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