WEBVTT
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PROFESSOR: Hi.
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Welcome back to recitation.
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We've been talking about
antidifferentiation
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or integration by substitution
and also by a method
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that Professor Jerison
called "advanced guessing".
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So I have a few problems
up on the board behind me.
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Three antiderivatives
for you to compute.
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So the first one is e to the 2x
times cosine of the quantity 1
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minus e to the 2x dx.
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The second one is 4x times the
quantity 5 x squared minus 1
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raised to the 1/3 power.
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And the third one
is tan of x dx.
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So why that you
spend a few minutes,
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try to compute those
antiderivatives yourself,
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come back and we
can see how you did.
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All right, so welcome back.
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We have these three
antiderivative,
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so let's take them in order.
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So this first one that I wrote
is the antiderivative of e
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to the 2x times cosine
of the quantity 1
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minus e to the 2x dx.
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So this problem seems to
me like a good candidate
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for a substitution.
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So we have this clear sort of
nested function thing going on.
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We have cosine of 1
minus e to the 2x.
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And then out front
we have something
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that looks a lot like the
derivative of this 1 minus
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e to the 2x.
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So I'm going to try this
with substitution then.
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And I think there are,
you know, a few choices
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of substitution
but a natural one
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is to sort of find the most
complicated inside piece.
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So in this case, that's
this whole thing, 1 minus e
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to the 2x.
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So I'm going to take,
for my substitution,
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I'm going to take u equals
1 minus e to the 2x.
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And so that means du is equal
to minus 2 e to the 2x dx.
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OK, so that's my substitution.
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And when I put my substitution
into this integral,
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what do I get?
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Well, so I have
cosine of 1 minus e
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to the 2x just
becomes cosine of u.
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Now e to the 2x dx, that's very,
very close to this du here.
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So what's different is
that here I have a minus 2.
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So actually e to the 2x dx
is du divided by minus 2.
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So this is cosine u times
du divided by minus 2.
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Now another way to
get to this point
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is you could solve
this equation for dx
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and substitute it in and also
solve this equation for e
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to the 2x and substitute
it in and things
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should work out more or less
the same if you try that out.
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So actually you won't
even need to make
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that second substitution.
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You'll just get some
nice cancellation there.
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It's even simpler
than what I just said.
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OK, so we do this
antiderivative,
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we've made this substitution.
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So now we have just
the antiderivative
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of a cosine function.
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All right, well
that's not that bad.
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Because we know that the
derivative of sine is cosine.
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So the antiderivative
of cosine is sine.
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So this is equal to-- so that
minus 2, that 1 over minus 2
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is just going to stick around.
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So it's 1 over minus 2 sine of
u plus a constant of course,
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plus an arbitrary constant.
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And OK, and so, but my
original function was
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in terms of x, so I want
to bring everything back
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in terms of x.
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And so I need to substitute
back in, get rid of u
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and replace it with x.
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So here that's a--
I'll just go back
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to what my substitution was and
I replace all my u's with it.
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So this is minus 1/2
sine the quantity 1 minus
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e to the 2x plus a constant.
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All right, so this
is the antiderivative
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of this first expression here.
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OK, so now, how
about the second one?
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So the second one, we could
also do it with a substitution.
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This is also sort of a prime
suspect for advanced guessing.
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So we see here that we have
some, this polynomial raised
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to some power.
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So this is 5 x squared
minus 1 to the 1/3.
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So how can we get from a
derivative, something like 5
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x squared minus 1
quantity to the 1/3?
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Well if you started off with
5 x squared minus 1 to the 4/3
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and took a derivative, you would
have this 5 x squared minus 1
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to the 1/3 coming out.
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And you would also have some
stuff coming out in front
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by the chain rule.
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Well what kind of stuff?
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Well you know, it would
be some derivative
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of this quadratic
polynomial, which would
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be some linear polynomial.
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And indeed, that kind of
matches what we have out front.
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So a good guess for
advanced guessing,
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is that we can look at-- so d
over dx of 5x squared minus 1,
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quantity, to the 4/3.
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So OK, so this derivative we
can compute by the chain rule.
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So this is 4/3 times 5 x squared
minus 1 to the 1/3 times-- OK,
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so now I need to
do the chain rule,
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I need to take the
derivative of the inside.
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Well that's-- OK, so the minus
1 gets killed by the derivative,
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so 5 x squared,
that becomes 10x.
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So I can rewrite this
as 40x over 3 times 5 x
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squared minus 1 to the 1/3.
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So this looks very
much like the thing
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that we were interested in.
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Right?
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We were where'd it go?
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Oh, here it is.
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And the thing we were
trying to antidifferentiate.
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So the difference is just
this constant out front
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is a little bit different.
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So here I have 4, whereas
when I took this derivative
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I had 40/3.
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So I need to correct for that.
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And the correction
is just to say,
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instead of starting with this
5x squared minus 1 to the 4/3,
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I need to start with
some multiple of it
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to make the constant work
out right in the end.
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So in this case I was off
by a multiple of 10/3,
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so I need to correct by
multiplying through by 3/10.
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So we get that the
antiderivative that we want.
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The antiderivative of 4x
times the quantity 5 x squared
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minus 1 to the 1/3
dx is equal to,
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well it's equal to 3/10 of this.
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5x squared minus 1 to the 4/3.
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OK, so that's our second
antiderivative, which
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we got by advanced guessing.
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Now let's look at the third one.
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So the third one is tan x.
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Now I sort of promised you
by asking this question
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in this section on substitution
that there's, you know,
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some substitution you can make.
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But it's not sort
of obvious just
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from looking at tan x what
should be substituted where.
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At least it isn't obvious to me.
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But one thing that
can help often
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when you don't immediately
see a substitution,
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is to try rewriting
things in equivalent ways.
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So sometimes you
can do some algebra
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or some other manipulation.
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In this case there's a very
simple sort of rewriting
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that you can do.
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Which is that tangent of x
can be expressed in terms
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of sine and cosine of x.
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So we can rewrite the
antiderivative of tan
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x dx as integral sine
x over cosine x dx.
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OK, so now what do we see?
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So, I see in the
denominator a cosine of x.
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And then up top I
have a sine x dx.
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So sine x dx, that's really
close to the differential
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of cosine of x.
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So I'm going to try
this substitution.
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And I'm going to try the
substitution u equals cosine x.
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So if I make the
substitution I get
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du is equal to minus sine x dx.
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Which is, OK, so now
if I plug these values
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in with this substitution, this
integral becomes the integral,
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well it's minus 1
or minus du over u.
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That's a nice simple
antiderivative to have.
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So we've seen this before.
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So this is just a logarithm.
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So the minus sine
comes out front.
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So this is minus ln
of the absolute value
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of u plus a constant.
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And now we had this,
that u was cosine of x.
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So this is minus ln
of the absolute value
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of cosine of x plus a constant.
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Now this should look
a little bit familiar.
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Because in one of
Christine's recitations
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earlier, she had you compute the
derivative of ln of cosine x.
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And in that case you saw
that that derivative was
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equal to minus tangent of x.
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Just like it should be.
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So, all right, so there we go.
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That was three examples
of antidifferentiation
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by substitution and
advanced guessing.
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So I'll leave you with that.