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CHRISTINE BREINER: Welcome
back to recitation.

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I want us to work a little
more on finding

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anti-derivatives.

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In particular, in this video
we want to find an

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anti-derivative of a
trigonometric function, a

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power of the tangent function.

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So I would like you to find an
anti-derivative of tangent

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theta quantity to the fourth.

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And the hint I will give you is
that you're going to need

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some fairly familiar,
hopefully, by now,

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trigonometric identities
to get this to work.

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And then you will need some
other strategies that you've

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also been developing.

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So I'll give you a while to work
on it and then I'll be

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back and I'll show
you how I did it.

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OK.

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Welcome back.

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We want to, again, we want to
find an anti-derivative for

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tangent theta quantity
to the fourth.

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And I mentioned that what
we're going to need is a

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particular trigonometric--

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well, I didn't say particular,
sorry-- but we will need some

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trigonometric identities
to make this work.

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And the one in particular I'll
be exploiting is a certain

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one, which I'll write down here,
which is, 1 plus tangent

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squared theta is equal to
secant squared theta.

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We've seen that, I think, a fair
amount now, but just to

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remind ourselves, it is
important, and this is the one

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we're going to use.

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So let me show you how this
works, how this identity will

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be very useful in here.

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And the idea is that we can
break up this tangent to the

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fourth theta into two,
the product of two

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tangent squared thetas.

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So I can rewrite this integral
above as the integral of

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tangent squared theta times
another tangent squared theta.

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But instead of that, I'm going
to use this identity.

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So I'm going to write
it as secant squared

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theta minus 1 d theta.

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So just let me make sure
everybody follows what I did.

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I had tangent to the fourth
theta as my initial integral,

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so then I wrote it as tangent
squared times tangent squared.

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And this is actually equal
to tangent squared.

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You notice I just subtracted
1 from both sides

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of the starred identity.

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So that's my other
tangent squared.

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So these two integrals
are actually equal.

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So I haven't changed anything
fundamentally

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at all in the problem.

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All right.

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Now let's look at what
we get here.

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We get, if I distribute this,
I get integral tan squared

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theta secant squared theta d
theta minus the integral of

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tan squared theta d theta.

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Now, you should start to see
that maybe even powers of

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tangent theta are nice
to deal with.

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Because this kind of stuff is
going to happen, this is going

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to happen every time with, you
know, n minus 2, the power n

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minus 2, here, for any
n I have up here.

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And the reason the even is--

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actually, I guess the even
doesn't even really matter.

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I could just have any n
and this would be n

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minus 2 down here.

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This is an easy integral
to deal with.

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Why is that?

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Because what's the derivative
of the tangent function?

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It's secant squared.

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Right?

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So this is actually a straight
up u-substitution, or

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substitution type problem.

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So this, I can finish and I will
later, but this will be

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substitution.

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So I'll finish that
in a little bit.

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But what about this?

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Now I have tan squared
theta d theta.

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That's, you know, we don't
have any, we don't have a

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secant here.

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We don't have secant squared
here, which would make it

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obviously nice-- that's
what we had here.

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So we need to do something
else with this.

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Well, what I'm going to do is,
I'm again going to use the

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trigonometric identity.

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I'm going to replace this
tangent squares theta by

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secant squared theta minus 1.

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Let's think about,
why is that good?

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Well, that's good because what
I end up with is, if I have

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secant squared theta minus 1--

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is what this will equal--

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secant squared theta is
easy to integrate.

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Because it's the derivative of a
trig function we know-- it's

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the derivative of tangent.

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And 1, I think, is pretty
easy to integrate, too.

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So we have two functions we
can integrate very easily.

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So I'm going to bring this back
up on the next line, I'm

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going to do the replacement
here, and then we'll finish

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the problem.

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So let me write this down.

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OK.

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Now I'm going to do my
replacement and minus the

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quantity the integral secant
squared theta minus 1 d theta.

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Let's make sure I didn't
make any mistakes.

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So I had tan squared theta
secant squared theta d theta.

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That looks good.

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And then I'm subtracting tan
squared theta, the integral of

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tan squared theta d theta.

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And that's that.

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So I'm OK.

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So this one, again, I mentioned
that this is going

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to be a substitution.

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If you need to write it out
explicitly, this is u equals

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tan theta, so du is equal to
secant squared theta d theta.

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So this is the integral
of u squared.

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Right?

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If I substitute in I
get u squared du.

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So it's the integral of
u squared, which is

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u cubed over 3.

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So that first part is going to
be tan cubed theta over 3.

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That's my first term.

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That's a straight up
substitution pretty similar to

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what you've seen.

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Now I have two things
to integrate.

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I have to integrate secant
squared theta, and I have to

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integrate the 1.

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Well, derivative of tangent
is secant squared.

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So the integral of secant
squared theta is

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just tangent theta.

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So I have to subtract, so
there's a minus sine of

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tangent theta.

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And then I have a minus, minus,
so when I integrate 1 d

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theta, I'm going to
get a plus theta.

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And then, obviously, because
it's a family of possible

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solutions, I can add a constant
there, plus c.

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So again, where did
these come from?

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This first one was
a u-substitution

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on the first integral.

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And then over here I have
another integral with two

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terms inside.

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The first one is just
the, I just need to

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integrate secant squared.

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I get tangent theta.

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The second one, I just need to
integrate the 1, and so I have

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a negative, negative.

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That makes it a positive
theta.

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And then I have to
add my constant.

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So let's come back
and just remind

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ourselves where we started.

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We started with this
trig function that

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was a power of tangent.

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And what we ultimately did is
we took two of the powers of

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tangent, we made a substitution
with the

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appropriate trigonometric
identity to make this an

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easier problem to solve.

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And actually-- yeah--

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if you take two of the powers
of tangent away and replace

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them by this, then you're always
going to end up with

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something of this form, tangent
to the power 2 less

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times secant squared theta d
theta, which you can always

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handle by a u-substitution.

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And you're going to end
up with an integral--

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now here's where it gets a
little tough-- here, you

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wouldn't have tangent squared.

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I think I might have said that
incorrectly earlier.

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If this was any power,
here, you wouldn't

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have tangent squared.

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You would have had whatever--

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if this was power n, this would
be power n minus 2, so

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this would be power n minus 2.

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Right?

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So if this was power 8, when
we do the substitution,

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this'll be power 6, so this
would be power 6, so this

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would be power 6.

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So you'd have to do
the process again.

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This should remind you
of the reduction

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formulas you've seen.

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So it's good of it's even,
because if this is power 6,

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you do the problem again and
you end up with, the second

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term has a power 4.

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You do the problem again, the
second term has a power 2,

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and, oh, we know how
to deal with those.

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So we like it when it's
an even power.

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So that's kind of how these even
powers of tangent, you

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can take, you can find
anti-derivatives of the even

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powers of tangent by this
strategy that winds up, you

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could actually get a reduction
formula out of this.

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But I think that's where I
should stop with this problem,

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and I hope you enjoyed it.