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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.

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