Hi.
Welcome back to recitation.
In lecture, you've learned a whole bunch of different
integration techniques.
So I have here a couple of problems that will give you an
opportunity to apply those techniques, and figure out
which ones to apply.
So in particular I have one definite integral, the
integral 0 to 2 of the quantity x times e to the 1
minus x squared with respect to x.
And the second one is an indefinite integral of 2 arc
tan x divided by x squared with respect to x.
So why don't you pause the video, take some time to work
these out, come back, and we can work them out together.
Welcome back.
Hopefully you had some luck working on these integrals.
Let's get started.
We can do the first one first, and then we'll go on to the
second one.
So here we want to compute the integral from 0 to 2 of x e to
the 1 minus x squared dx.
So in order to do this, we have to, you know--
well, we could look at it and say, do I know the answer
immediately, off the top of my head?
And this is not one of the, you know, ones that we have
immediately already memorized a formula for.
So then you say, OK.
So now I need to do something in order to try
to integrate it.
And so, you know, you have a whole bunch of techniques.
And for this one, when you look at it, the technique that
is going to work for you is just the simplest technique
that you have, which is just a straightforward substitution.
And one way to figure that out, is that you can look at
this and you can see that here we have a composition of
functions, this e to the 1 minus x squared, and so
whenever you have a composition of functions,
there's the opportunity for there to have been a chain
rule somewhere.
And in this case, we see that this composition of functions
is multiplied by x, and x is very closely related to the
derivative of 1 minus x squared.
So that's what's suggesting for us a nice substitution.
And so the substitution we're going to try, then, is that
we're going to put this inside function, this 1 minus x
squared, we're going to set u equal to that.
So, all right.
So this is equal to-- so I'm going to say, u
equal 1 minus x squared.
So now if I take a differential, if u equals 1
minus x squared, that means that du is minus 2x dx.
And since this is a definite integral, I also need to
change the bounds of integration.
So when x is equal to 0, that means that u is equal to,
well, 1 minus 0 squared is 1, so that's the lower bound.
And when x is equal to 2, that means that u is equal to 1
minus 2 squared, which is negative 3.
So we make these substitutions.
OK.
So what does our integral become?
So this becomes the integral--
so now the lower bound becomes u equals 1, and the upper
bound is u equals negative 3.
OK.
So now e to the 1 minus x squared, that's
just e to the u.
And x dx--
well, that's almost du.
It's du divided by negative 2.
So du divided by negative 2.
So we make the substitution.
We transform our integral to this form.
Now, this is a very nice, simple integral, right?
This is just the integral of e to the u with a, you know, a
constant multiple.
So, OK.
So we can do this right away.
So this is--
I'll pull the minus 1/2 out in front.
And now fundamental theorem of calculus.
Antiderivative of the e to the u is just e to the u.
And I have to take that between my bounds.
So between u equals 1 and u equals minus 3.
So notice that because we changed bounds, we don't have
to go back and convert everything back to x.
We can just plug in directly.
So OK.
So we plug in directly.
What do I get?
So first I get minus 1/2 times--
all right, so I'll just pull the constant out in front.
So the first term is e to the minus third minus e to the 1.
OK.
And if I wanted to, I could rewrite this as e minus e to
the minus 3 over 2, or in a variety of other ways.
All right.
So this one was a pretty straightforward integration by
substitution, definite integral, change bounds, and
at the end you get this answer.
So let's take a look at the second one.
So the second one is a bit trickier, I think, and it
requires a bit more work.
So here it is.
Let me rewrite it.
So it's the antiderivative of 2 arc tan x over x squared dx.
So we have here an antiderivative
that we have to compute.
And this is a kind of messy-looking expression that
we're working with.
All right?
So this is not very pleasant.
So there are a couple of different things we could
think to do.
So one thing that you could think to do is you could say,
I really don't like arc tan.
I'm going to try and get rid of arc tan.
And the way I'm going to get rid of arc tan is by doing a
substitution, u equals arc tan x, or x equals tan u.
So if you do that, that'll get rid of your inverse
trigonometric function, but it'll introduce, in the
bottom, you have this x squared, so that'll introduce
like a tan u squared, and dx will give you
some more trig functions.
So what you'll be left with, then-- so this
will be u on top.
So you'll have u times a product of trig functions.
So I guess it's up to you whether you think that a
polynomial, or a power of u times a bunch of trig
functions, is simpler than an inverse trig
function times a power.
It's not clear to me that it's actually that much simpler.
And in any case, we wouldn't sort of be
finished at that stage.
There would still be a fair amount more work to do.
So we can think about, what other things could we do?
That's kind of the only obvious substitution here.
But another thing that suggests itself, is that we
could try an integration by parts.
And one reason is, this is a product of things.
You know, here it's arc 10x times 1 over x squared.
And arc 10x is something that works nicely with integration
by parts if you can differentiate it.
Right?
So inverse trigonometric functions behave like
logarithms with respect to integration by parts.
By which I mean, they really like to be differentiated.
Because when you differentiate an arc tan, you get something
much simpler.
Well, simpler, anyhow.
Maybe not much simpler.
You get 1 over 1 plus x squared.
So if you can apply integration by parts and
differentiate the arc tan, that makes your life nice.
And so, then, in order for that to work, you need for
everything else to be something you can
antidifferentiate, and here everything else is just 2 over
x squared, and so that's something that
we know how to integrate.
It's easy to do.
So, OK.
So let's give that a try, then.
So we're going to do integration by parts, so I'm
going to set you equal to arc tan x.
So that means u prime is equal to 1 over 1 plus x squared.
And I'm going to let v prime equal, well, I guess I have to
pick up the 2 somewhere, so I might as well pick it up here.
2 over x squared.
And so then v, so I integrate 2 over x squared, so that
gives me x to the minus 1, with a minus sign.
So it's minus 2 over x.
So these are u, u prime, v and v prime.
So now OK, so now I apply integration by parts, and that
tells me this is equal to--
well, let's see.
So it's uv, which is minus 2 arc tan x over x.
That's uv. Minus the integral of, now what the second part
is, v u prime dx.
So here v u prime is--
well, OK.
So they're minus 2.
I'm going to pull that out front.
So it's plus 2--
all right.
And then the rest of it is 1 over x times 1
plus x squared dx.
So if you apply integration by parts, you have the uv stuff
that gets kicked out out front, and then what you're
left with is this second integral.
Now this is, to me this is simpler-looking than what we
started with.
It's still not simple, but it's simpler.
So it seems to me that, you know, having reached this
stage, we can say, OK.
We've made some progress, and now we have to keep going.
Right?
So it's not obvious to me what the antiderivative of this
expression should be.
So how can I figure that out?
Well, there are a couple of different options here.
One thing you might look at this and see,
is you might see--
again, I have a 1 plus x squared in the denominator,
and so that might make you tempted to do a trig
substitution again.
And in fact, you could do a trig substitution here.
You could put x equal tan theta, and you would be able
to solve the question like this.
But I don't think it's the simplest way to go.
Because in addition--
well, on the one hand, this has this 1 plus x squared in
the denominator, but on the other hand, this is just a
rational function.
Yeah?
It's a ratio of two polynomials.
The polynomial on top is just 1, and the polynomial on the
bottom is this product, x times 1 plus x squared.
And so whenever you have a rational function that you're
trying to integrate, you can also always use--
what's it called--
partial fraction decomposition.
So we can try to use partial fractions on the second
expression.
And so I think I'd like to do it that way.
So let's do that.
Let's do it down here.
So partial fractions on the quantity 1 over x
times 1 plus x squared.
So if you remember your partial fractions here, this
is completely factored.
This quadratic doesn't factor.
It doesn't have any real roots, doesn't factor.
So OK, so we have a single linear term, and a single
non-factorable quadratic term.
So partial fractions tells us that we can write this in the
form a over x plus-- and now because the second term is
quadratic, it needs to be bx plus c over 1 plus x squared.
Right?
So when you have a quadratic in the bottom, you get two
constants up at the top, a linear polynomial.
And then you can always check.
You have three constants here, and the degree of the
denominator is 3, so that's one way of checking that
you're not completely off base, that the degree down
here should always agree with the number of constants that
you're solving for.
OK.
Good.
So now we need to solve this for a, b, and c.
And so OK, because we have this nice single linear
factor, we can do the cover up method there.
So we cover up x, and we cover up over x, and we cover up
everything else.
So on the right hand side, we just end up with a, and on the
left hand side, well, so x.
So that's what we need, we get whatever when we
plug in x equals 0.
So on this side, we get 1 over 1 plus 0 squared,
so that's just 1.
So the cover up method gives us that a is equal to 1.
And now we need to solve for b and c.
And probably the most straightforward way, in this
case, to solve for b and c, is you can always
just multiply through.
And if you multiply through--
OK.
So we'll multiply through on the left hand.
We'll clear denominator, we'll multiply through by x times 1
plus x squared.
So on the left we'll get just 1.
On the right--
OK.
So we multiply a over x times x times 1 plus x squared, so
that gives us a times 1 plus x squared, but we know that a is
1, so this is-- the first term becomes 1 plus x squared, and
the second term we multiply through by x times 1 plus x
squared, and the 1 plus x squareds cancel, and we're
left with plus x times bx plus c.
Now, if you, maybe you could at this point, there are a
number of different ways to finish.
But one, for example, is you could see, all right, if you
subtract the 1 plus x squared over to the other side, you
have minus x squared, and over here, you have bx
squared plus cx.
So for those things to be equal, you need b to be
negative 1, and you need c to be 0.
Right?
So OK.
So good.
So we have b equals minus 1, c equals 0.
So the partial fraction decomposition, then, if we
plug this in, is 1 over x minus x over 1 plus x squared.
So this is the partial fraction decomposition of this
rational function.
So you apply that partial fraction decomposition.
What do you do?
All right, So now you carry that back upstairs to the
integral we were working on.
So our integral that we started with is equal to,
now-- well, so this constant is still out in front.
So it's minus 2 arc tan x over x plus--
so now we've got 2 times--
all right.
So first, this is just algebra that we've done, so we can
just substitute it in.
So we replace this 1 over x times 1 plus x squared by 1
over x minus x over 1 plus x squared dx.
OK.
And now using the properties of integration, this is a
difference of two--
an integral of a difference is the
difference of the two integrals.
So we just integrate them separately, right?
So this is equal to-- so OK, the part
out front never changes.
2 arc tan x over x, plus 2 times--
OK.
So you integrate 1 over x, and that just gives you ln of
absolute value of x.
And OK, so you integrate, so the second part is x over 1
plus x squared.
Now, in the worst case, we would need to do some more
work on this term and split things up.
And one of them, we might have to complete
the square or something.
But in this case, it actually has worked out kind of nicely.
This x over 1 plus x squared is a simple thing to handle.
That's just ln of 1 plus x squared.
Well, actually, 1/2 ln of 1 plus x squared.
So this is minus 1/2 ln of 1 plus x squared.
I could write absolute value, but 1 plus x squared is always
positive, so it doesn't matter if I do or not.
And if I wanted, I could rewrite this a bit.
I could rewrite this as-- just to finish it, I could write 2
arc tan x over x plus 2 ln absolute value of x minus ln
of 1 plus x squared.
And I was doing an antiderivative.
So at some point, whenever I finished the last one, I
should have added a constant.
So this should have had a plus c here, and it should have a
plus c there.
Right, good.
So OK, so there's my answer.
And let's just recap quickly how we arrived at it.
So we started with this integral.
And we saw, we recognized it as a good candidate for
integration by parts, and so we applied integration by
parts, seeing that the arc tan part was the part that really
wanted to be differentiated, and that the 1 over x squared
part, you know, it could have been differentiated, it
could've been integrated, but since the arc tan wanted to be
differentiated, this got integrated.
Good.
So we did integration by parts, and then we're left
with a rational function as the piece of the integral we
had left to compute.
So when you have a rational function, one method that
always works is that you can do partial fractions.
Now, in this case, this was a fairly simple partial
fractions to do.
First of all, the degree of the numerator was smaller than
the degree of the denominator, so we didn't have to divide by
anything, you know, we didn't have to do long division.
And the denominator had a fairly simple form, so we
could just do the cover up method, and then multiply
through and be done fairly quickly.
But in any case.
So we did, we applied partial fractions, so we got this
expression.
So then we carried that back up to our integral, and
integrated.
OK, so then we had this integral with the partial
fraction expression in it, and that was easy to integrate.
So in this case, quite easy to integrate.
Sometimes it's a little messier, but we came out
pretty luckily this time.
So OK.
And so there you go.
So that's going to be our final answer.
Now, this was long and complicated.
And so sometimes, when you do a long and complicated
computation, you worry, did I make any stupid
mistakes along the way?
One way to check, always, when you've done an antiderivative,
is you could always check, you could take the derivative of
this expression.
So that'll require you to use some product rule and some
chain rule.
IBut, you know, arithmetically a little difficult, but not
sort of intellectually, right?
You just are applying rules sort of automatically.
So you can take the derivative of this expression, and check
to make sure that it actually agrees with the thing that we
started with.
So good.
I'll end there.