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Welcome back to recitation.

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In this video, I want us to
practice doing integration

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again.

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And so what we're going
to do is two problems.

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One is a definite integral,
one is indefinite.

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So this integral,
we're going to have,

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find the value of the integral
minus pi over 4 to pi over 4

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of secant cubed u du,
and then this one,

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I just want you to actually just
find an antiderivative, then,

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for 1 over 2 sine u cosine u.

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So why don't you take a while to
work on this, pause the video,

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and then when you're
ready, restart the video,

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and I will come back and
show you how I did it.

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OK.

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Welcome back.

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So again, we're going
to work on finding,

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in this one, a specific
number, and in this one,

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an antiderivative.

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So we'll start off
with the integral

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from minus pi over 4 to pi
over 4 of secant cubed u du.

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And the first thing
I'm going to do,

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is I'm going to drop
the bounds and just find

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the function I need, and then
I'll bring the bounds back in.

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I don't want to write the
bounds down at every step.

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I'm not doing any
substitution, I don't think.

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So I don't need to worry about
if I'm changing the bounds

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or not.

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So I'll keep the
bounds the same,

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but I won't write them down
again until near the end.

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All right.

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So let's look at this integral.

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So the easiest way, I think,
to start an integral like this,

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is to split it up into secant
u and then secant squared u.

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Maybe there are some
other ways you did it,

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but when I was looking at
this problem, the way I saw

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was to start off and write this
as secant u times 1-- oops,

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let me make sure, yes-- times
1 plus tan squared u du.

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Because secant squared
u is equal to 1

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plus tangent squared u.

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And what does it do for us?

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Well, it partially
answers the question,

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but not completely, we'll see.

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So what it does first,
is it gives us something

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we can find pretty easily.

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The first one, we just get
the integral of secant u du.

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We know that one.

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The second one is
a little harder,

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because we get the integral of
secant u tangent squared u du.

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And this has
promise, but it's not

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going to work for us right away.

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Because, you know, the
derivative of tangent

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is secant squared, and
the derivative of secant

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is secant tangent.

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So a substitution won't
work in this case,

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because neither one
of these functions

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is a derivative
of the other one.

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If I even move off
a tangent, and I

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have secant tangent
times another tangent,

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it's not the right derivative.

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A tangent's derivative
is secant squared.

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So my main point here is that
a u-substitution doesn't work,

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or the substitution
strategy doesn't work.

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So what we're going to
do, is we're actually

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going to take this part, and
make it into an integration

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by parts problem.

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So this is going
to be, I'm going

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to stop writing
equals signs, I'm

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going to figure out the
integral of this one.

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So this is a little
sidebar, and I'm

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going to look at the integral
of secant u tangent squared u.

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And the way I'm
going to write that,

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is I'm going to write
that as secant u tangent

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u will be one function I want.

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We'll call that w--
I guess, v prime.

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We can use v prime.

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And then the usual thing
we call u, we'll call w.

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So tangent u we'll call w.

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And I want to write down,
I want to explain to you

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why I'm picking these,
and sort of where

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this is going to get us.

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So this is an easy thing to
take an antiderivative of.

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Right?

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I can take this.

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I know v is going
to be secant u.

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Because the derivative of
secant u is secant u tangent u.

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And then w is tangent u, w
prime is secant squared u.

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So if you think about, what does
an integration by parts have?

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It's going to have v*w minus
the integral of v w prime.

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That's a lot to take in.

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But the point is
that that integral

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is going to have the
antiderivative of this, which

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is secant, and the derivative of
this, which is secant squared.

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So it's going to
have a secant cubed.

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Which might seem weird,
because now we're

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getting back to what
we started with.

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But the sign on the secant
cubed is going to be opposite.

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Again, this is a lot of talking.

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But let's figure out now.

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I just want to show you
where we're headed with this,

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why I picked the things I did.

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So as I mentioned-- let me
just write these down-- secant

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u is equal to v, and secant
squared u is equal to w prime.

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Sorry if that looks a little
weird, but that's a u.

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OK.

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So now let's write this with
an integration by parts.

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I get v*w, which is secant u
tangent u minus the integral

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of v*dw, which is
secant cubed u du.

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So this is what I was trying to
show you we were anticipating.

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So when I put this all together,
I replace this integral

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by these two things.

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And what's the point there?

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Notice what I have.

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If I actually look at
the pieces, I have,

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up here, I have
a secant cubed u.

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It's going to equal
secant u plus this--

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I have to evaluate that at
the bounds-- minus this.

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So I have the same
thing on this side

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that I had on the other
side, but with a minus sign.

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So if I add it to
the other side,

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I'm going to get two of them.

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So this was sort of
where we're headed.

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Now let's put it all together.

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Let's take it back.

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So this stuff here
is that, right?

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That's what we did.

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I'm going to write just the
important stuff right here.

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I've got the integral
of secant cubed u du is

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equal to secant u tangent u.

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Plus that secant-- I
forgot to put that one in,

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so let me write in that one.

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Plus the integral of secant
u du minus secant cubed u du.

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The integral of
secant cubed u du.

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OK?

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And now I'm going
to work some magic.

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I'm actually going to
erase something and move it

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to the other side.

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So let me sneak an
eraser off screen.

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I'm going to add this
to the other side.

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And what's going to happen?

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I come over here and
I get two of them.

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Right?

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There was one on this
side, with a minus sign.

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I added it, and now
I have two of them.

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And now what's the magic?

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Well, I just divide
everything by 2.

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And so this is
going to be over 2,

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and this is going to be over 2.

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And now I know what
an antiderivative is.

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Notice I haven't
put in the plus c,

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because I'm about to
put in some bounds.

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All right?

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And by the way, I know an
antiderivative of secant u.

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So we'll get to
that in one second.

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But hopefully everyone follows.

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I had an integral of secant
cubed u on this side.

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I had a minus integral of
secant cubed u over here.

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I added it to the other side.

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It gave me two of them, so
then I just divided by 2.

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All right?

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And now let me just explicitly
write down the last thing

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we need.

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We still need this one.

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And this is going to be 1/2
natural log absolute value

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secant u plus tangent u.

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OK.

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So now we just have to evaluate
everywhere and we're done.

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So now we have to
evaluate all of this.

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Remember, I said I
left out the bounds.

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The bounds are pi over
4, minus pi over 4;

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pi over 4 minus pi over 4.

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All right.

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So I'm going to give myself
a little cheat sheet up here,

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and then I'm going to write
down the numbers I get here.

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So my cheat sheet
is to remind myself

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that secant of plus
or minus pi over 4

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is equal to square root 2.

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Let me just make
sure that's right.

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Cosine pi over 4 is
1 over square root 2.

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Cosine is even, so plus or minus
pi over 4 will be the same.

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Secant is 1 over
that, so I'm good.

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Tangent of pi over 4.

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Well, tangent is odd, so I
should say plus or minus here,

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tangent is odd, so it's
going to be, they're going

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to have two different signs.

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Tangent pi over 4 is sine pi
over 4 over cosine pi over 4.

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It's the same value there.

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So it's 1.

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So tangent plus or minus pi
over 4 is plus or minus 1.

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So that's what
we're working with.

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So now let's start plugging in.

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Secant pi over 4
tangent pi over 4

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gives me root 2 times 1 over 2.

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So I get root 2 over
2, is the first thing.

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So I have to write--
this equals is from here.

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So I have root 2 over 2.

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And then secant minus
pi over 4 is again

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root 2, tangent minus
pi over 4 is minus 1.

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So I have minus, I
have a negative here,

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and then I have a 1 here, or a
root 2 here, negative 1 over 2.

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So I get another negative,
so I have a plus.

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So one negative came from,
I was using the lower bound,

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and one negative came
from the tangent.

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That gave me a plus.

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And then I have plus 1/2.

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Well, again.

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Natural log of
secant pi over 4 is

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going to be-- so I'm going
to have natural log of root

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2 plus 1, and I'm going
to have a natural log

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of root 2 minus 1.

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And I'm going to have a
negative in between them.

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So I'm going to work
a little magic here.

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It's going to be natural log of
2 plus 1 over root 2 minus 1.

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00:10:23,730 --> 00:10:26,750
Now, just to point out,
where did that come from?

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That came from putting in root
2 for both of the pi over 4's

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00:10:30,310 --> 00:10:31,830
and minus pi over 4.

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00:10:31,830 --> 00:10:35,170
Tangent pi over
4 was the plus 1.

217
00:10:35,170 --> 00:10:37,590
Tangent of negative pi
over 4 was the minus 1.

218
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How do I get this division?

219
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I had natural log of
something minus natural log

220
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of something else.

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So in the end, I get root 2
plus 1/2 natural log absolute

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root 2 plus 1 over
root 2 minus 1.

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All right.

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That is part (a).

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00:11:00,030 --> 00:11:03,550
So part (a), let me just
remind you, what did we do?

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We had secant cubed u.

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We did a substitution for one
of the, for secant squared.

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We got something
we could deal with,

229
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and something that
didn't look so promising,

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but once we did an
integration by parts,

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we were back to what we started
with, with a different sign.

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So we moved it to
the other side.

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We were basically able to
solve for secant cubed u.

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So we got all the
way through, and then

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we just had to evaluate.

236
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So the big step was,
once you were here,

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knowing an integration by
parts actually would save you.

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00:11:32,890 --> 00:11:34,680
That's sort of the
hard thing to see.

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00:11:34,680 --> 00:11:37,010
Takes a little while
to see that one, maybe.

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00:11:37,010 --> 00:11:37,510
All right.

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00:11:37,510 --> 00:11:38,780
So now the next one.

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00:11:38,780 --> 00:11:40,920
If we come back
here, we're trying

243
00:11:40,920 --> 00:11:44,530
to find an antiderivative
of 1 over 2 sine u cosine u.

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And there may be some
other ways to do this,

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but actually, this
problem, part of the reason

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I wanted to do
this problem, was I

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00:11:51,460 --> 00:11:53,460
wanted to remind you that
it's good to know some

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00:11:53,460 --> 00:11:55,370
of the basic
trigonometric identities,

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because it'll make
your life a lot easier.

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00:11:57,670 --> 00:12:00,610
So this integral is
actually the same

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00:12:00,610 --> 00:12:04,470
as, is the integral
of du over sine 2u.

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00:12:04,470 --> 00:12:08,010
And the reason is, there's a
trigonometric identity that

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says, 2 sine u cosine
u is equal to sine 2u.

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00:12:12,060 --> 00:12:15,900
So I wanted to give
you a reason for why

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00:12:15,900 --> 00:12:19,100
we know those, why we know
some of those identities,

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00:12:19,100 --> 00:12:21,075
and you end up in
these situations.

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00:12:21,075 --> 00:12:22,950
There might be other
ways to handle this one,

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00:12:22,950 --> 00:12:27,120
but the easiest, most direct
way is if you do this.

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00:12:27,120 --> 00:12:29,900
You change it so that (b)
is actually the same thing

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00:12:29,900 --> 00:12:32,860
as integral of du over sine 2u.

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00:12:32,860 --> 00:12:36,230
So it's just a straight
up double angle formula,

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00:12:36,230 --> 00:12:39,690
you can call it if
you need a fancy name.

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00:12:39,690 --> 00:12:40,670
And now what is this?

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00:12:40,670 --> 00:12:44,430
Well, this is equal to
the integral of-- what's

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00:12:44,430 --> 00:12:49,130
1 over sine, is
going to be cosecant.

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00:12:49,130 --> 00:12:51,930
Cosecant 2u du.

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00:12:51,930 --> 00:12:54,910
And we know the
antiderivative to cosecant u.

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00:12:54,910 --> 00:12:57,680
We know that that's going
to be negative natural log

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00:12:57,680 --> 00:13:00,500
of cosecant u plus cotangent u.

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00:13:00,500 --> 00:13:03,880
But the problem is, when there's
a 2 there, what do we do?

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00:13:03,880 --> 00:13:06,940
Well, just think about it as,
if you had the antiderivative,

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00:13:06,940 --> 00:13:10,157
you know by the chain
rule, if you just

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00:13:10,157 --> 00:13:11,990
put in two u's everywhere
there was a u when

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00:13:11,990 --> 00:13:14,580
you took the derivative, you
would end up with an extra 2

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00:13:14,580 --> 00:13:15,220
in front.

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00:13:15,220 --> 00:13:16,670
So you have to,
basically you have

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00:13:16,670 --> 00:13:19,830
to just put in 1/2 in front.

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00:13:19,830 --> 00:13:22,270
You could do a
substitution to check this,

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00:13:22,270 --> 00:13:24,890
but it's really
straightforward to say,

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00:13:24,890 --> 00:13:30,490
this antiderivative is equal
to negative 1/2 natural log

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00:13:30,490 --> 00:13:35,860
absolute cosecant 2u
plus cotangent 2u.

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00:13:38,331 --> 00:13:40,830
Now that I have that written
out, I'll just point out again,

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00:13:40,830 --> 00:13:44,210
if there was no 2 here,
the 1/2 wouldn't be here,

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00:13:44,210 --> 00:13:47,720
and we'd just have cosecant
u cotangent u inside here.

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00:13:47,720 --> 00:13:52,720
But once we have to have
the 2 to get a cosecant 2u

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00:13:52,720 --> 00:13:56,660
in the end, we also need
to divide by 2 to kill it

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00:13:56,660 --> 00:13:58,220
off when we take a derivative.

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00:13:58,220 --> 00:14:00,070
The chain rule would
give us a 2 in front,

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00:14:00,070 --> 00:14:01,940
so the 1/2 kills it off.

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00:14:01,940 --> 00:14:04,670
So we don't end up with, you
know, with this not here,

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00:14:04,670 --> 00:14:07,590
the derivative of
this is 2 cosecant 2u.

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00:14:07,590 --> 00:14:09,980
So we divide by
2, then we get it.

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00:14:09,980 --> 00:14:11,190
We get the right answer.

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00:14:11,190 --> 00:14:13,910
So this one-- you know,
I'm not intentionally

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00:14:13,910 --> 00:14:14,850
trying to trick you.

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00:14:14,850 --> 00:14:17,704
I just want to
point out that it's

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00:14:17,704 --> 00:14:19,870
good to know some of these
trigonometric identities.

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00:14:19,870 --> 00:14:22,470
It makes solving these problems
a lot easier to deal with.

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00:14:22,470 --> 00:14:24,670
So the main point of
this one was actually

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00:14:24,670 --> 00:14:26,630
knowing the
identity, in my mind.

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00:14:26,630 --> 00:14:29,490
Maybe you found
another way to do it.

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00:14:29,490 --> 00:14:31,810
Probably it didn't
take two lines, though.

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00:14:31,810 --> 00:14:35,350
So if you found other way to
do it, actually, it's good.

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00:14:35,350 --> 00:14:36,220
It's creative.

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00:14:36,220 --> 00:14:37,250
And I like that.

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00:14:37,250 --> 00:14:40,300
But I was hoping to convince
you that sometimes it's

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00:14:40,300 --> 00:14:42,950
simple to know a few
of these identities.

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00:14:42,950 --> 00:14:45,052
And that is where I will stop.