WEBVTT

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Hi.

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Welcome back to recitation.

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In lecture, you've learned a
whole bunch of different

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integration techniques.

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So I have here a couple of
problems that will give you an

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opportunity to apply those
techniques, and figure out

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which ones to apply.

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So in particular I have one
definite integral, the

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integral 0 to 2 of the quantity
x times e to the 1

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minus x squared with
respect to x.

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And the second one is an
indefinite integral of 2 arc

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tan x divided by x squared
with respect to x.

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So why don't you pause the
video, take some time to work

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these out, come back, and we
can work them out together.

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Welcome back.

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Hopefully you had some luck
working on these integrals.

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Let's get started.

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We can do the first one first,
and then we'll go on to the

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second one.

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So here we want to compute the
integral from 0 to 2 of x e to

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the 1 minus x squared dx.

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So in order to do this,
we have to, you know--

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well, we could look at it and
say, do I know the answer

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immediately, off the
top of my head?

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And this is not one of the, you
know, ones that we have

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immediately already memorized
a formula for.

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So then you say, OK.

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So now I need to do something
in order to try

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to integrate it.

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And so, you know, you have a
whole bunch of techniques.

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And for this one, when you look
at it, the technique that

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is going to work for you is just
the simplest technique

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that you have, which is just a
straightforward substitution.

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And one way to figure that out,
is that you can look at

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this and you can see that here
we have a composition of

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functions, this e to the 1
minus x squared, and so

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whenever you have a composition
of functions,

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there's the opportunity for
there to have been a chain

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rule somewhere.

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And in this case, we see that
this composition of functions

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is multiplied by x, and x is
very closely related to the

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derivative of 1 minus
x squared.

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So that's what's suggesting for
us a nice substitution.

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And so the substitution we're
going to try, then, is that

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we're going to put this inside
function, this 1 minus x

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squared, we're going to
set u equal to that.

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So, all right.

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So this is equal to-- so
I'm going to say, u

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equal 1 minus x squared.

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So now if I take a differential,
if u equals 1

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minus x squared, that means
that du is minus 2x dx.

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And since this is a definite
integral, I also need to

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change the bounds
of integration.

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So when x is equal to 0, that
means that u is equal to,

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well, 1 minus 0 squared is 1,
so that's the lower bound.

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And when x is equal to 2, that
means that u is equal to 1

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minus 2 squared, which
is negative 3.

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So we make these
substitutions.

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OK.

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So what does our integral
become?

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So this becomes the integral--

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so now the lower bound becomes
u equals 1, and the upper

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bound is u equals negative 3.

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OK.

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So now e to the 1 minus
x squared, that's

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just e to the u.

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And x dx--

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well, that's almost du.

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It's du divided by negative 2.

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So du divided by negative 2.

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So we make the substitution.

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We transform our integral
to this form.

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Now, this is a very nice,
simple integral, right?

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This is just the integral of e
to the u with a, you know, a

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constant multiple.

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So, OK.

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So we can do this right away.

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So this is--

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I'll pull the minus
1/2 out in front.

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And now fundamental theorem
of calculus.

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Antiderivative of the e to
the u is just e to the u.

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And I have to take that
between my bounds.

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So between u equals 1 and
u equals minus 3.

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So notice that because we
changed bounds, we don't have

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to go back and convert
everything back to x.

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We can just plug in directly.

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So OK.

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So we plug in directly.

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What do I get?

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So first I get minus
1/2 times--

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all right, so I'll just pull
the constant out in front.

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So the first term is e to the
minus third minus e to the 1.

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OK.

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And if I wanted to, I could
rewrite this as e minus e to

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the minus 3 over 2, or in
a variety of other ways.

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All right.

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So this one was a pretty
straightforward integration by

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substitution, definite integral,
change bounds, and

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at the end you get
this answer.

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So let's take a look
at the second one.

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So the second one is a bit
trickier, I think, and it

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requires a bit more work.

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So here it is.

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Let me rewrite it.

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So it's the antiderivative of 2
arc tan x over x squared dx.

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So we have here an
antiderivative

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that we have to compute.

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And this is a kind of
messy-looking expression that

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we're working with.

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All right?

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So this is not very pleasant.

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So there are a couple of
different things we could

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think to do.

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So one thing that you could
think to do is you could say,

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I really don't like arc tan.

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I'm going to try and
get rid of arc tan.

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And the way I'm going to get rid
of arc tan is by doing a

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substitution, u equals arc
tan x, or x equals tan u.

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So if you do that, that'll
get rid of your inverse

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trigonometric function, but
it'll introduce, in the

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bottom, you have this x squared,
so that'll introduce

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like a tan u squared,
and dx will give you

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some more trig functions.

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So what you'll be left
with, then-- so this

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will be u on top.

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So you'll have u times a product
of trig functions.

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So I guess it's up to you
whether you think that a

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polynomial, or a power of
u times a bunch of trig

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functions, is simpler
than an inverse trig

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function times a power.

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It's not clear to me that it's
actually that much simpler.

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And in any case, we wouldn't
sort of be

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finished at that stage.

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There would still be a fair
amount more work to do.

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So we can think about, what
other things could we do?

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That's kind of the only obvious
substitution here.

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But another thing that suggests
itself, is that we

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could try an integration
by parts.

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And one reason is, this is
a product of things.

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You know, here it's arc 10x
times 1 over x squared.

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And arc 10x is something that
works nicely with integration

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by parts if you can
differentiate it.

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Right?

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So inverse trigonometric
functions behave like

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logarithms with respect to
integration by parts.

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By which I mean, they really
like to be differentiated.

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Because when you differentiate
an arc tan, you get something

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much simpler.

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Well, simpler, anyhow.

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Maybe not much simpler.

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You get 1 over 1
plus x squared.

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So if you can apply integration
by parts and

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differentiate the arc tan, that
makes your life nice.

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And so, then, in order for that
to work, you need for

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everything else to be
something you can

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antidifferentiate, and here
everything else is just 2 over

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x squared, and so that's
something that

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we know how to integrate.

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It's easy to do.

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So, OK.

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So let's give that
a try, then.

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So we're going to do integration
by parts, so I'm

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going to set you equal
to arc tan x.

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So that means u prime is equal
to 1 over 1 plus x squared.

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And I'm going to let v prime
equal, well, I guess I have to

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pick up the 2 somewhere, so I
might as well pick it up here.

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2 over x squared.

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And so then v, so I integrate
2 over x squared, so that

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gives me x to the minus
1, with a minus sign.

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So it's minus 2 over x.

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So these are u, u prime,
v and v prime.

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So now OK, so now I apply
integration by parts, and that

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tells me this is equal to--

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well, let's see.

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So it's uv, which is minus
2 arc tan x over x.

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That's uv. Minus the integral
of, now what the second part

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is, v u prime dx.

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So here v u prime is--

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well, OK.

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So they're minus 2.

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I'm going to pull
that out front.

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So it's plus 2--

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all right.

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And then the rest of it
is 1 over x times 1

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plus x squared dx.

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So if you apply integration by
parts, you have the uv stuff

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that gets kicked out out front,
and then what you're

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left with is this
second integral.

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Now this is, to me this is
simpler-looking than what we

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started with.

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It's still not simple,
but it's simpler.

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So it seems to me that, you
know, having reached this

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stage, we can say, OK.

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We've made some progress, and
now we have to keep going.

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Right?

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So it's not obvious to me what
the antiderivative of this

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expression should be.

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So how can I figure that out?

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Well, there are a couple of
different options here.

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One thing you might look
at this and see,

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is you might see--

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again, I have a 1 plus x squared
in the denominator,

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and so that might make you
tempted to do a trig

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substitution again.

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And in fact, you could do a
trig substitution here.

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You could put x equal tan theta,
and you would be able

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to solve the question
like this.

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But I don't think it's the
simplest way to go.

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Because in addition--

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well, on the one hand, this has
this 1 plus x squared in

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the denominator, but on the
other hand, this is just a

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rational function.

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Yeah?

00:09:19.690 --> 00:09:21.580
It's a ratio of two
polynomials.

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The polynomial on top is just
1, and the polynomial on the

00:09:23.950 --> 00:09:26.950
bottom is this product, x
times 1 plus x squared.

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And so whenever you have a
rational function that you're

00:09:29.450 --> 00:09:32.580
trying to integrate, you
can also always use--

00:09:32.580 --> 00:09:34.150
what's it called--

00:09:34.150 --> 00:09:35.950
partial fraction
decomposition.

00:09:35.950 --> 00:09:38.110
So we can try to use partial
fractions on the second

00:09:38.110 --> 00:09:39.290
expression.

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And so I think I'd like
to do it that way.

00:09:43.140 --> 00:09:44.760
So let's do that.

00:09:44.760 --> 00:09:45.760
Let's do it down here.

00:09:45.760 --> 00:09:58.900
So partial fractions on
the quantity 1 over x

00:09:58.900 --> 00:10:02.080
times 1 plus x squared.

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So if you remember your partial
fractions here, this

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is completely factored.

00:10:05.090 --> 00:10:07.080
This quadratic doesn't factor.

00:10:07.080 --> 00:10:10.150
It doesn't have any real
roots, doesn't factor.

00:10:10.150 --> 00:10:14.580
So OK, so we have a single
linear term, and a single

00:10:14.580 --> 00:10:16.340
non-factorable quadratic term.

00:10:16.340 --> 00:10:19.230
So partial fractions tells us
that we can write this in the

00:10:19.230 --> 00:10:24.880
form a over x plus-- and now
because the second term is

00:10:24.880 --> 00:10:31.450
quadratic, it needs to be bx
plus c over 1 plus x squared.

00:10:31.450 --> 00:10:31.710
Right?

00:10:31.710 --> 00:10:35.610
So when you have a quadratic
in the bottom, you get two

00:10:35.610 --> 00:10:38.870
constants up at the top,
a linear polynomial.

00:10:38.870 --> 00:10:39.760
And then you can always check.

00:10:39.760 --> 00:10:43.330
You have three constants here,
and the degree of the

00:10:43.330 --> 00:10:45.530
denominator is 3, so that's
one way of checking that

00:10:45.530 --> 00:10:48.860
you're not completely off base,
that the degree down

00:10:48.860 --> 00:10:50.990
here should always agree with
the number of constants that

00:10:50.990 --> 00:10:52.390
you're solving for.

00:10:52.390 --> 00:10:53.530
OK.

00:10:53.530 --> 00:10:53.880
Good.

00:10:53.880 --> 00:10:57.310
So now we need to solve
this for a, b, and c.

00:10:57.310 --> 00:11:00.310
And so OK, because we have
this nice single linear

00:11:00.310 --> 00:11:02.730
factor, we can do the cover
up method there.

00:11:02.730 --> 00:11:08.360
So we cover up x, and we cover
up over x, and we cover up

00:11:08.360 --> 00:11:09.430
everything else.

00:11:09.430 --> 00:11:12.660
So on the right hand side, we
just end up with a, and on the

00:11:12.660 --> 00:11:14.290
left hand side, well, so x.

00:11:14.290 --> 00:11:16.500
So that's what we need, we
get whatever when we

00:11:16.500 --> 00:11:17.960
plug in x equals 0.

00:11:17.960 --> 00:11:21.900
So on this side, we get 1
over 1 plus 0 squared,

00:11:21.900 --> 00:11:25.060
so that's just 1.

00:11:25.060 --> 00:11:29.770
So the cover up method gives
us that a is equal to 1.

00:11:29.770 --> 00:11:31.630
And now we need to solve
for b and c.

00:11:31.630 --> 00:11:33.910
And probably the most
straightforward way, in this

00:11:33.910 --> 00:11:36.830
case, to solve for b and
c, is you can always

00:11:36.830 --> 00:11:39.340
just multiply through.

00:11:39.340 --> 00:11:40.990
And if you multiply through--

00:11:40.990 --> 00:11:41.310
OK.

00:11:41.310 --> 00:11:42.830
So we'll multiply through
on the left hand.

00:11:42.830 --> 00:11:45.850
We'll clear denominator, we'll
multiply through by x times 1

00:11:45.850 --> 00:11:46.500
plus x squared.

00:11:46.500 --> 00:11:49.240
So on the left we'll
get just 1.

00:11:49.240 --> 00:11:49.920
On the right--

00:11:49.920 --> 00:11:50.350
OK.

00:11:50.350 --> 00:11:56.945
So we multiply a over x times
x times 1 plus x squared, so

00:11:56.945 --> 00:11:59.780
that gives us a times 1 plus x
squared, but we know that a is

00:11:59.780 --> 00:12:05.220
1, so this is-- the first term
becomes 1 plus x squared, and

00:12:05.220 --> 00:12:08.560
the second term we multiply
through by x times 1 plus x

00:12:08.560 --> 00:12:10.920
squared, and the 1 plus x
squareds cancel, and we're

00:12:10.920 --> 00:12:18.850
left with plus x times
bx plus c.

00:12:18.850 --> 00:12:23.540
Now, if you, maybe you could
at this point, there are a

00:12:23.540 --> 00:12:24.810
number of different
ways to finish.

00:12:24.810 --> 00:12:27.230
But one, for example, is you
could see, all right, if you

00:12:27.230 --> 00:12:31.180
subtract the 1 plus x squared
over to the other side, you

00:12:31.180 --> 00:12:36.430
have minus x squared, and
over here, you have bx

00:12:36.430 --> 00:12:38.540
squared plus cx.

00:12:38.540 --> 00:12:41.850
So for those things to be
equal, you need b to be

00:12:41.850 --> 00:12:45.910
negative 1, and you
need c to be 0.

00:12:45.910 --> 00:12:47.490
Right?

00:12:47.490 --> 00:12:47.756
So OK.

00:12:47.756 --> 00:12:51.700
So good.

00:12:51.700 --> 00:12:58.240
So we have b equals minus
1, c equals 0.

00:12:58.240 --> 00:13:00.630
So the partial fraction
decomposition, then, if we

00:13:00.630 --> 00:13:12.300
plug this in, is 1 over x minus
x over 1 plus x squared.

00:13:12.300 --> 00:13:17.250
So this is the partial fraction
decomposition of this

00:13:17.250 --> 00:13:18.890
rational function.

00:13:18.890 --> 00:13:20.980
So you apply that partial
fraction decomposition.

00:13:20.980 --> 00:13:21.400
What do you do?

00:13:21.400 --> 00:13:24.230
All right, So now you carry
that back upstairs to the

00:13:24.230 --> 00:13:25.660
integral we were working on.

00:13:29.170 --> 00:13:32.470
So our integral that we started
with is equal to,

00:13:32.470 --> 00:13:34.910
now-- well, so this constant
is still out in front.

00:13:34.910 --> 00:13:45.280
So it's minus 2 arc tan
x over x plus--

00:13:45.280 --> 00:13:47.110
so now we've got 2 times--

00:13:47.110 --> 00:13:47.370
all right.

00:13:47.370 --> 00:13:49.990
So first, this is just algebra
that we've done, so we can

00:13:49.990 --> 00:13:51.210
just substitute it in.

00:13:51.210 --> 00:13:56.440
So we replace this 1 over x
times 1 plus x squared by 1

00:13:56.440 --> 00:14:04.910
over x minus x over 1
plus x squared dx.

00:14:04.910 --> 00:14:05.410
OK.

00:14:05.410 --> 00:14:08.290
And now using the properties
of integration, this is a

00:14:08.290 --> 00:14:09.420
difference of two--

00:14:09.420 --> 00:14:11.180
an integral of a difference
is the

00:14:11.180 --> 00:14:13.440
difference of the two integrals.

00:14:13.440 --> 00:14:15.780
So we just integrate them
separately, right?

00:14:15.780 --> 00:14:17.440
So this is equal to--
so OK, the part

00:14:17.440 --> 00:14:19.460
out front never changes.

00:14:19.460 --> 00:14:26.330
2 arc tan x over x,
plus 2 times--

00:14:26.330 --> 00:14:26.580
OK.

00:14:26.580 --> 00:14:32.190
So you integrate 1 over x, and
that just gives you ln of

00:14:32.190 --> 00:14:34.180
absolute value of x.

00:14:34.180 --> 00:14:38.280
And OK, so you integrate, so
the second part is x over 1

00:14:38.280 --> 00:14:39.300
plus x squared.

00:14:39.300 --> 00:14:42.720
Now, in the worst case, we would
need to do some more

00:14:42.720 --> 00:14:44.420
work on this term and
split things up.

00:14:44.420 --> 00:14:45.820
And one of them, we might
have to complete

00:14:45.820 --> 00:14:46.820
the square or something.

00:14:46.820 --> 00:14:49.160
But in this case, it actually
has worked out kind of nicely.

00:14:49.160 --> 00:14:54.850
This x over 1 plus x squared is
a simple thing to handle.

00:14:54.850 --> 00:14:56.860
That's just ln of 1
plus x squared.

00:14:56.860 --> 00:15:00.410
Well, actually, 1/2 ln
of 1 plus x squared.

00:15:00.410 --> 00:15:09.500
So this is minus 1/2 ln
of 1 plus x squared.

00:15:09.500 --> 00:15:11.950
I could write absolute value,
but 1 plus x squared is always

00:15:11.950 --> 00:15:15.580
positive, so it doesn't
matter if I do or not.

00:15:15.580 --> 00:15:19.330
And if I wanted, I could
rewrite this a bit.

00:15:19.330 --> 00:15:22.290
I could rewrite this as-- just
to finish it, I could write 2

00:15:22.290 --> 00:15:30.780
arc tan x over x plus 2 ln
absolute value of x minus ln

00:15:30.780 --> 00:15:36.550
of 1 plus x squared.

00:15:36.550 --> 00:15:39.020
And I was doing an
antiderivative.

00:15:39.020 --> 00:15:40.930
So at some point, whenever I
finished the last one, I

00:15:40.930 --> 00:15:42.130
should have added a constant.

00:15:42.130 --> 00:15:45.350
So this should have had a plus
c here, and it should have a

00:15:45.350 --> 00:15:46.586
plus c there.

00:15:46.586 --> 00:15:47.836
Right, good.

00:15:51.140 --> 00:15:54.350
So OK, so there's my answer.

00:15:54.350 --> 00:15:56.940
And let's just recap quickly
how we arrived at it.

00:15:56.940 --> 00:16:00.720
So we started with
this integral.

00:16:00.720 --> 00:16:04.150
And we saw, we recognized it
as a good candidate for

00:16:04.150 --> 00:16:06.260
integration by parts, and so
we applied integration by

00:16:06.260 --> 00:16:10.350
parts, seeing that the arc tan
part was the part that really

00:16:10.350 --> 00:16:13.500
wanted to be differentiated, and
that the 1 over x squared

00:16:13.500 --> 00:16:15.760
part, you know, it could have
been differentiated, it

00:16:15.760 --> 00:16:20.080
could've been integrated, but
since the arc tan wanted to be

00:16:20.080 --> 00:16:21.600
differentiated, this
got integrated.

00:16:24.350 --> 00:16:24.560
Good.

00:16:24.560 --> 00:16:26.800
So we did integration by parts,
and then we're left

00:16:26.800 --> 00:16:30.250
with a rational function as the
piece of the integral we

00:16:30.250 --> 00:16:31.320
had left to compute.

00:16:31.320 --> 00:16:33.800
So when you have a rational
function, one method that

00:16:33.800 --> 00:16:36.690
always works is that you can
do partial fractions.

00:16:36.690 --> 00:16:38.890
Now, in this case, this was
a fairly simple partial

00:16:38.890 --> 00:16:39.910
fractions to do.

00:16:39.910 --> 00:16:44.260
First of all, the degree of the
numerator was smaller than

00:16:44.260 --> 00:16:46.550
the degree of the denominator,
so we didn't have to divide by

00:16:46.550 --> 00:16:48.920
anything, you know, we didn't
have to do long division.

00:16:48.920 --> 00:16:52.710
And the denominator had a fairly
simple form, so we

00:16:52.710 --> 00:16:55.650
could just do the cover up
method, and then multiply

00:16:55.650 --> 00:16:57.430
through and be done
fairly quickly.

00:16:57.430 --> 00:16:58.050
But in any case.

00:16:58.050 --> 00:17:01.410
So we did, we applied partial
fractions, so we got this

00:17:01.410 --> 00:17:02.110
expression.

00:17:02.110 --> 00:17:04.320
So then we carried that back
up to our integral, and

00:17:04.320 --> 00:17:05.950
integrated.

00:17:05.950 --> 00:17:08.690
OK, so then we had this integral
with the partial

00:17:08.690 --> 00:17:13.190
fraction expression in it, and
that was easy to integrate.

00:17:13.190 --> 00:17:16.540
So in this case, quite
easy to integrate.

00:17:16.540 --> 00:17:19.240
Sometimes it's a little messier,
but we came out

00:17:19.240 --> 00:17:20.586
pretty luckily this time.

00:17:20.586 --> 00:17:22.320
So OK.

00:17:22.320 --> 00:17:23.170
And so there you go.

00:17:23.170 --> 00:17:25.150
So that's going to be
our final answer.

00:17:25.150 --> 00:17:27.240
Now, this was long
and complicated.

00:17:27.240 --> 00:17:29.610
And so sometimes, when you
do a long and complicated

00:17:29.610 --> 00:17:31.530
computation, you worry,
did I make any stupid

00:17:31.530 --> 00:17:33.490
mistakes along the way?

00:17:33.490 --> 00:17:36.800
One way to check, always, when
you've done an antiderivative,

00:17:36.800 --> 00:17:38.990
is you could always check, you
could take the derivative of

00:17:38.990 --> 00:17:39.870
this expression.

00:17:39.870 --> 00:17:43.520
So that'll require you to use
some product rule and some

00:17:43.520 --> 00:17:45.240
chain rule.

00:17:45.240 --> 00:17:51.420
IBut, you know, arithmetically
a little difficult, but not

00:17:51.420 --> 00:17:52.760
sort of intellectually, right?

00:17:52.760 --> 00:17:56.470
You just are applying rules
sort of automatically.

00:17:56.470 --> 00:17:59.310
So you can take the derivative
of this expression, and check

00:17:59.310 --> 00:18:02.120
to make sure that it actually
agrees with the thing that we

00:18:02.120 --> 00:18:04.330
started with.

00:18:04.330 --> 00:18:05.820
So good.

00:18:05.820 --> 00:18:07.520
I'll end there.