing
can be rewritten

00:03:48.850 --> 00:03:51.870
as 1/3-- that's
just the 3 over 9--

00:03:51.870 --> 00:03:56.090
times the indefinite integral
of just d theta, which

00:03:56.090 --> 00:04:01.510
is equal to 1/3 theta plus c.

00:04:01.510 --> 00:04:04.130
And now we just have to
put things in terms of x.

00:04:04.130 --> 00:04:07.560
And we see theta is equal
to arctangent of x over 3.

00:04:07.560 --> 00:04:13.930
So this is going to be
equal to 1/3 arctangent of x

00:04:13.930 --> 00:04:16.480
over 3 plus c.

00:04:16.480 --> 00:04:19.680
And we are done.

00:04:19.680 --> 00:04:23.180
So now we know how to deal with
cases where we see something

00:04:23.180 --> 00:04:25.580
like an a squared
minus an x squared

00:04:25.580 --> 00:04:27.090
and an a squared
plus an x squared.

00:04:27.090 --> 00:04:28.850
It won't always
work, but it might

00:04:28.850 --> 00:04:33.600
be a useful-- it'll definitely
allow you to do this thing.

00:04:33.600 --> 00:04:36.100
It might not always make
the integral solvable,

00:04:36.100 --> 00:04:38.530
but it's not a bad thing to try.

00:04:38.530 --> 00:04:40.970
When it looks like
u-substitution isn't working,

00:04:40.970 --> 00:04:42.600
then you can look
for these patterns

00:04:42.600 --> 00:04:44.802
and try some trig substitution.


Trig and U substitution together (part 2)

WEBVTT
Kind: captions
Language: en

00:00:00.499 --> 00:00:03.320
In the last video,
in order to evaluate

00:00:03.320 --> 00:00:05.790
this indefinite integral, we
first made the substitution

00:00:05.790 --> 00:00:08.500
that x is equal to 3 sine theta.

00:00:08.500 --> 00:00:11.010
And then this got us to
an integral of this form.

00:00:11.010 --> 00:00:13.500
Then we were able to break
up these sines and cosines

00:00:13.500 --> 00:00:15.350
and use a little bit
of our trig identities.

00:00:15.350 --> 00:00:17.820
To get it into the form where
we could do u substitution,

00:00:17.820 --> 00:00:19.260
we did another
substitution where

00:00:19.260 --> 00:00:22.234
we said that u is equal
to cosine of theta.

00:00:22.234 --> 00:00:23.900
And then finally, we
were able to get it

00:00:23.900 --> 00:00:26.560
into a form using that
second round of substitution.

00:00:26.560 --> 00:00:28.204
And this time, it
was u substitution.

00:00:28.204 --> 00:00:29.620
We were able to
get it into a form

00:00:29.620 --> 00:00:31.800
that we could actually
take the antiderivative.

00:00:31.800 --> 00:00:34.033
And we got the final
answer here in terms of u.

00:00:34.033 --> 00:00:36.520
But now we've got to
go and undo everything.

00:00:36.520 --> 00:00:38.690
We have to undo
the substitutions.

00:00:38.690 --> 00:00:40.984
So the last substitution
we had done--

00:00:40.984 --> 00:00:42.650
we're now going to
go in reverse order--

00:00:42.650 --> 00:00:46.240
was that u was equal
to cosine theta.

00:00:46.240 --> 00:00:49.780
So you might just want to
substitute u with cosine theta

00:00:49.780 --> 00:00:50.280
here.

00:00:50.280 --> 00:00:51.904
But then we're going
to have everything

00:00:51.904 --> 00:00:54.850
in terms of cosine theta, which
still doesn't get us to x.

00:00:54.850 --> 00:00:59.970
So the ideal is if I can
somehow express u in terms of x.

00:00:59.970 --> 00:01:01.380
So let's think how we can do it.

00:01:01.380 --> 00:01:04.150
We know that u is
equal to cosine theta.

00:01:04.150 --> 00:01:07.590
We know the relationship between
x and theta is right over here.

00:01:07.590 --> 00:01:09.590
x is equal to 3 sine theta.

00:01:09.590 --> 00:01:10.920
So let's write that over here.

00:01:10.920 --> 00:01:14.030
So we know that x-- let
me write it over here--

00:01:14.030 --> 00:01:21.840
we know that x is
equal to 3 sine theta.

00:01:21.840 --> 00:01:23.470
So if we could
somehow write cosine--

00:01:23.470 --> 00:01:25.440
let me rewrite this a
little bit differently.

00:01:25.440 --> 00:01:29.850
Or we could also say that x
over 3 is equal to sine theta.

00:01:29.850 --> 00:01:32.060
I just divided both sides by 3.

00:01:32.060 --> 00:01:33.650
So if we could
somehow re-express

00:01:33.650 --> 00:01:35.780
this in terms of
sine theta, then we

00:01:35.780 --> 00:01:38.960
can replace all the sine
thetas with x over 3's, and we

00:01:38.960 --> 00:01:39.940
are done.

00:01:39.940 --> 00:01:40.889
So how can we do that?

00:01:40.889 --> 00:01:43.180
And I'll actually show you
two techniques for doing it.

00:01:43.180 --> 00:01:45.360
So the first one is to
make the realization, OK, u

00:01:45.360 --> 00:01:47.210
is equal to cosine of theta.

00:01:47.210 --> 00:01:49.550
If I want to write this
in terms of sine of theta,

00:01:49.550 --> 00:01:52.280
I can just say that this
is equal to-- straight up,

00:01:52.280 --> 00:01:54.880
this is the most fundamental
trigonometric identity.

00:01:54.880 --> 00:02:00.760
Cosine theta is the square root
of 1 minus sine squared theta.

00:02:00.760 --> 00:02:03.020
And we see sine of theta
is equal to x over 3.

00:02:03.020 --> 00:02:12.290
So this is the square root
of 1 minus x over 3 squared.

00:02:12.290 --> 00:02:14.362
So this is u in terms of x.

00:02:14.362 --> 00:02:15.820
So everywhere we
see a u up here we

00:02:15.820 --> 00:02:17.610
can replace it with
this expression.

00:02:17.610 --> 00:02:18.910
And we are essentially done.

00:02:18.910 --> 00:02:21.217
We would have written
this in terms of x.

00:02:21.217 --> 00:02:23.300
Now, there's another
technique you might sometimes

00:02:23.300 --> 00:02:25.383
see in a calculus class
where someone says, OK, we

00:02:25.383 --> 00:02:27.350
know that u is equal
to cosine theta.

00:02:27.350 --> 00:02:28.570
We know this relationship.

00:02:28.570 --> 00:02:30.860
How can we express
u in terms of x?

00:02:30.860 --> 00:02:33.400
And we'll say, let's
draw a right triangle.

00:02:33.400 --> 00:02:36.400
They'll draw a right
triangle like this.

00:02:36.400 --> 00:02:39.150
They'll draw a right triangle,
and they'll say, OK, look,

00:02:39.150 --> 00:02:40.620
sine of theta is x over 3.

00:02:40.620 --> 00:02:43.640
So if we say that this is theta
right over here, sine of theta

00:02:43.640 --> 00:02:45.980
is the same thing as
opposite over hypotenuse.

00:02:45.980 --> 00:02:49.620
Opposite over hypotenuse
is equal to x over 3.

00:02:49.620 --> 00:02:52.660
So let's say that this is x and
then this right over here is 3.

00:02:52.660 --> 00:02:54.760
Then the sine of theta
will be x over 3.

00:02:54.760 --> 00:02:57.065
So we look at that first
substitution right over here.

00:02:57.065 --> 00:02:59.190
But in order to figure out
what u is in terms of x,

00:02:59.190 --> 00:03:01.550
we need to figure out
what cosine of theta is.

00:03:01.550 --> 00:03:04.160
Well, cosine is adjacent
over hypotenuse.

00:03:04.160 --> 00:03:06.540
So we have to figure out
what this adjacent side is.

00:03:06.540 --> 00:03:09.046
Well, we can just use the
Pythagorean theorem for that.

00:03:09.046 --> 00:03:10.420
Pythagorean theorem
would tell us

00:03:10.420 --> 00:03:12.086
that this is going
to be the square root

00:03:12.086 --> 00:03:14.520
of the hypotenuse
squared, which is

00:03:14.520 --> 00:03:19.550
9, minus the other side
squared, minus x squared.

00:03:19.550 --> 00:03:22.340
So from this, we fully
solved the right triangle

00:03:22.340 --> 00:03:23.420
in terms of x.

00:03:23.420 --> 00:03:25.740
We can realize that
cosine of theta

00:03:25.740 --> 00:03:29.860
is going to be equal to the
adjacent side, square root of 9

00:03:29.860 --> 00:03:32.800
minus x squared,
over the hypotenuse,

00:03:32.800 --> 00:03:37.700
over 3, which is the same thing
as 1/3 times the square root

00:03:37.700 --> 00:03:40.780
of 9 minus x squared,
which is the same thing

00:03:40.780 --> 00:03:43.980
if we square 1/3 and
put it into the radical.

00:03:43.980 --> 00:03:46.220
So we're essentially going
to take the square-- 1/3

00:03:46.220 --> 00:03:49.490
is the same thing as
the square root of 1/9.

00:03:49.490 --> 00:03:54.830
So can rewrite this as the
square root of 1/9 times

00:03:54.830 --> 00:03:56.460
9 minus x squared.

00:03:56.460 --> 00:03:59.160
Essentially, we just brought
the 1/3 third into the radical.

00:03:59.160 --> 00:04:00.400
Now it's 1/9.

00:04:00.400 --> 00:04:02.590
And so now this is going
to be the same thing

00:04:02.590 --> 00:04:12.720
as the square root of 1
minus x squared over 9,

00:04:12.720 --> 00:04:14.760
which is exactly this
thing right over here. x

00:04:14.760 --> 00:04:18.120
squared over 9 is the same
thing as x over 3 squared.

00:04:18.120 --> 00:04:19.970
So either way, you
get the same result.

00:04:19.970 --> 00:04:22.940
I find using the trig
identity right over here

00:04:22.940 --> 00:04:25.220
to express cosine of theta
in terms of sine theta

00:04:25.220 --> 00:04:26.970
and then just do
the substitution

00:04:26.970 --> 00:04:28.889
to be a little bit
more straightforward.

00:04:28.889 --> 00:04:31.180
But now we can just substitute
into the original thing.

00:04:31.180 --> 00:04:34.000
So either of these-- I can
write it as either way--

00:04:34.000 --> 00:04:36.700
this thing right over here,
this is the same thing as 1

00:04:36.700 --> 00:04:40.570
minus x squared over
9 to the 1/2 power.

00:04:40.570 --> 00:04:42.130
That's what u is equal to.

00:04:42.130 --> 00:04:45.040
And everywhere we see u, we just
substitute it with this thing.

00:04:45.040 --> 00:04:48.080
So our final answer
in terms of x

00:04:48.080 --> 00:04:53.000
is going to be equal to 243
times u to the fifth, this

00:04:53.000 --> 00:04:54.780
to the fifth power over 5.

00:04:54.780 --> 00:05:00.590
This to the fifth power is
1 minus x squared over 9.

00:05:00.590 --> 00:05:03.050
It was to the 1/2, but if we
raise that to the fifth power,

00:05:03.050 --> 00:05:05.560
it's now going to
be to the 5/2 power

00:05:05.560 --> 00:05:12.750
over 5 minus this to the
third power, 1 minus x squared

00:05:12.750 --> 00:05:16.800
over 9 to the 3/2 raising this
to the third power-- that's

00:05:16.800 --> 00:05:21.765
this right over here-- over 3,
and then all of that plus c.

00:05:21.765 --> 00:05:22.390
And we're done.

00:05:22.390 --> 00:05:26.570
It's messy, but using first
trig substitution then

00:05:26.570 --> 00:05:29.830
u substitution, or trig
substitution then rearranging

00:05:29.830 --> 00:05:32.780
using a couple of our
techniques for manipulating

00:05:32.780 --> 00:05:35.505
these powers of trig functions,
we're able to get into a form

00:05:35.505 --> 00:05:37.410
where we could use
u substitution,

00:05:37.410 --> 00:05:40.230
and then we were able to
unwind all the substitutions

00:05:40.230 --> 00:05:44.420
and actually evaluate
the indefinite integral.


Trig and U substitution together (part 1)

WEBVTT
Kind: captions
Language: en

00:00:00.530 --> 00:00:03.130
Let's attempt to take
the antiderivative,

00:00:03.130 --> 00:00:06.580
or the indefinite integral,
of x to the third times

00:00:06.580 --> 00:00:11.539
the square root of 9
minus x squared dx.

00:00:11.539 --> 00:00:13.080
And you might attempt
to do something

00:00:13.080 --> 00:00:14.500
like use substitution,
but you'll

00:00:14.500 --> 00:00:16.580
find that you're not
getting very far.

00:00:16.580 --> 00:00:17.890
And we have a big clue here.

00:00:17.890 --> 00:00:20.320
We have something
of the form-- so 9

00:00:20.320 --> 00:00:24.190
minus x squared, that can
be viewed as the same thing

00:00:24.190 --> 00:00:26.270
as 3 squared minus x squared.

00:00:26.270 --> 00:00:30.760
And any time you have the form
a squared minus x squared,

00:00:30.760 --> 00:00:33.910
it might be useful to
make the substitution

00:00:33.910 --> 00:00:39.472
that x is equal to a sine theta.

00:00:39.472 --> 00:00:40.430
Now, why would that be?

00:00:40.430 --> 00:00:42.230
Well, then a squared
minus x squared

00:00:42.230 --> 00:00:49.300
would become a squared
minus a squared sine squared

00:00:49.300 --> 00:00:52.300
theta, which is the same
thing as a squared times 1

00:00:52.300 --> 00:00:53.644
minus sine squared theta.

00:00:53.644 --> 00:00:55.310
And I think you see
where this is going.

00:00:55.310 --> 00:00:57.200
This is a squared
times cosine squared

00:00:57.200 --> 00:01:00.209
of theta, which might be
a useful simplification.

00:01:00.209 --> 00:01:01.250
So let's do it over here.

00:01:01.250 --> 00:01:03.860
In this case, our a is 3.

00:01:03.860 --> 00:01:07.070
So let's make the
substitution that x

00:01:07.070 --> 00:01:13.310
is going to be equal to a sine
theta, or three times sine

00:01:13.310 --> 00:01:14.600
of theta.

00:01:14.600 --> 00:01:16.283
And then we're
going to also have

00:01:16.283 --> 00:01:17.730
to figure out what
dx is equal to.

00:01:17.730 --> 00:01:21.340
So if you take the
derivative, we will get dx.

00:01:21.340 --> 00:01:24.800
We could have dx d theta
is equal to 3 cosine theta.

00:01:24.800 --> 00:01:28.820
Or if we wanted to write
it in differential form,

00:01:28.820 --> 00:01:32.020
we could write that
dx is equal to 3 times

00:01:32.020 --> 00:01:34.350
cosine theta d theta.

00:01:34.350 --> 00:01:38.540
This is just the derivative
of this with respect to theta.

00:01:38.540 --> 00:01:40.300
And we're ready to
substitute back.

00:01:40.300 --> 00:01:44.490
Our original expression
now becomes--

00:01:44.490 --> 00:01:49.030
I'll write it in that
original green-- 3 sine theta

00:01:49.030 --> 00:01:53.130
to the third power, which
is the same thing as 27

00:01:53.130 --> 00:01:54.880
sine-- actually,
let me color code it

00:01:54.880 --> 00:01:56.910
just so you know
what parts I'm doing.

00:01:56.910 --> 00:01:59.900
So this part right over
here, x to the third,

00:01:59.900 --> 00:02:05.500
is now going to become 27 sine
to the third power of theta,

00:02:05.500 --> 00:02:07.910
or sine theta to
the third power.

00:02:07.910 --> 00:02:12.120
And then all of
this business, this

00:02:12.120 --> 00:02:18.130
is going to be the square
root of 9 minus x squared,

00:02:18.130 --> 00:02:23.450
so minus 9 sine squared theta.

00:02:23.450 --> 00:02:26.850
And then dx-- let me do
this in a new color-- dx

00:02:26.850 --> 00:02:28.880
right over here is going
to be equal to-- that's

00:02:28.880 --> 00:02:30.230
not a new color.

00:02:30.230 --> 00:02:34.980
dx is going to be equal
to all of this business,

00:02:34.980 --> 00:02:42.480
so times 3 cosine theta d theta.

00:02:42.480 --> 00:02:44.540
And now let's see
if we can simplify

00:02:44.540 --> 00:02:46.690
this business a little bit.

00:02:46.690 --> 00:02:49.190
Let me do this over to the side.

00:02:49.190 --> 00:02:53.700
This thing right over here
can be written as 9 times 1

00:02:53.700 --> 00:02:57.240
minus sine squared
theta, which is

00:02:57.240 --> 00:03:03.490
equal to the square root of
9 times cosine squared theta.

00:03:03.490 --> 00:03:09.170
And we can assume that
cosine theta is positive,

00:03:09.170 --> 00:03:10.830
as we did in the last video.

00:03:10.830 --> 00:03:20.620
And so this is going to be equal
to 3 cosine theta in orange.

00:03:20.620 --> 00:03:26.000
So this right over
here is 3 cosine theta.

00:03:26.000 --> 00:03:28.350
And so what does
this simplify to?

00:03:28.350 --> 00:03:38.810
We have a 27 times 3 is 81
times 3 is going to be 243.

00:03:38.810 --> 00:03:41.350
So this is going to
be 243-- I'll put it

00:03:41.350 --> 00:03:45.414
out front-- times the
integral of-- let's see,

00:03:45.414 --> 00:03:46.955
we're going to have
sine cubed theta.

00:03:50.839 --> 00:03:53.130
And then you're going to have
cosine theta times cosine

00:03:53.130 --> 00:03:55.310
theta, or we could say
cosine squared theta.

00:03:58.250 --> 00:04:00.690
That's this term right over
here and this term right

00:04:00.690 --> 00:04:04.450
over there, and of
course, d theta.

00:04:04.450 --> 00:04:06.477
I think I've taken
care of everything.

00:04:06.477 --> 00:04:08.310
So it might not look
like I've simplified it

00:04:08.310 --> 00:04:09.851
a lot, because, hey,
look, this still

00:04:09.851 --> 00:04:14.010
doesn't seem like a trivially
easy problem to solve,

00:04:14.010 --> 00:04:16.380
but we are getting closer.

00:04:16.380 --> 00:04:21.160
Now this turns into just a
classic u substitution problem.

00:04:21.160 --> 00:04:22.730
And it's not obvious just yet.

00:04:22.730 --> 00:04:24.350
There's actually a little
bit of a layer of a technique

00:04:24.350 --> 00:04:25.183
to figure out first.

00:04:25.183 --> 00:04:28.840
How do you do u substitution
right over here?

00:04:28.840 --> 00:04:31.830
And the key when you have
powers of trig functions,

00:04:31.830 --> 00:04:33.330
especially when you
have one of them

00:04:33.330 --> 00:04:35.140
as an odd power,
what you want to do

00:04:35.140 --> 00:04:37.280
is separate one of
those odd powers out

00:04:37.280 --> 00:04:39.830
so you can kind of construct
a u substitution problem.

00:04:39.830 --> 00:04:40.890
So let's do that.

00:04:40.890 --> 00:04:44.800
So this is going to
be equal to 243 times

00:04:44.800 --> 00:04:47.370
the integral of--
sine cubed theta

00:04:47.370 --> 00:04:50.100
I can rewrite as sine
of theta-- actually,

00:04:50.100 --> 00:04:52.100
let me write it
this way-- as sine

00:04:52.100 --> 00:04:56.210
squared theta cosine
squared theta.

00:04:58.740 --> 00:05:06.130
And then I still have 1 sine
theta right over here, d theta.

00:05:06.130 --> 00:05:08.840
And what I'm trying to do
is turn this expression

00:05:08.840 --> 00:05:10.879
into something where I
can do u substitution,

00:05:10.879 --> 00:05:12.420
and as you could
imagine, maybe where

00:05:12.420 --> 00:05:15.400
the du has to deal with
sine theta d theta.

00:05:15.400 --> 00:05:18.770
And it would if I can get my
u being equal to cosine theta.

00:05:18.770 --> 00:05:22.350
Then my du is going to be
negative sine theta d theta.

00:05:22.350 --> 00:05:24.610
So let me see if I can do that.

00:05:24.610 --> 00:05:28.760
So if I say that
sine squared theta

00:05:28.760 --> 00:05:32.110
is the same thing as 1
minus cosine squared theta,

00:05:32.110 --> 00:05:38.580
then this whole thing becomes
243 times the integral of 1

00:05:38.580 --> 00:05:46.600
minus cosine squared theta
times cosine squared theta times

00:05:46.600 --> 00:05:47.980
sine theta d theta.

00:05:50.429 --> 00:05:52.470
And I want to be very
clear what I did over here.

00:05:52.470 --> 00:05:54.060
So we used some
trig substitution

00:05:54.060 --> 00:05:55.761
to get to this point
right over here.

00:05:55.761 --> 00:05:57.760
And at this point, I took
one of the sine thetas

00:05:57.760 --> 00:05:59.601
out-- I separated
it right over here--

00:05:59.601 --> 00:06:01.350
and then I converted
this to an expression

00:06:01.350 --> 00:06:03.340
in terms of cosine theta.

00:06:03.340 --> 00:06:07.220
Now, the whole reason why I
did this is right over here I

00:06:07.220 --> 00:06:09.770
have a function of
cosine theta, and then I

00:06:09.770 --> 00:06:12.010
have something that's pretty
close to the derivative

00:06:12.010 --> 00:06:14.050
of cosine theta right over here.

00:06:14.050 --> 00:06:18.500
So this is now ripe
for u substitution.

00:06:18.500 --> 00:06:20.210
So let's do u substitution.

00:06:20.210 --> 00:06:22.080
If I have a function
of something

00:06:22.080 --> 00:06:23.990
and then I have this
derivative, maybe u

00:06:23.990 --> 00:06:25.460
should be equal
to that something.

00:06:25.460 --> 00:06:31.130
So let me set u as being
equal to cosine of theta.

00:06:31.130 --> 00:06:37.360
Then du is going to be equal to
negative sine of theta d theta.

00:06:37.360 --> 00:06:39.710
Well, I have a
sine theta d theta.

00:06:39.710 --> 00:06:42.239
I can multiply that
times a negative

00:06:42.239 --> 00:06:43.780
as long as I put a
negative out here.

00:06:43.780 --> 00:06:45.310
I'm multiplying by
a negative twice.

00:06:45.310 --> 00:06:47.110
I'm not changing the value.

00:06:47.110 --> 00:06:49.950
So notice, now this
right over here is du,

00:06:49.950 --> 00:06:51.970
and this right over
here is a function of u.

00:06:51.970 --> 00:06:53.380
So let's write it that way.

00:06:53.380 --> 00:06:59.620
All of this business is going to
be equal to negative 243 times

00:06:59.620 --> 00:07:09.060
the integral of 1 minus u
squared times u squared.

00:07:12.480 --> 00:07:16.730
And then this right over
here is just our du.

00:07:16.730 --> 00:07:19.201
So this is just du.

00:07:19.201 --> 00:07:21.990
Now, this is pretty
straightforward.

00:07:21.990 --> 00:07:26.250
We can just multiply
our u out, and this

00:07:26.250 --> 00:07:31.240
will become negative 243.

00:07:31.240 --> 00:07:33.700
So this is all equal
to negative 243 times

00:07:33.700 --> 00:07:37.830
the indefinite
integral of u squared

00:07:37.830 --> 00:07:43.060
minus u to the fourth--
I'm just distributing

00:07:43.060 --> 00:07:45.590
the u squared-- du.

00:07:47.667 --> 00:07:49.125
Now, this is pretty
straightforward

00:07:49.125 --> 00:07:50.490
to take the antiderivative of.

00:07:50.490 --> 00:07:56.090
This is negative 243 times the
antiderivative of u squared

00:07:56.090 --> 00:07:59.060
is u to the third over 3.

00:07:59.060 --> 00:08:01.220
Antiderivative of
u to the fourth

00:08:01.220 --> 00:08:05.670
is u to the fifth over 5.

00:08:05.670 --> 00:08:08.239
And of course, we're going
to have a plus c out here.

00:08:08.239 --> 00:08:10.280
And just so that we can
get rid of this negative,

00:08:10.280 --> 00:08:11.180
we can swap this.

00:08:11.180 --> 00:08:12.830
We can distribute
the negative sign.

00:08:12.830 --> 00:08:14.730
So this one becomes negative.

00:08:14.730 --> 00:08:16.420
That one becomes positive.

00:08:16.420 --> 00:08:26.640
And we get all of this business
as being equal to 243 times u

00:08:26.640 --> 00:08:32.970
to the fifth over 5 minus u
to the third over 3 plus c.

00:08:32.970 --> 00:08:35.190
And you might say, wow,
finally, we are done.

00:08:35.190 --> 00:08:36.090
But we aren't done.

00:08:36.090 --> 00:08:39.320
We have everything in terms of
u while our original integral

00:08:39.320 --> 00:08:41.250
was in terms of x.

00:08:41.250 --> 00:08:44.340
So the next video, we're going
to unwind all the substitution.

00:08:44.340 --> 00:08:46.548
We're going to try to take
this expression right over

00:08:46.548 --> 00:08:47.960
here and write it in terms of x.

00:08:47.960 --> 00:08:50.890
So we're going to have to
go from u to theta to x,

00:08:50.890 --> 00:08:53.744
because we've done two
rounds of substitutions.


Fundamental theorem of calculus (Part 2)

WEBVTT
Kind: captions
Language: en

00:00:00.690 --> 00:00:02.510
Let's say we've
got some function

00:00:02.510 --> 00:00:06.259
f that is continuous over
the interval between c and d.

00:00:06.259 --> 00:00:08.550
And the reason why I'm using
c and d instead of a and b

00:00:08.550 --> 00:00:11.370
is so I can use a
and b for later.

00:00:11.370 --> 00:00:13.890
And let's say we set up
some function capital

00:00:13.890 --> 00:00:17.490
F of x which is
defined as the area

00:00:17.490 --> 00:00:22.450
under the curve between
c and some value x,

00:00:22.450 --> 00:00:26.320
where x is in this interval
where f is continuous,

00:00:26.320 --> 00:00:28.980
under the curve-- so it's the
area under the curve between c

00:00:28.980 --> 00:00:33.590
and x-- so if this is x right
over here-- under the curve

00:00:33.590 --> 00:00:35.550
f of t dt.

00:00:35.550 --> 00:00:39.580
So this right over here,
F of x, is that area.

00:00:39.580 --> 00:00:43.130
That right over there
is what F of x is.

00:00:43.130 --> 00:00:46.150
Now, the fundamental
theorem of calculus

00:00:46.150 --> 00:00:53.100
tells us that if f is continuous
over this interval, then F of x

00:00:53.100 --> 00:00:56.430
is differentiable at
every x in the interval,

00:00:56.430 --> 00:01:02.370
and the derivative of capital
F of x-- and let me be clear.

00:01:02.370 --> 00:01:06.440
Capital F of x is differentiable
at every possible x

00:01:06.440 --> 00:01:10.250
between c and d, and the
derivative of capital F

00:01:10.250 --> 00:01:15.640
of x is going to be equal
to lowercase f of x.

00:01:15.640 --> 00:01:17.350
Fair enough.

00:01:17.350 --> 00:01:18.850
Now, what I want
to do in this video

00:01:18.850 --> 00:01:22.190
is connect the first
fundamental theorem of calculus

00:01:22.190 --> 00:01:24.890
to the second part, or the
second fundamental theorem

00:01:24.890 --> 00:01:28.860
of calculus, which we tend
to use to actually evaluate

00:01:28.860 --> 00:01:30.670
definite integrals.

00:01:30.670 --> 00:01:41.940
So let's think about what F of
b minus F of a is, what this is,

00:01:41.940 --> 00:01:45.290
where both b and a are
also in this interval.

00:01:45.290 --> 00:01:48.460
So F of b-- and we're going to
assume that b is larger than a.

00:01:48.460 --> 00:01:50.790
So let's say that b is
this right over here.

00:01:50.790 --> 00:01:53.380
And we'll do that
in the same color.

00:01:53.380 --> 00:01:56.810
So let's say that b
is right over here.

00:01:56.810 --> 00:02:01.005
F of b is going to be equal to--
we just literally replace the b

00:02:01.005 --> 00:02:02.760
where you see the
x-- it's going to be

00:02:02.760 --> 00:02:09.530
equal to the definite integral
between c and b of f of t dt,

00:02:09.530 --> 00:02:12.400
which is just
another way of saying

00:02:12.400 --> 00:02:15.750
the area under the
curve between c and b.

00:02:15.750 --> 00:02:19.550
So this F of b,
capital F of b, is

00:02:19.550 --> 00:02:23.880
all of this business
right over here.

00:02:23.880 --> 00:02:25.830
And from that, we
are going to want

00:02:25.830 --> 00:02:29.660
to subtract capital
F of a, which

00:02:29.660 --> 00:02:33.990
is just the integral between
c and lowercase a of lowercase

00:02:33.990 --> 00:02:36.070
f of t dt.

00:02:36.070 --> 00:02:41.380
So let's say that this
is a right over here.

00:02:41.380 --> 00:02:43.610
Capital F of a is just
literally the area

00:02:43.610 --> 00:02:47.370
between c and a under the
curve lowercase f of t.

00:02:47.370 --> 00:02:49.640
So it's this right over here.

00:02:49.640 --> 00:02:54.890
It's all of this
business right over here.

00:02:54.890 --> 00:02:57.840
So if you have this blue
area, which is all of this,

00:02:57.840 --> 00:03:02.000
and you subtract out
to this magenta area,

00:03:02.000 --> 00:03:03.340
what are you left with?

00:03:03.340 --> 00:03:08.810
Well, you're left with this
green area right over here.

00:03:08.810 --> 00:03:10.380
And how would we represent that?

00:03:10.380 --> 00:03:11.790
How would we denote that?

00:03:11.790 --> 00:03:14.420
Well, we could denote that
as the definite integral

00:03:14.420 --> 00:03:21.540
between a and b of f of t dt.

00:03:21.540 --> 00:03:22.700
And there you have it.

00:03:22.700 --> 00:03:25.110
This right over here is the
second fundamental theorem

00:03:25.110 --> 00:03:25.660
of calculus.

00:03:25.660 --> 00:03:29.700
It tells us that if f is
continuous on the interval,

00:03:29.700 --> 00:03:33.420
that this is going to be
equal to the antiderivative,

00:03:33.420 --> 00:03:35.070
or an antiderivative, of f.

00:03:35.070 --> 00:03:36.860
And we see right over
here that capital F

00:03:36.860 --> 00:03:39.090
is the antiderivative of f.

00:03:39.090 --> 00:03:45.280
So we could view this as
capital F antiderivative--

00:03:45.280 --> 00:03:48.301
this is how we defined capital
F-- the antiderivative-- or we

00:03:48.301 --> 00:03:50.550
didn't define it that way,
but the fundamental theorem

00:03:50.550 --> 00:03:52.990
of calculus tells
us that capital F is

00:03:52.990 --> 00:03:56.790
an antiderivative
of lowercase f.

00:03:56.790 --> 00:03:58.630
So right over here,
this tells you,

00:03:58.630 --> 00:04:01.200
if you have a definite
integral like this,

00:04:01.200 --> 00:04:05.740
it's completely equivalent to an
antiderivative of it evaluated

00:04:05.740 --> 00:04:11.275
at b, and from that, you
subtract it evaluated at a.

00:04:11.275 --> 00:04:13.110
So normally it looks like this.

00:04:13.110 --> 00:04:14.810
I've just switched the order.

00:04:14.810 --> 00:04:19.029
The definite integral
from a to b of f of t dt

00:04:19.029 --> 00:04:23.820
is equal to an antiderivative
of f, so capital F, evaluated

00:04:23.820 --> 00:04:30.420
at b, and from that, subtract
the antiderivative evaluated

00:04:30.420 --> 00:04:31.301
at a.

00:04:31.301 --> 00:04:33.550
And this is the second part
of the fundamental theorem

00:04:33.550 --> 00:04:35.960
of calculus, or the
second fundamental theorem

00:04:35.960 --> 00:04:36.850
of calculus.

00:04:36.850 --> 00:04:39.830
And it's really the core of
an integral calculus class,

00:04:39.830 --> 00:04:43.989
because it's how you actually
evaluate definite integrals.


Finding derivative with fundamental theorem of calculus: x is on both bounds

WEBVTT
Kind: captions
Language: en

00:00:00.830 --> 00:00:03.670
So let's see if we can take the
derivative of this expression

00:00:03.670 --> 00:00:07.790
right over here, if we can
find capital F prime of x.

00:00:07.790 --> 00:00:09.630
And once again,
it looks like you

00:00:09.630 --> 00:00:11.670
might be able to use
the fundamental theorem

00:00:11.670 --> 00:00:12.250
of calculus.

00:00:12.250 --> 00:00:13.810
A big giveaway is
that you're taking

00:00:13.810 --> 00:00:15.710
the derivative of a
definite integral that

00:00:15.710 --> 00:00:17.980
gives you a function of x.

00:00:17.980 --> 00:00:21.144
But here I have x on both the
upper and the lower boundary,

00:00:21.144 --> 00:00:22.810
and the fundamental
theorem of calculus,

00:00:22.810 --> 00:00:24.530
is at least from
what we've seen,

00:00:24.530 --> 00:00:27.616
is when we have x's only
on the upper boundary.

00:00:27.616 --> 00:00:29.240
And then, of course,
it's an x squared,

00:00:29.240 --> 00:00:30.970
but we've seen examples
of that already

00:00:30.970 --> 00:00:32.770
when we used the
chain rule to do it.

00:00:32.770 --> 00:00:34.320
But how can we break
this up and put

00:00:34.320 --> 00:00:36.700
this in a form that's a little
bit closer to what we're

00:00:36.700 --> 00:00:39.370
familiar with when we apply
the fundamental theorem

00:00:39.370 --> 00:00:40.450
of calculus?

00:00:40.450 --> 00:00:41.920
And to realize
that, we really just

00:00:41.920 --> 00:00:45.350
have to attempt to graph
what this is representing.

00:00:45.350 --> 00:00:47.380
So let's say that
this is our lowercase

00:00:47.380 --> 00:00:50.040
f of x, or I should say f of t.

00:00:50.040 --> 00:00:52.980
So let's call this
lowercase f of t.

00:00:52.980 --> 00:00:54.700
And let's graph it
over the interval

00:00:54.700 --> 00:00:57.070
between x and x squared.

00:00:57.070 --> 00:01:01.490
So let's say this is my y-axis.

00:01:01.490 --> 00:01:04.360
This is my t-axis.

00:01:04.360 --> 00:01:06.340
And let's say that
this right over here

00:01:06.340 --> 00:01:09.600
is y is equal to f of t.

00:01:09.600 --> 00:01:10.880
I'm drawing it generally.

00:01:10.880 --> 00:01:12.671
I don't know what this
exactly looks like.

00:01:12.671 --> 00:01:14.420
And we're going to
talk about the interval

00:01:14.420 --> 00:01:16.530
between x and x squared.

00:01:16.530 --> 00:01:19.870
So if we're going to talk about
the interval between x, which

00:01:19.870 --> 00:01:21.970
is right over here,
it's the lower bound,

00:01:21.970 --> 00:01:23.224
so x and x squared.

00:01:23.224 --> 00:01:25.640
It's the lower bound, at least
for this definite integral.

00:01:25.640 --> 00:01:26.370
We don't know for sure.

00:01:26.370 --> 00:01:28.703
It depends on what x you
choose on which one is actually

00:01:28.703 --> 00:01:29.470
smaller.

00:01:29.470 --> 00:01:31.670
But let's just say that for
the sake of visualizing,

00:01:31.670 --> 00:01:34.040
we'll draw x right
over here, and we

00:01:34.040 --> 00:01:37.600
will draw x squared
right over here.

00:01:40.340 --> 00:01:43.730
So this whole expression,
this entire definite integral,

00:01:43.730 --> 00:01:46.490
is essentially asking for,
is essentially representing

00:01:46.490 --> 00:01:51.820
this entire area, the
entire area under the curve.

00:01:51.820 --> 00:01:54.520
But what we could do
is introduce a constant

00:01:54.520 --> 00:01:56.610
that's someplace in
between x and x squared.

00:01:56.610 --> 00:02:00.230
Let's say that constant
is c, and break this area

00:02:00.230 --> 00:02:03.210
into two different areas
with c as the divider.

00:02:03.210 --> 00:02:05.150
So that same exact
whole area we can now

00:02:05.150 --> 00:02:06.940
write it as two
separate integrals.

00:02:06.940 --> 00:02:10.360
So one integral that
represents this area right over

00:02:10.360 --> 00:02:14.562
here, and then another integral
that represents this area

00:02:14.562 --> 00:02:16.020
right over there,
and where we just

00:02:16.020 --> 00:02:19.520
say c is some constant
between x and x squared.

00:02:19.520 --> 00:02:22.269
Well, how can we denote
this area in purple?

00:02:22.269 --> 00:02:24.310
Well, that's going to be--
So this thing is going

00:02:24.310 --> 00:02:26.250
to be equal to the sum
of these two areas.

00:02:26.250 --> 00:02:30.500
The purple area we can show
is the definite integral

00:02:30.500 --> 00:02:39.240
from x to c of our function
of t, cosine t over t dt.

00:02:39.240 --> 00:02:42.100
And then to that we're
going to add the green area.

00:02:42.100 --> 00:02:44.100
And then we'll get
the original area.

00:02:44.100 --> 00:02:48.020
So for the green area, our
lower bound of integration

00:02:48.020 --> 00:02:51.590
is now our constant c, and
our upper bound of integration

00:02:51.590 --> 00:03:00.630
is x squared, and it's going
to be of cosine t over t dt.

00:03:00.630 --> 00:03:03.280
And this is a form where, if
we know how to apply the chain

00:03:03.280 --> 00:03:05.660
rule, we can apply the
fundamental theorem

00:03:05.660 --> 00:03:06.690
of calculus.

00:03:06.690 --> 00:03:08.260
And this is almost in a form.

00:03:08.260 --> 00:03:10.552
We're used to seeing it where
the x is the upper bound.

00:03:10.552 --> 00:03:12.218
And, well, we already
know what happens.

00:03:12.218 --> 00:03:14.210
We can swap these two
bounds, but it'll just

00:03:14.210 --> 00:03:15.760
be the negative
of that integral.

00:03:15.760 --> 00:03:18.850
So this is going to be equal
to-- let me rewrite it--

00:03:18.850 --> 00:03:24.360
the negative of the
definite integral from c

00:03:24.360 --> 00:03:30.170
to x of cosine t over t dt.

00:03:30.170 --> 00:03:33.330
And then we have plus
the definite integral

00:03:33.330 --> 00:03:41.497
that goes from c to x squared
of cosine t over t dt.

00:03:41.497 --> 00:03:43.830
So all we've done is we've
rewritten this thing in a way

00:03:43.830 --> 00:03:46.930
that we're used to applying
the fundamental theorem

00:03:46.930 --> 00:03:48.090
of calculus.

00:03:48.090 --> 00:03:52.579
So if we want to find
F prime of x, well,

00:03:52.579 --> 00:03:54.370
applying the derivative
operator over here,

00:03:54.370 --> 00:03:56.078
we're going to have
a negative out front.

00:03:56.078 --> 00:04:00.530
It's going to be equal to
negative cosine x over x.

00:04:03.520 --> 00:04:06.360
Once again, just the
fundamental theorem of calculus.

00:04:06.360 --> 00:04:08.830
And then plus--
we're first going

00:04:08.830 --> 00:04:12.000
to take the derivative of
this thing with respect

00:04:12.000 --> 00:04:14.790
to x squared, and
that's going to give you

00:04:14.790 --> 00:04:18.480
cosine of x squared
over x squared.

00:04:18.480 --> 00:04:21.057
Wherever you saw t, you
replace it with an x squared.

00:04:21.057 --> 00:04:22.890
And then you're going
to multiply that times

00:04:22.890 --> 00:04:25.530
the derivative of x
squared with respect to x.

00:04:25.530 --> 00:04:27.420
So that's just going
to be-- derivative

00:04:27.420 --> 00:04:31.075
of x squared with
respect to x is just 2x.

00:04:31.075 --> 00:04:31.700
And we're done.

00:04:31.700 --> 00:04:33.350
We just need to
simplify this thing.

00:04:33.350 --> 00:04:37.400
So all of this is going to
be equal to negative cosine

00:04:37.400 --> 00:04:41.750
x over x plus-- well, this
is going to cancel out

00:04:41.750 --> 00:04:48.650
with just one of those-- plus
2 cosine x squared over x.

00:04:48.650 --> 00:04:51.540
And I guess we could simplify
it even more as being equal to--

00:04:51.540 --> 00:04:56.590
and we can swap these--
everything over x 2 cosine

00:04:56.590 --> 00:05:01.730
of x squared minus cosine of x.

00:05:01.730 --> 00:05:03.790
And we are done.


Finding derivative with fundamental theorem of calculus: x is on lower bound

WEBVTT
Kind: captions
Language: en

00:00:00.510 --> 00:00:03.000
We want to find the
derivative with respect

00:00:03.000 --> 00:00:05.920
to x of all of this
business right over here.

00:00:05.920 --> 00:00:08.880
And you might guess-- and this
is definitely a function of x.

00:00:08.880 --> 00:00:11.130
x is one of the
boundaries of integration

00:00:11.130 --> 00:00:12.660
for this definite integral.

00:00:12.660 --> 00:00:14.170
And you might say,
well, it looks

00:00:14.170 --> 00:00:17.080
like the fundamental theorem
of calculus might apply,

00:00:17.080 --> 00:00:19.150
but I'm used to seeing
the x, or the function

00:00:19.150 --> 00:00:22.320
x, as the upper bound,
not as the lower bound.

00:00:22.320 --> 00:00:25.120
How do I deal with this?

00:00:25.120 --> 00:00:28.550
And the key realization
is to realize

00:00:28.550 --> 00:00:32.000
what happens when you switch
bounds for a definite integral.

00:00:32.000 --> 00:00:34.730
And I'll do a little bit
of an aside to review that.

00:00:34.730 --> 00:00:41.610
So if I'm taking the definite
integral from a to b of f of t,

00:00:41.610 --> 00:00:46.260
dt, we know that this is capital
F, the antiderivative of f,

00:00:46.260 --> 00:00:50.710
evaluated at b minus the
antiderivative of F evaluated

00:00:50.710 --> 00:00:52.035
at a.

00:00:52.035 --> 00:00:54.100
This is corollary to
the fundamental theorem,

00:00:54.100 --> 00:00:55.808
or it's the fundamental
theorem part two,

00:00:55.808 --> 00:00:58.500
or the second fundamental
theorem of calculus.

00:00:58.500 --> 00:01:01.910
This is how we evaluate
definite integrals.

00:01:01.910 --> 00:01:04.819
Now, let's think about what
the negative of this is.

00:01:04.819 --> 00:01:09.610
So the negative of that--
of a to b of f of t,

00:01:09.610 --> 00:01:13.770
dt, is just going to be equal
to the negative of this, which

00:01:13.770 --> 00:01:18.160
is equal to-- so it's the
negative of f of b minus f

00:01:18.160 --> 00:01:25.020
of a, which is equal to capital
F of a minus capital F of b.

00:01:25.020 --> 00:01:26.930
All I did is distribute
the negative sign

00:01:26.930 --> 00:01:28.750
and then switch the two terms.

00:01:28.750 --> 00:01:32.170
But this right over here is
equal to the definite integral

00:01:32.170 --> 00:01:41.430
from, instead of a to b, but
from b to a of f of t, dt.

00:01:41.430 --> 00:01:43.370
So notice, when
you put a negative,

00:01:43.370 --> 00:01:44.980
that's just like
switching the signs

00:01:44.980 --> 00:01:46.160
or switching the boundaries.

00:01:46.160 --> 00:01:47.701
Or if you switch
the boundaries, they

00:01:47.701 --> 00:01:49.340
are the negatives of each other.

00:01:49.340 --> 00:01:51.880
So we can go back to
our original problem.

00:01:51.880 --> 00:01:54.730
We can rewrite this as being
equal to the derivative with

00:01:54.730 --> 00:02:00.460
respect to x of--
instead of this,

00:02:00.460 --> 00:02:03.950
it'll be the negative of the
same definite integral but with

00:02:03.950 --> 00:02:08.479
the boundaries switched-- the
negative of x with the upper

00:02:08.479 --> 00:02:11.950
boundary is x, the lower
bound is 3 of the square root

00:02:11.950 --> 00:02:15.700
of the absolute value
of cosine t, dt,

00:02:15.700 --> 00:02:18.130
which is equal to--we can
take the negative out front--

00:02:18.130 --> 00:02:22.600
negative times the derivative
with respect to x of all

00:02:22.600 --> 00:02:23.410
of this business.

00:02:23.410 --> 00:02:25.420
I should just copy
and paste that,

00:02:25.420 --> 00:02:28.000
so I'll just copy and paste.

00:02:28.000 --> 00:02:29.690
Let me-- and paste it.

00:02:29.690 --> 00:02:33.270
So times the derivative with
respect to x of all that,

00:02:33.270 --> 00:02:37.360
and now the fundamental theorem
of calculus directly applies.

00:02:37.360 --> 00:02:40.699
This is going to be equal to--
we deserve a drum roll now.

00:02:40.699 --> 00:02:42.490
This is going to be
equal to the negative--

00:02:42.490 --> 00:02:43.410
can't forget the negative.

00:02:43.410 --> 00:02:44.900
And the fundamental
theorem of calculus

00:02:44.900 --> 00:02:46.490
tells us that that's
just going to be

00:02:46.490 --> 00:02:49.440
this function as
a function of x.

00:02:49.440 --> 00:02:52.240
So it's going to be
negative square root

00:02:52.240 --> 00:02:57.580
of the absolute value of
cosine of not t anymore, but x.

00:02:57.580 --> 00:02:59.860
And we are done.


Worked example: Finding derivative with fundamental theorem of calculus

WEBVTT
Kind: captions
Language: en

00:00:00.660 --> 00:00:03.210
Let's say that we
have capital F of x

00:00:03.210 --> 00:00:06.420
as being equal to the
definite integral from pi

00:00:06.420 --> 00:00:11.400
to x of cotangent
squared of t dt.

00:00:11.400 --> 00:00:13.450
And what we're
curious about finding

00:00:13.450 --> 00:00:15.750
is the derivative
of this business.

00:00:15.750 --> 00:00:21.000
We want to figure out what
F prime of x is equal to.

00:00:21.000 --> 00:00:22.790
Well this is a
direct application

00:00:22.790 --> 00:00:24.704
of the fundamental
theorem of calculus.

00:00:24.704 --> 00:00:26.870
This is just going to be
the derivative with respect

00:00:26.870 --> 00:00:29.140
to x of all of this craziness.

00:00:29.140 --> 00:00:30.910
So let me just copy
and paste that.

00:00:33.624 --> 00:00:35.790
So it's just going to be
the derivative with respect

00:00:35.790 --> 00:00:36.580
to that.

00:00:36.580 --> 00:00:38.740
And the fundamental
theorem of calculus

00:00:38.740 --> 00:00:46.229
tells us that this is going to
be equal to just this function,

00:00:46.229 --> 00:00:48.270
but it's not going to be
a function of t anymore.

00:00:48.270 --> 00:00:51.210
It's going to be
a function of x.

00:00:51.210 --> 00:00:56.330
So it's going to be equal
to cotangent squared of x.

00:00:56.330 --> 00:00:58.100
And you'll often see
problems like this

00:00:58.100 --> 00:01:00.170
if you go to calculus
competitions or things

00:01:00.170 --> 00:01:01.895
like that and, or
maybe certain exams.

00:01:01.895 --> 00:01:04.790
And It's like, oh my god, I
have to take the antiderivative

00:01:04.790 --> 00:01:06.859
of all of this
business, evaluate it

00:01:06.859 --> 00:01:08.400
at the different
boundaries, and then

00:01:08.400 --> 00:01:09.470
I got to take the derivative.

00:01:09.470 --> 00:01:09.970
No.

00:01:09.970 --> 00:01:12.510
You just apply the fundamental
theorem of calculus,

00:01:12.510 --> 00:01:15.520
and it's actually a very
straightforward and a very fast

00:01:15.520 --> 00:01:16.800
thing to do.

00:01:16.800 --> 00:01:19.650
Now let's mix it
up a little bit.

00:01:19.650 --> 00:01:23.260
Let's say that you
had the expression

00:01:23.260 --> 00:01:26.240
the definite integral from
pi-- instead of from pi

00:01:26.240 --> 00:01:28.654
to x, let's say it's
from pi to x squared--

00:01:28.654 --> 00:01:30.820
and let me write that x
squared in a different color

00:01:30.820 --> 00:01:38.040
just to make it clear what I'm
doing-- from pi to x squared

00:01:38.040 --> 00:01:43.720
of cotangent squared of t dt.

00:01:43.720 --> 00:01:47.380
And you wanted to take the
derivative of this business.

00:01:47.380 --> 00:01:50.050
So you want to take the
derivative with respect

00:01:50.050 --> 00:01:54.020
to x of this business.

00:01:54.020 --> 00:01:55.430
How would you do it?

00:01:55.430 --> 00:02:00.070
Well the recognition here is
that this is capital-- capital

00:02:00.070 --> 00:02:01.910
F of x was defined as this.

00:02:01.910 --> 00:02:03.970
Now instead of an x,
you have an x squared.

00:02:03.970 --> 00:02:06.303
So this is the exact same
thing as taking the derivative

00:02:06.303 --> 00:02:12.740
with respect to x of
capital F of-- not x.

00:02:12.740 --> 00:02:13.970
That would be this.

00:02:13.970 --> 00:02:16.190
Instead, we have
capital F of x squared.

00:02:16.190 --> 00:02:20.100
Where we had an x here before,
we now have an x squared.

00:02:20.100 --> 00:02:20.910
You can verify.

00:02:20.910 --> 00:02:23.140
If you had capital
F of x squared,

00:02:23.140 --> 00:02:25.720
every place where you saw the
x here would be an x squared.

00:02:25.720 --> 00:02:28.370
So it would look
exactly like this.

00:02:28.370 --> 00:02:31.580
And so we just have to
apply the chain rule.

00:02:31.580 --> 00:02:38.840
So this is going to be
equal to the derivative of F

00:02:38.840 --> 00:02:42.341
with respect to x squared.

00:02:42.341 --> 00:02:44.340
And this is just straight
out of the chain rule.

00:02:44.340 --> 00:02:51.760
It's going to be F
prime of x squared

00:02:51.760 --> 00:02:58.840
times the derivative with
respect to x of x squared.

00:02:58.840 --> 00:03:02.490
This is the derivative of
F with respect to x squared

00:03:02.490 --> 00:03:05.300
and then times the derivative
of x squared with respect to x.

00:03:05.300 --> 00:03:08.610
This is just the chain
rule right over here.

00:03:08.610 --> 00:03:11.510
So what's F prime of x
squared-- the derivative of F

00:03:11.510 --> 00:03:13.310
with respect to x squared?

00:03:13.310 --> 00:03:16.710
Well, if you evaluate F prime
at x squared, instead of just

00:03:16.710 --> 00:03:18.930
being cotangent
squared of x, it's

00:03:18.930 --> 00:03:21.820
going to be cotangent
squared of x squared.

00:03:21.820 --> 00:03:24.890
So this part right over here,
this business right over here,

00:03:24.890 --> 00:03:28.753
is going to be cotangent
squared of x squared.

00:03:31.270 --> 00:03:34.770
So this is the derivative of all
of this business with respect

00:03:34.770 --> 00:03:39.430
to x squared, and then you
have to multiply that times

00:03:39.430 --> 00:03:43.720
the derivative of x squared with
respect to x, which is just 2x.

00:03:43.720 --> 00:03:45.000
So there you have it.

00:03:45.000 --> 00:03:46.940
The derivative of this,
all this craziness,

00:03:46.940 --> 00:03:52.743
is equal to 2x cotangent
squared of x squared.


Proof of fundamental theorem of calculus

WEBVTT
Kind: captions
Language: en

00:00:00.560 --> 00:00:02.940
Let's say that we've
got some function

00:00:02.940 --> 00:00:09.410
f that is continuous
on the interval a to b.

00:00:09.410 --> 00:00:12.480
So let's try to see if
we can visualize that.

00:00:12.480 --> 00:00:14.816
So this is my y-axis.

00:00:17.680 --> 00:00:20.600
This right over here, I
want to make it my t-axis.

00:00:20.600 --> 00:00:23.050
We'll use x little bit later.

00:00:23.050 --> 00:00:24.820
So I'll call this my t-axis.

00:00:24.820 --> 00:00:26.720
And then let's say that
this right over here

00:00:26.720 --> 00:00:29.025
is the graph of y
is equal to f of t.

00:00:31.830 --> 00:00:34.670
And we're saying it's continuous
on the interval from a to b.

00:00:34.670 --> 00:00:36.880
So this is t is equal to a.

00:00:36.880 --> 00:00:38.800
This is t is equal to b.

00:00:38.800 --> 00:00:42.320
So we're saying that
it is continuous

00:00:42.320 --> 00:00:45.350
over this whole interval.

00:00:45.350 --> 00:00:51.580
Now, for fun, let's define
a function capital F of x.

00:00:51.580 --> 00:00:53.670
And I will do it in blue.

00:00:53.670 --> 00:00:58.810
Let's define capital F of x as
equal to the definite integral

00:00:58.810 --> 00:01:06.630
from a is a lower
bound to x of f of t--

00:01:06.630 --> 00:01:12.460
let me do that in--
of f of t dt, where

00:01:12.460 --> 00:01:19.500
x is in this interval, where
a is less than or equal to x

00:01:19.500 --> 00:01:21.371
is less than or equal to b.

00:01:21.371 --> 00:01:22.870
Or that's just
another way of saying

00:01:22.870 --> 00:01:26.390
that x is in this
interval right over here.

00:01:26.390 --> 00:01:28.360
Now, when you see this,
you might say, oh,

00:01:28.360 --> 00:01:31.017
the definite integral, this
has to do with differentiation

00:01:31.017 --> 00:01:32.350
an antiderivatives and all that.

00:01:32.350 --> 00:01:33.910
But we don't know that yet.

00:01:33.910 --> 00:01:36.890
All we know right
now is that this

00:01:36.890 --> 00:01:43.150
is the area under the curve f
between a and x, so between a,

00:01:43.150 --> 00:01:47.250
and let's say this
right over here is x.

00:01:47.250 --> 00:01:53.620
So f of x is just this
area right over here.

00:01:53.620 --> 00:01:55.660
That's all we know about it.

00:01:55.660 --> 00:01:58.560
We don't know it has anything
to do with antiderivatives just

00:01:58.560 --> 00:01:59.060
yet.

00:01:59.060 --> 00:02:02.800
That's what we're going to
try to prove in this video.

00:02:02.800 --> 00:02:06.312
So just for fun, let's
take the derivative of f.

00:02:06.312 --> 00:02:07.770
And we're going to
do it just using

00:02:07.770 --> 00:02:09.949
the definition of
derivatives and see

00:02:09.949 --> 00:02:12.520
what we get when we take
the derivative using

00:02:12.520 --> 00:02:15.950
the definition of derivatives.

00:02:15.950 --> 00:02:20.230
So the derivative, f prime
of x-- well this definition

00:02:20.230 --> 00:02:24.790
of derivatives, it's the
limit as delta x approaches

00:02:24.790 --> 00:02:31.940
0 of capital F of x
plus delta x minus f

00:02:31.940 --> 00:02:37.620
of x, all of that over delta x.

00:02:37.620 --> 00:02:41.450
This is just the definition
of the derivative.

00:02:41.450 --> 00:02:43.611
Now, what is this equal to?

00:02:43.611 --> 00:02:46.110
Well, let me rewrite it using
these integrals right up here.

00:02:46.110 --> 00:02:51.620
This is going to be
equal to the limit

00:02:51.620 --> 00:02:58.790
as delta x approaches 0 of--
what's f of x plus delta x?

00:02:58.790 --> 00:03:00.670
Well, put x in right over here.

00:03:00.670 --> 00:03:05.330
You're going to get the definite
integral from a to x plus delta

00:03:05.330 --> 00:03:09.360
x of f of t dt.

00:03:09.360 --> 00:03:10.990
And then from
that, you are going

00:03:10.990 --> 00:03:17.450
to subtract this business,
f of x, which we've already

00:03:17.450 --> 00:03:24.870
written as the definite integral
from a to x of f of t dt,

00:03:24.870 --> 00:03:26.700
and then all of that
is over delta x.

00:03:32.740 --> 00:03:34.866
Now, what does this represent?

00:03:34.866 --> 00:03:37.240
Remember, we don't know anything
about definite integrals

00:03:37.240 --> 00:03:38.560
or somehow dealing
with something

00:03:38.560 --> 00:03:40.060
with an antiderivative
and all that.

00:03:40.060 --> 00:03:41.870
We just know this is
another way of saying

00:03:41.870 --> 00:03:45.630
the area under the curve f
between a and x plus delta x.

00:03:53.630 --> 00:04:00.620
So it's this entire
area right over here.

00:04:00.620 --> 00:04:02.460
So that's this part.

00:04:02.460 --> 00:04:06.090
We already know what
this blue stuff is.

00:04:06.090 --> 00:04:08.830
Let me do it in that
same shade of blue.

00:04:08.830 --> 00:04:11.420
So this blue stuff
right over here,

00:04:11.420 --> 00:04:14.650
this is equal to all
of this business.

00:04:14.650 --> 00:04:15.920
We've already shaded this in.

00:04:15.920 --> 00:04:19.070
It's equal to all of this
business right over here.

00:04:19.070 --> 00:04:22.130
So if you were to take all
of this green area, which

00:04:22.130 --> 00:04:24.730
is from a to x plus
delta x, and subtract out

00:04:24.730 --> 00:04:26.500
the blue area, which
is exactly what we're

00:04:26.500 --> 00:04:29.420
doing in the numerator
what are you're left with?

00:04:29.420 --> 00:04:32.450
Well, you're going to
be left with-- what

00:04:32.450 --> 00:04:34.280
color have I not used yet?

00:04:34.280 --> 00:04:36.381
Maybe I will use
this pink color.

00:04:36.381 --> 00:04:37.630
Well, no, I already used that.

00:04:37.630 --> 00:04:38.754
I'll use this purple color.

00:04:38.754 --> 00:04:43.340
You're going to be left with
this area right over here.

00:04:43.340 --> 00:04:45.490
So what's another
way of writing that?

00:04:45.490 --> 00:04:48.450
Well, another way of writing
this area right over here

00:04:48.450 --> 00:04:51.870
is the definite
integral between x and x

00:04:51.870 --> 00:04:58.550
plus delta x of f of t dt.

00:04:58.550 --> 00:05:01.350
So we can rewrite this entire
expression, the derivative

00:05:01.350 --> 00:05:04.020
of capital F of x--
this is capital F

00:05:04.020 --> 00:05:08.960
prime of x-- we can rewrite it
now as being equal to the limit

00:05:08.960 --> 00:05:13.530
as delta x approaches
0-- this I can write as 1

00:05:13.530 --> 00:05:16.932
over delta x times
the numerator.

00:05:16.932 --> 00:05:18.640
And we already figured
out the numerator.

00:05:18.640 --> 00:05:22.170
The green area minus the blue
area is just the purple area,

00:05:22.170 --> 00:05:24.510
and another way of
denoting that area

00:05:24.510 --> 00:05:26.520
is this expression
right over here.

00:05:26.520 --> 00:05:28.680
So 1 over delta x times
the definite integral

00:05:28.680 --> 00:05:38.200
from x to x plus
delta x of f of t dt.

00:05:38.200 --> 00:05:40.650
Now, this expression
is interesting.

00:05:40.650 --> 00:05:43.680
This might look familiar
from the mean value

00:05:43.680 --> 00:05:46.220
theorem of definite integrals.

00:05:46.220 --> 00:05:48.570
The mean value theorem
of definite integrals

00:05:48.570 --> 00:06:10.360
tells us there exists
a c in the interval

00:06:10.360 --> 00:06:12.990
see where-- I'll
write it this way--

00:06:12.990 --> 00:06:18.394
where a is less than or equal
to c, which is less than--

00:06:18.394 --> 00:06:19.810
or actually, let
me make it clear.

00:06:19.810 --> 00:06:22.110
The interval that we care
about is between x and x

00:06:22.110 --> 00:06:26.620
plus delta x--
where x is less than

00:06:26.620 --> 00:06:28.370
or equal to c,
which is less than

00:06:28.370 --> 00:06:36.940
or equal to x plus delta x,
such that the function evaluated

00:06:36.940 --> 00:06:40.710
at c-- so let me draw this c.

00:06:40.710 --> 00:06:43.210
So there's a c
someplace over here--

00:06:43.210 --> 00:06:46.240
so if I were take the
function evaluated at this c--

00:06:46.240 --> 00:06:48.999
so that's f of c
right over here.

00:06:48.999 --> 00:06:50.790
So if I were to take
the function evaluated

00:06:50.790 --> 00:06:52.206
at this c, which
would essentially

00:06:52.206 --> 00:06:55.520
be the height of this line,
and I multiply times the base,

00:06:55.520 --> 00:06:57.960
this interval, if I multiply
it times the interval--

00:06:57.960 --> 00:06:59.860
and this interval
is just delta x. x

00:06:59.860 --> 00:07:02.420
plus delta x minus
x is just delta x.

00:07:02.420 --> 00:07:07.140
So if we just multiply
the height times the base,

00:07:07.140 --> 00:07:14.180
this is going to be equal to
the area under the curve, which

00:07:14.180 --> 00:07:24.010
is the definite integral from x
to x plus delta x of f of t dt.

00:07:24.010 --> 00:07:27.320
This is what the mean value
theorem of integrals tells us.

00:07:27.320 --> 00:07:29.160
If f is a continuous
function, there

00:07:29.160 --> 00:07:34.460
exists a c in this interval
between our two end points

00:07:34.460 --> 00:07:37.820
where the function evaluated
at the c is essentially,

00:07:37.820 --> 00:07:39.530
you could view it
as the mean height.

00:07:39.530 --> 00:07:41.650
And if you take that mean
value of the function

00:07:41.650 --> 00:07:43.180
and you multiply
it times the base,

00:07:43.180 --> 00:07:44.810
you're going to get
the area of the curve.

00:07:44.810 --> 00:07:46.184
Or another way of
rewriting this,

00:07:46.184 --> 00:07:49.770
you could say that there exists
a c in that interval where

00:07:49.770 --> 00:07:53.450
f of c is equal to 1 over
delta x-- I'm just dividing

00:07:53.450 --> 00:07:57.840
both sides by delta x-- times
the definite integral from x

00:07:57.840 --> 00:08:02.530
to x plus delta x of f of t dt.

00:08:02.530 --> 00:08:05.024
And this is often
viewed as the mean value

00:08:05.024 --> 00:08:06.440
of the function
over the interval.

00:08:06.440 --> 00:08:07.410
Why is that?

00:08:07.410 --> 00:08:11.410
Well, this part right over
here gives you the area,

00:08:11.410 --> 00:08:13.170
and then you divide
the area by the base,

00:08:13.170 --> 00:08:14.892
and you get the mean height.

00:08:14.892 --> 00:08:16.350
Or another way you
could say it is,

00:08:16.350 --> 00:08:18.850
if you were to take the height
right over here, multiply it

00:08:18.850 --> 00:08:20.790
times the base, you
get a rectangle that

00:08:20.790 --> 00:08:24.560
has the exact same area as
the area under the curve.

00:08:24.560 --> 00:08:26.330
Well, this is
useful, because this

00:08:26.330 --> 00:08:30.430
is exactly what we got as the
derivative of f prime of x.

00:08:30.430 --> 00:08:34.330
So there must exist
a c such that f of c

00:08:34.330 --> 00:08:35.370
is equal to this stuff.

00:08:35.370 --> 00:08:38.770
Or we could say that the limit--
and let me rewrite all of this

00:08:38.770 --> 00:08:39.919
now in a new color.

00:08:39.919 --> 00:08:47.640
So there exists a
c in the interval

00:08:47.640 --> 00:08:56.250
x to x plus delta x where
f prime of x, which we know

00:08:56.250 --> 00:09:00.330
is equal to this, we can now
say is now equal to the limit

00:09:00.330 --> 00:09:02.920
as delta x approaches 0.

00:09:02.920 --> 00:09:04.420
And instead of
writing this, we know

00:09:04.420 --> 00:09:07.920
that there's some c that's
equal to all of this business,

00:09:07.920 --> 00:09:10.420
of f of c.

00:09:10.420 --> 00:09:12.370
Now we're in the home stretch.

00:09:12.370 --> 00:09:14.870
We just have to figure out
what the limit as delta x

00:09:14.870 --> 00:09:17.790
approaches 0 of f of c is.

00:09:17.790 --> 00:09:20.570
And the main realization is
this part right over here.

00:09:20.570 --> 00:09:24.690
We know that c is always
sandwiched in between x and x

00:09:24.690 --> 00:09:25.960
plus delta x.

00:09:25.960 --> 00:09:27.750
And intuitively, you
can tell that, look,

00:09:27.750 --> 00:09:33.450
as delta x approaches 0, as
this green line right over here

00:09:33.450 --> 00:09:36.800
moves more and more to
the left, as it approaches

00:09:36.800 --> 00:09:43.750
this blue line, the c
has to be in between,

00:09:43.750 --> 00:09:46.280
and so the c is
going to approach x.

00:09:46.280 --> 00:09:51.180
So we know intuitively
that c approaches

00:09:51.180 --> 00:09:56.890
x as delta x approaches 0.

00:09:56.890 --> 00:10:00.120
Or another way of
saying it is that f of c

00:10:00.120 --> 00:10:07.200
is going to approach f of
x as delta x approaches 0.

00:10:07.200 --> 00:10:09.350
And so intuitively,
we could say that this

00:10:09.350 --> 00:10:14.129
is going to be equal to f of x.

00:10:14.129 --> 00:10:15.920
Now, you might say,
OK, that's intuitively,

00:10:15.920 --> 00:10:18.086
but we're kind of working
on a little bit of a proof

00:10:18.086 --> 00:10:18.880
here, Sal.

00:10:18.880 --> 00:10:21.310
Let me know for sure that
x is going to approach c.

00:10:21.310 --> 00:10:23.846
Don't just do this little thing
where you drew this diagram

00:10:23.846 --> 00:10:25.220
and it makes sense
that c's going

00:10:25.220 --> 00:10:26.950
to have to get closer
and closer to x.

00:10:26.950 --> 00:10:28.650
And if you want
that, you could just

00:10:28.650 --> 00:10:30.449
resort to the squeeze theorem.

00:10:30.449 --> 00:10:31.990
And to resort to
the squeeze theorem,

00:10:31.990 --> 00:10:34.610
you just have to view c
as a function of delta x.

00:10:34.610 --> 00:10:35.340
And it really is.

00:10:35.340 --> 00:10:37.923
Depending on your delta x, c's
going to be further to the left

00:10:37.923 --> 00:10:39.470
or to the right, possibly.

00:10:39.470 --> 00:10:41.450
And so I can just
rewrite this expression

00:10:41.450 --> 00:10:46.980
as x is less than or equal to c
as a function of delta x, which

00:10:46.980 --> 00:10:50.030
is less than or equal
to x plus delta x.

00:10:50.030 --> 00:10:53.020
So now you see that c is always
sandwiched between x and x

00:10:53.020 --> 00:10:54.320
plus delta x.

00:10:54.320 --> 00:10:58.775
But what's the limit of x
as delta x approaches 0?

00:10:58.775 --> 00:11:01.040
Well, x isn't dependent
on delta x in any way,

00:11:01.040 --> 00:11:03.500
so this is just going
to be equal to x.

00:11:03.500 --> 00:11:10.600
What's the limit of x plus
delta x as delta x approaches 0?

00:11:10.600 --> 00:11:12.200
Well, as delta x
approaches 0, this

00:11:12.200 --> 00:11:14.150
is just going to be equal to x.

00:11:14.150 --> 00:11:17.370
So if this approaches
as delta x approaches 0,

00:11:17.370 --> 00:11:19.190
and it's less than
this function,

00:11:19.190 --> 00:11:22.080
and if this approaches x
as delta x approaches 0,

00:11:22.080 --> 00:11:23.950
and it's always
greater than this,

00:11:23.950 --> 00:11:27.000
then we know from the squeeze
theorem or the sandwich theorem

00:11:27.000 --> 00:11:29.860
that the limit as
delta x approaches

00:11:29.860 --> 00:11:34.180
0 of c as a function
of delta x is

00:11:34.180 --> 00:11:38.480
going to be equal to x as well.

00:11:38.480 --> 00:11:40.940
It has to approach the same
thing that that and that is.

00:11:40.940 --> 00:11:42.950
It's sandwiched in between.

00:11:42.950 --> 00:11:45.160
And so that's a slight--
we resort to the sandwich

00:11:45.160 --> 00:11:46.951
theorem-- it's a little
bit more rigorous--

00:11:46.951 --> 00:11:48.580
to get to this exact result.

00:11:48.580 --> 00:11:52.820
As delta x approaches
0, c approaches x.

00:11:52.820 --> 00:11:57.910
If c is approaching x, then f of
c is going to approach f of x.

00:11:57.910 --> 00:12:00.850
And then we essentially
have our proof.

00:12:00.850 --> 00:12:02.250
F is a continuous function.

00:12:02.250 --> 00:12:06.820
We defined capital
F in this way,

00:12:06.820 --> 00:12:10.500
and we were able to use just
the definition of the derivative

00:12:10.500 --> 00:12:14.130
to figure out that the
derivative of capital F of x

00:12:14.130 --> 00:12:21.500
is equal to f of x.

00:12:21.500 --> 00:12:25.080
And once again, why
is this a big deal?

00:12:25.080 --> 00:12:27.770
Well, it tells you that if you
have any continuous function

00:12:27.770 --> 00:12:29.090
f-- and that's what we assume.

00:12:29.090 --> 00:12:32.880
We assume that f is continuous
over the interval-- there

00:12:32.880 --> 00:12:35.460
exists some function--
you can just

00:12:35.460 --> 00:12:37.340
define the function
this way as the area

00:12:37.340 --> 00:12:41.170
under the curve between some
endpoint, or the beginning

00:12:41.170 --> 00:12:43.450
of the interval, and
sum x-- if you define

00:12:43.450 --> 00:12:46.190
a function in that way, the
derivative of this function

00:12:46.190 --> 00:12:49.130
is going to be equal to
your continuous function.

00:12:49.130 --> 00:12:51.610
Or another way of saying
it is that you always

00:12:51.610 --> 00:12:55.020
have an antiderivative, that
any continuous function has

00:12:55.020 --> 00:12:56.440
an antiderivative.

00:12:56.440 --> 00:12:58.070
And so it's a couple
of cool things.

00:12:58.070 --> 00:13:00.250
Any continuous function
has an antiderivative.

00:13:00.250 --> 00:13:03.925
It's going to be
that capital F of x.

00:13:03.925 --> 00:13:06.050
And this is why it's called
the fundamental theorem

00:13:06.050 --> 00:13:07.090
of calculus.

00:13:07.090 --> 00:13:09.500
It ties together
these two ideas.

00:13:09.500 --> 00:13:11.250
And you have
differential calculus.

00:13:11.250 --> 00:13:14.760
You have the idea
of a derivative.

00:13:14.760 --> 00:13:16.210
And then in integral
calculus, you

00:13:16.210 --> 00:13:17.910
have the idea of an integral.

00:13:17.910 --> 00:13:20.670
Before this proof, all
we viewed an integral

00:13:20.670 --> 00:13:22.596
as is the area under the curve.

00:13:22.596 --> 00:13:24.220
It was just literally
a notation to say

00:13:24.220 --> 00:13:26.330
the area under the curve.

00:13:26.330 --> 00:13:29.050
But now we've been able to
make a connection that there's

00:13:29.050 --> 00:13:31.670
a connection between the
integral and the derivative,

00:13:31.670 --> 00:13:34.680
or a connection between the
integral and the antiderivative

00:13:34.680 --> 00:13:35.510
in particular.

00:13:35.510 --> 00:13:40.170
So it connects all of calculus
together in a very, very, very

00:13:40.170 --> 00:13:42.490
powerful-- and we're
so used to it now,

00:13:42.490 --> 00:13:44.694
and now we can say almost
a somewhat obvious way,

00:13:44.694 --> 00:13:45.610
but it wasn't obvious.

00:13:45.610 --> 00:13:47.060
Remember, we always
think of integrals

00:13:47.060 --> 00:13:48.518
as somehow doing
an antiderivative,

00:13:48.518 --> 00:13:49.620
but it wasn't clear.

00:13:49.620 --> 00:13:51.662
If you just viewed an
integral as only an area,

00:13:51.662 --> 00:13:53.370
you would have to go
through this process

00:13:53.370 --> 00:13:57.600
to say, wow, no, it's connected
to the process of taking

00:13:57.600 --> 00:13:59.450
a derivative.