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PROFESSOR: Hi.

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Welcome back to recitation.

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We've been talking about
antidifferentiation or

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integration by substitution
and also by a method that

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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

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quantity 1 minus
e to the 2x dx.

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The second one is 4x times the
quantity 5x 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, try to compute those

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antiderivatives yourself,
come back and we

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can see how you did.

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All right, so welcome back.

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We have these three
antiderivative, so let's take

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them in order.

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So this first one that I wrote
is the antiderivative of e to

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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 for a

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substitution.

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So we have this clear
sort of nested

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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 that looks a lot

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like the derivative of this
1 minus 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 of

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substitution but a natural one
is to sort of find the most

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complicated inside piece.

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So in this case, that's
this whole thing, 1

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minus e to the 2x.

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So I'm going to take for my
substitution, I'm going to

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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

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integral, what do I get?

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Well, so I have cosine of 1
minus e to the 2x just becomes

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cosine of u.

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Now e to the 2x dx,
that's very, very

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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 is you could solve this

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equation for dx and substitute
it in and also solve this

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equation for e 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 that second

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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, we've made

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this substitution.

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So now we have just the

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antiderivative 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 in terms

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of x, so I want to bring
everything back in terms of x.

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And so I need to substitute
back in, get rid of u and

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replace it with x.

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So here that's a, I'll just go
back to what my substitution

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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 e to the 2x

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plus a constant.

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All right, so this is the
antiderivative of this first

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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

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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 5x squared
minus 1 to the 1/3.

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So how can we get from a
derivative, something like 5x

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squared minus 1 quantity
to the 1/3?

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Well if you started off with 5x
squared minus 1 to the 4/3

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thirds and took a derivative,
you would have this 5x squared

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minus 1 to the 1/3 coming out.

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And you would also have some
stuff coming out in front by

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the chain rule.

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Well what kind of stuff?

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Well you know, it would be
some derivative of this

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quadratic polynomial, which
would be some linear

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polynomial.

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And indeed, that
kind of matches

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what we have out front.

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So a good guess for advanced
guessing, is that we can look

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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
5x squared minus

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1 to the 1/3 times--

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OK, so now I need to do the
chain rule, I need to take the

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derivative of the inside.

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Well that's--

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OK, so the minus 1 gets killed
by the derivative, so 5x

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squared, that becomes 10x.

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So I can rewrite this as 40x
over 3 times 5x squared minus

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1 to the 1/3.

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So this looks very much like
the thing that we were

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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 is a little

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bit different.

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So here I have 4, whereas
when I took this

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derivative 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, instead of starting with

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this 5x squared minus 1 to the
4/3, I need to start with some

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multiple of it to make
the constant work out

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right in the end.

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So in this case I was off by a
multiple of 10/3, so I need to

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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 5x squared minus

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1 to the 1/3 dx is equal
to, well it's

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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 we got

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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 in this

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section on substitution that
there's, you know, some

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substitution you can make.

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But it's not sort of obvious
just from looking at tan x

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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 when you don't

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immediately see a substitution,
is to try

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rewriting things in
equivalent ways.

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So sometimes you can do some
algebra or some other

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manipulation.

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In this case there's a very
simple sort of rewriting that

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you can do.

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Which is that tangent of x can
be expressed in terms of sine

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and cosine of x.

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So we can rewrite the
antiderivative of tan x dx as

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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 of

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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

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equals cosine x.

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So if I make the substitution
I get du is equal to

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minus sine x dx.

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Which is, OK, so now if I plug
these values in with this

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substitution, this integral
becomes the integral, well

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it's minus 1 or minus
du over u.

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That's a nice simple
antiderivative to have. So

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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

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value 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 of cosine of x

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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 earlier, she had

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you compute the derivative
of ln of cosine x.

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And in that case you saw that
that derivative was equal to

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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 by

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substitution and advanced
guessing.

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So I'll leave you with that.

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