WEBVTT

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Welcome back to recitation.

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In this video, I want
us to practice

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doing integration again.

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

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definite integral, one
is indefinite.

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So this integral, we're going to
have, find the value of the

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integral minus pi over 4 to pi
over 4 of secant cubed u du,

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and then this one, I just want
you to actually just find an

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antiderivative, then, 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

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

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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, in this one, a

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specific number, and in this
one, an antiderivative.

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So we'll start off with the
integral from minus pi over 4

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to pi over 4 of secant
cubed u du.

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And the first thing I'm going
to do, is I'm going to drop

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the bounds and just find the
function I need, and then I'll

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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

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the bounds or not.

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So I'll keep the bounds the
same, but I won't write them

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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

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this, is to split it up
into secant u and then

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secant squared u.

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Maybe there are some other ways
you did it, but when I

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was looking at this problem, the
way I saw was to start off

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and write this as secant
u times 1--

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oops, let me make sure, yes--

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times 1 plus tan squared u du.

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Because secant squared
u is equal to 1 plus

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tangent squared u.

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And what does it do for us?

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Well, it partially answers
the question, but not

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completely, we'll see.

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So what it does first, is it
gives us something we can find

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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, because we get the

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integral of secant u tangent
squared u du.

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And this has promise, but
it's not going to work

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for us right away.

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Because, you know, the
derivative of tangent is

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secant squared, and
the derivative of

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secant is secant tangent.

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So a substitution won't work in
this case, because neither

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one of these functions is a
derivative of the other one.

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If I even move off of tangent,
and I have secant tangent

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times another tangent, 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 the substitution
strategy doesn't work.

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

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this part, and make it into an
integration by parts problem.

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So this is going to be, I'm
going to stop writing equals

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signs, I'm going to figure out
the integral of this one.

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So this is a little sidebar, and
I'm going to look at the

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integral of secant u
tangent squared u.

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And the way I'm going to write
that, is I'm going to write

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that as secant u tangent 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 what 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 why I'm

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picking these, and sort of where
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

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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

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integration by parts have?

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It's going to have vw 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 is going to have the

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antiderivative of this, which is
secant, and the derivative

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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 getting back

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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, why I

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picked the things I did.

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

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secant u is equal to v,
and secant squared u

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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 vw, which is secant u
tangent u minus the integral

00:05:18.350 --> 00:05:22.830
of vdw, which is secant
cubed u du.

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

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anticipating.

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So when I put this all together,
I replace this

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integral by these two things.

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And what's the point there?

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Notice what I have. If I
actually look at the pieces, I

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have, 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
as the bounds--

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minus this.

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So I have the same thing on this
side that I had on the

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other side, but with
a minus sign.

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So if I add it to the other
side, I'm going

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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 equal to

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secant u tangent u.

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Plus that secant--

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I forgot to put that one in, 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 to the

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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?

00:07:07.660 --> 00:07:09.270
There was one on this side,
with a minus sign.

00:07:09.270 --> 00:07:11.330
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, and this is

00:07:16.940 --> 00:07:19.100
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, because I'm about to

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put in some bounds.

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All right?

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

00:07:30.700 --> 00:07:33.140
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

00:07:48.890 --> 00:07:50.140
last thing 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

00:08:00.180 --> 00:08:03.910
secant u plus tangent u.

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OK.

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So now we just have to evaluate

00:08:05.780 --> 00:08:07.540
everywhere and we're done.

00:08:07.540 --> 00:08:09.490
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, pi over 4

00:08:18.830 --> 00:08:21.760
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 that secant of plus or

00:08:34.630 --> 00:08:38.775
minus pi over 4 is equal
to square root 2.

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Let me just make sure
that's right.

00:08:41.060 --> 00:08:44.440
Cosine pi over 4 is 1
over square root 2.

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Cosine is even, so plus
or minus pi over

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4 will be the same.

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

00:08:51.160 --> 00:08:54.490
Tangent of pi over 4.

00:08:54.490 --> 00:08:57.190
Well, tangent is odd, so I
should say plus or minus here,

00:08:57.190 --> 00:08:59.950
tangent is odd, so it's going
to be, they're going to have

00:08:59.950 --> 00:09:01.620
two different signs.

00:09:01.620 --> 00:09:04.920
Tangent pi over 4 is sine pi
over 4 over cosine pi over 4.

00:09:04.920 --> 00:09:06.530
It's the same value there.

00:09:06.530 --> 00:09:07.430
So it's 1.

00:09:07.430 --> 00:09:10.720
So tangent plus or minus pi
over 4 is plus or minus 1.

00:09:10.720 --> 00:09:12.530
So that's what we're
working with.

00:09:12.530 --> 00:09:15.530
So now let's start
plugging in.

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Secant pi over 4 tangent pi
over 4 gives me root 2

00:09:19.530 --> 00:09:20.860
times 1 over 2.

00:09:20.860 --> 00:09:23.860
So I get root 2 over 2,
is the first thing.

00:09:23.860 --> 00:09:25.360
So I have to write--

00:09:25.360 --> 00:09:27.760
this equals is from here.

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So I have root 2 over 2.

00:09:30.490 --> 00:09:36.200
And then secant minus pi over
4 is again root 2, tangent

00:09:36.200 --> 00:09:38.040
minus pi over 4 is minus 1.

00:09:38.040 --> 00:09:43.310
So I have minus, I have a
negative here, and then I have

00:09:43.310 --> 00:09:47.740
a 1 here, or a root 2 here,
negative 1 over 2.

00:09:47.740 --> 00:09:51.800
So I get another negative,
so I have a plus.

00:09:51.800 --> 00:09:54.380
So one negative came from, I was
using the lower bound, and

00:09:54.380 --> 00:09:56.830
one negative came from
the tangent.

00:09:56.830 --> 00:09:58.740
That gave me a plus.

00:09:58.740 --> 00:10:00.960
And then I have plus 1/2.

00:10:00.960 --> 00:10:02.190
Well, again.

00:10:02.190 --> 00:10:06.920
Natural log of secant pi over
4 is going to be-- so I'm

00:10:06.920 --> 00:10:10.140
going to have natural log of
root 2 plus 1, and I'm going

00:10:10.140 --> 00:10:13.470
to have a natural log
of root 2 minus 1.

00:10:13.470 --> 00:10:15.510
And I'm going to have a negative
in between them.

00:10:15.510 --> 00:10:17.005
So I'm going to work a
little magic here.

00:10:17.005 --> 00:10:21.350
It's going to be natural
log of 2 plus 1 over

00:10:21.350 --> 00:10:23.730
root 2 minus 1.

00:10:23.730 --> 00:10:26.750
Now, just to point out, where
did that come from?

00:10:26.750 --> 00:10:30.310
That came from putting in root
2 for both of the pi over 4s

00:10:30.310 --> 00:10:31.830
and minus pi over 4.

00:10:31.830 --> 00:10:35.170
Tangent pi over 4
was the plus 1.

00:10:35.170 --> 00:10:37.590
Tangent of negative pi over
4 was the minus 1.

00:10:37.590 --> 00:10:39.110
How do I get this division?

00:10:39.110 --> 00:10:41.500
I had natural log of something
minus natural log

00:10:41.500 --> 00:10:42.860
of something else.

00:10:42.860 --> 00:10:49.630
So in the end, I get root 2 plus
1/2 natural log absolute

00:10:49.630 --> 00:10:54.330
root 2 plus 1 over
root 2 minus 1.

00:10:54.330 --> 00:10:56.230
All right.

00:10:56.230 --> 00:11:01.470
That is part A. So part
A, let me just remind

00:11:01.470 --> 00:11:03.550
you, what did we do?

00:11:03.550 --> 00:11:06.060
We had secant cubed u.

00:11:06.060 --> 00:11:10.280
We did a substitution for one
of the, for secant squared.

00:11:10.280 --> 00:11:12.800
We got something we could deal
with, and something that

00:11:12.800 --> 00:11:15.270
didn't look so promising, but
once we did an integration by

00:11:15.270 --> 00:11:17.730
parts, we were back to
what we started with,

00:11:17.730 --> 00:11:19.780
with a different sign.

00:11:19.780 --> 00:11:21.810
So we moved it to
the other side.

00:11:21.810 --> 00:11:24.920
We were basically able to solve
for secant cubed u.

00:11:24.920 --> 00:11:26.120
So we got all the way
through, and then

00:11:26.120 --> 00:11:27.150
we just had to evaluate.

00:11:27.150 --> 00:11:30.190
So the big step was, once you
were here, knowing an

00:11:30.190 --> 00:11:32.890
integration by parts actually
would save you.

00:11:32.890 --> 00:11:34.680
That's sort of the hard
thing to see.

00:11:34.680 --> 00:11:37.300
Takes a little while to
see that one, maybe.

00:11:37.300 --> 00:11:37.510
All right.

00:11:37.510 --> 00:11:38.780
So now the next one.

00:11:38.780 --> 00:11:41.390
If we come back here, we're
trying to find an

00:11:41.390 --> 00:11:44.530
antiderivative of 1 over
2 sine u cosine u.

00:11:44.530 --> 00:11:48.720
And there may be some other ways
to do this, but actually,

00:11:48.720 --> 00:11:51.080
this problem, part of the reason
I wanted to do this

00:11:51.080 --> 00:11:52.760
problem, was I wanted to remind
you that it's good to

00:11:52.760 --> 00:11:55.370
know some of the basic
trigonometric identities,

00:11:55.370 --> 00:11:57.670
because it'll make your
life a lot easier.

00:11:57.670 --> 00:12:02.510
So this integral is actually the
same as is the integral of

00:12:02.510 --> 00:12:04.470
du over sine 2u.

00:12:04.470 --> 00:12:08.010
And the reason is, there's a
trigonometric identity that

00:12:08.010 --> 00:12:12.060
says, 2 sine u cosine u
is equal to sine 2u.

00:12:12.060 --> 00:12:17.570
So I wanted to give you a reason
for why we know those,

00:12:17.570 --> 00:12:19.640
why we know some of those
identities, and you end up in

00:12:19.640 --> 00:12:21.300
these situations.

00:12:21.300 --> 00:12:23.220
There might be other ways to
handle this one, but the

00:12:23.220 --> 00:12:27.120
easiest, most direct way
is if you do this.

00:12:27.120 --> 00:12:30.650
You change it so that b is
actually the same thing as

00:12:30.650 --> 00:12:32.860
integral of du over sine 2u.

00:12:32.860 --> 00:12:37.230
So it's just a straight up
double angle formula, you can

00:12:37.230 --> 00:12:39.690
call it if you need
a fancy name.

00:12:39.690 --> 00:12:40.670
And now what is this?

00:12:40.670 --> 00:12:44.960
Well, this is equal to the
integral of what's 1 over

00:12:44.960 --> 00:12:49.130
sine, is going to be cosecant.

00:12:49.130 --> 00:12:51.930
Cosecant 2u du.

00:12:51.930 --> 00:12:54.910
And we know the antiderivative
to cosecant u.

00:12:54.910 --> 00:12:58.300
We know that that's going to
be negative natural log of

00:12:58.300 --> 00:13:00.500
cosecant u plus cotangent u.

00:13:00.500 --> 00:13:02.980
But the problem is,
when there's a 2

00:13:02.980 --> 00:13:03.880
there, what do we do?

00:13:03.880 --> 00:13:06.940
Well, just think about it as, if
you had the antiderivative,

00:13:06.940 --> 00:13:10.950
you know by the chain rule,
if you just put in two u's

00:13:10.950 --> 00:13:13.570
everywhere there was a u when
you took the derivative, you

00:13:13.570 --> 00:13:15.220
would end up with an
extra 2 in front.

00:13:15.220 --> 00:13:16.950
So you have to, basically
you have to just

00:13:16.950 --> 00:13:19.830
put in 1/2 in front.

00:13:19.830 --> 00:13:22.830
You could do a substitution to
check this, but it's really

00:13:22.830 --> 00:13:26.820
straightforward to say, this
antiderivative is equal to

00:13:26.820 --> 00:13:35.020
negative 1/2 natural log
absolute cosecant 2u plus

00:13:35.020 --> 00:13:36.270
cotangent 2u.

00:13:38.400 --> 00:13:40.150
Now that I have that written
out, I'll just point out

00:13:40.150 --> 00:13:44.310
again, if there was no 2 here,
the 1/2 wouldn't be here, and

00:13:44.310 --> 00:13:47.720
we'd just have cosecant u
cotangent u inside here.

00:13:47.720 --> 00:13:52.960
But once we have to have the 2
to get a cosecant 2u in the

00:13:52.960 --> 00:13:57.310
end, we also need to divide by
2 to kill it off when we take

00:13:57.310 --> 00:13:58.220
a derivative.

00:13:58.220 --> 00:14:00.330
The chain rule would give
us a 2 in front, so the

00:14:00.330 --> 00:14:01.940
1/2 kills it off.

00:14:01.940 --> 00:14:04.800
So we don't end up with, you
know, with this not here, the

00:14:04.800 --> 00:14:07.590
derivative of this
is 2 cosecant 2u.

00:14:07.590 --> 00:14:09.980
So we divide by 2,
then we get it.

00:14:09.980 --> 00:14:11.190
We get the right answer.

00:14:11.190 --> 00:14:12.820
So this one--

00:14:12.820 --> 00:14:13.910
you know, I'm not intentionally

00:14:13.910 --> 00:14:14.850
trying to trick you.

00:14:14.850 --> 00:14:18.620
I just want to point out that
it's good to know some of

00:14:18.620 --> 00:14:19.870
these trigonometric
identities.

00:14:19.870 --> 00:14:21.150
It makes solving these
problems a lot

00:14:21.150 --> 00:14:22.470
easier to deal with.

00:14:22.470 --> 00:14:25.150
So the main point of this one
was actually knowing the

00:14:25.150 --> 00:14:26.630
identity, in my mind.

00:14:26.630 --> 00:14:29.490
Maybe you found another
way to do it.

00:14:29.490 --> 00:14:31.810
Probably it didn't take
two lines, though.

00:14:31.810 --> 00:14:35.350
So if you found other way to
do it, actually, it's good.

00:14:35.350 --> 00:14:36.220
It's creative.

00:14:36.220 --> 00:14:37.250
And I like that.

00:14:37.250 --> 00:14:40.590
But I was hoping to convince you
that sometimes it's simple

00:14:40.590 --> 00:14:42.950
to know a few of these
identities.

00:14:42.950 --> 00:14:45.450
And that is where I will stop.