Welcome back to recitation. In this video, I want us to
practice doing integration again. And so what we're going
to do is two problems. One is a definite integral,
one is indefinite. So this integral,
we're going to have, find the value of the integral
minus pi over 4 to pi over 4 of secant cubed u du,
and then this one, I just want you to actually just
find an antiderivative, then, for 1 over 2 sine u cosine u. So why don't you take a while to
work on this, pause the video, and then when you're
ready, restart the video, and I will come back and
show you how I did it. OK. Welcome back. So again, we're going
to work on finding, in this one, a specific
number, and in this one, an antiderivative. So we'll start off
with the integral from minus pi over 4 to pi
over 4 of secant cubed u du. And the first thing
I'm going to do, is I'm going to drop
the bounds and just find the function I need, and then
I'll bring the bounds back in. I don't want to write the
bounds down at every step. I'm not doing any
substitution, I don't think. So I don't need to worry about
if I'm changing the bounds or not. So I'll keep the
bounds the same, but I won't write them down
again until near the end. All right. So let's look at this integral. So the easiest way, I think,
to start an integral like this, is to split it up into secant
u and then secant squared u. Maybe there are some
other ways you did it, but when I was looking at
this problem, the way I saw was to start off and write this
as secant u times 1-- oops, let me make sure, yes-- times
1 plus tan squared u du. Because secant squared
u is equal to 1 plus tangent squared u. And what does it do for us? Well, it partially
answers the question, but not completely, we'll see. So what it does first,
is it gives us something we can find pretty easily. The first one, we just get
the integral of secant u du. We know that one. The second one is
a little harder, because we get the integral of
secant u tangent squared u du. And this has
promise, but it's not going to work for us right away. Because, you know, the
derivative of tangent is secant squared, and
the derivative of secant is secant tangent. So a substitution won't
work in this case, because neither one
of these functions is a derivative
of the other one. If I even move off
a tangent, and I have secant tangent
times another tangent, it's not the right derivative. A tangent's derivative
is secant squared. So my main point here is that
a u-substitution doesn't work, or the substitution
strategy doesn't work. So what we're going to
do, is we're actually going to take this part, and
make it into an integration by parts problem. So this is going
to be, I'm going to stop writing
equals signs, I'm going to figure out the
integral of this one. So this is a little
sidebar, and I'm going to look at the integral
of secant u tangent squared u. And the way I'm
going to write that, is I'm going to write
that as secant u tangent u will be one function I want. We'll call that w--
I guess, v prime. We can use v prime. And then the usual thing
we call u, we'll call w. So tangent u we'll call w. And I want to write down,
I want to explain to you why I'm picking these,
and sort of where this is going to get us. So this is an easy thing to
take an antiderivative of. Right? I can take this. I know v is going
to be secant u. Because the derivative of
secant u is secant u tangent u. And then w is tangent u, w
prime is secant squared u. So if you think about, what does
an integration by parts have? It's going to have v*w minus
the integral of v w prime. That's a lot to take in. But the point is
that that integral is going to have the
antiderivative of this, which is secant, and the derivative of
this, which is secant squared. So it's going to
have a secant cubed. Which might seem weird,
because now we're getting back to what
we started with. But the sign on the secant
cubed is going to be opposite. Again, this is a lot of talking. But let's figure out now. I just want to show you
where we're headed with this, why I picked the things I did. So as I mentioned-- let me
just write these down-- secant u is equal to v, and secant
squared u is equal to w prime. Sorry if that looks a little
weird, but that's a u. OK. So now let's write this with
an integration by parts. I get v*w, which is secant u
tangent u minus the integral of v*dw, which is
secant cubed u du. So this is what I was trying to
show you we were anticipating. So when I put this all together,
I replace this integral by these two things. And what's the point there? Notice what I have. If I actually look at
the pieces, I have, up here, I have
a secant cubed u. It's going to equal
secant u plus this-- I have to evaluate that at
the bounds-- minus this. So I have the same
thing on this side that I had on the other
side, but with a minus sign. So if I add it to
the other side, I'm going to get two of them. So this was sort of
where we're headed. Now let's put it all together. Let's take it back. So this stuff here
is that, right? That's what we did. I'm going to write just the
important stuff right here. I've got the integral
of secant cubed u du is equal to secant u tangent u. Plus that secant-- I
forgot to put that one in, so let me write in that one. Plus the integral of secant
u du minus secant cubed u du. The integral of
secant cubed u du. OK? And now I'm going
to work some magic. I'm actually going to
erase something and move it to the other side. So let me sneak an
eraser off screen. I'm going to add this
to the other side. And what's going to happen? I come over here and
I get two of them. Right? There was one on this
side, with a minus sign. I added it, and now
I have two of them. And now what's the magic? Well, I just divide
everything by 2. And so this is
going to be over 2, and this is going to be over 2. And now I know what
an antiderivative is. Notice I haven't
put in the plus c, because I'm about to
put in some bounds. All right? And by the way, I know an
antiderivative of secant u. So we'll get to
that in one second. But hopefully everyone follows. I had an integral of secant
cubed u on this side. I had a minus integral of
secant cubed u over here. I added it to the other side. It gave me two of them, so
then I just divided by 2. All right? And now let me just explicitly
write down the last thing we need. We still need this one. And this is going to be 1/2
natural log absolute value secant u plus tangent u. OK. So now we just have to evaluate
everywhere and we're done. So now we have to
evaluate all of this. Remember, I said I
left out the bounds. The bounds are pi over
4, minus pi over 4; pi over 4 minus pi over 4. All right. So I'm going to give myself
a little cheat sheet up here, and then I'm going to write
down the numbers I get here. So my cheat sheet
is to remind myself that secant of plus
or minus pi over 4 is equal to square root 2. Let me just make
sure that's right. Cosine pi over 4 is
1 over square root 2. Cosine is even, so plus or minus
pi over 4 will be the same. Secant is 1 over
that, so I'm good. Tangent of pi over 4. Well, tangent is odd, so I
should say plus or minus here, tangent is odd, so it's
going to be, they're going to have two different signs. Tangent pi over 4 is sine pi
over 4 over cosine pi over 4. It's the same value there. So it's 1. So tangent plus or minus pi
over 4 is plus or minus 1. So that's what
we're working with. So now let's start plugging in. Secant pi over 4
tangent pi over 4 gives me root 2 times 1 over 2. So I get root 2 over
2, is the first thing. So I have to write--
this equals is from here. So I have root 2 over 2. And then secant minus
pi over 4 is again root 2, tangent minus
pi over 4 is minus 1. So I have minus, I
have a negative here, and then I have a 1 here, or a
root 2 here, negative 1 over 2. So I get another negative,
so I have a plus. So one negative came from,
I was using the lower bound, and one negative came
from the tangent. That gave me a plus. And then I have plus 1/2. Well, again. Natural log of
secant pi over 4 is going to be-- so I'm going
to have natural log of root 2 plus 1, and I'm going
to have a natural log of root 2 minus 1. And I'm going to have a
negative in between them. So I'm going to work
a little magic here. It's going to be natural log of
2 plus 1 over root 2 minus 1. Now, just to point out,
where did that come from? That came from putting in root
2 for both of the pi over 4's and minus pi over 4. Tangent pi over
4 was the plus 1. Tangent of negative pi
over 4 was the minus 1. How do I get this division? I had natural log of
something minus natural log of something else. So in the end, I get root 2
plus 1/2 natural log absolute root 2 plus 1 over
root 2 minus 1. All right. That is part (a). So part (a), let me just
remind you, what did we do? We had secant cubed u. We did a substitution for one
of the, for secant squared. We got something
we could deal with, and something that
didn't look so promising, but once we did an
integration by parts, we were back to what we started
with, with a different sign. So we moved it to
the other side. We were basically able to
solve for secant cubed u. So we got all the
way through, and then we just had to evaluate. So the big step was,
once you were here, knowing an integration by
parts actually would save you. That's sort of the
hard thing to see. Takes a little while
to see that one, maybe. All right. So now the next one. If we come back
here, we're trying to find an antiderivative
of 1 over 2 sine u cosine u. And there may be some
other ways to do this, but actually, this
problem, part of the reason I wanted to do
this problem, was I wanted to remind you that
it's good to know some of the basic
trigonometric identities, because it'll make
your life a lot easier. So this integral is
actually the same as, is the integral
of du over sine 2u. And the reason is, there's a
trigonometric identity that says, 2 sine u cosine
u is equal to sine 2u. So I wanted to give
you a reason for why we know those, why we know
some of those identities, and you end up in
these situations. There might be other
ways to handle this one, but the easiest, most direct
way is if you do this. You change it so that (b)
is actually the same thing as integral of du over sine 2u. So it's just a straight
up double angle formula, you can call it if
you need a fancy name. And now what is this? Well, this is equal to
the integral of-- what's 1 over sine, is
going to be cosecant. Cosecant 2u du. And we know the
antiderivative to cosecant u. We know that that's going
to be negative natural log of cosecant u plus cotangent u. But the problem is, when there's
a 2 there, what do we do? Well, just think about it as,
if you had the antiderivative, you know by the chain
rule, if you just put in two u's everywhere
there was a u when you took the derivative, you
would end up with an extra 2 in front. So you have to,
basically you have to just put in 1/2 in front. You could do a
substitution to check this, but it's really
straightforward to say, this antiderivative is equal
to negative 1/2 natural log absolute cosecant 2u
plus cotangent 2u. Now that I have that written
out, I'll just point out again, if there was no 2 here,
the 1/2 wouldn't be here, and we'd just have cosecant
u cotangent u inside here. But once we have to have
the 2 to get a cosecant 2u in the end, we also need
to divide by 2 to kill it off when we take a derivative. The chain rule would
give us a 2 in front, so the 1/2 kills it off. So we don't end up with, you
know, with this not here, the derivative of
this is 2 cosecant 2u. So we divide by
2, then we get it. We get the right answer. So this one-- you know,
I'm not intentionally trying to trick you. I just want to
point out that it's good to know some of these
trigonometric identities. It makes solving these problems
a lot easier to deal with. So the main point of
this one was actually knowing the
identity, in my mind. Maybe you found
another way to do it. Probably it didn't
take two lines, though. So if you found other way to
do it, actually, it's good. It's creative. And I like that. But I was hoping to convince
you that sometimes it's simple to know a few
of these identities. And that is where I will stop.
