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CHRISTINE BREINER: Welcome
back to recitation.

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I want us to work a little more
on finding anti-derivatives.

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In particular, in
this video we want

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to find an anti-derivative of a
trigonometric function, a power

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of the tangent function.

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So I would like you to
find an anti-derivative

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of tangent theta
quantity to the fourth.

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And the hint I will
give you is that you're

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going to need 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

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that you've also
been developing.

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

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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

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an anti-derivative for tangent
theta quantity to the fourth.

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And I mentioned that
what we're going

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to need is a particular
trigonometric-- well,

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I didn't say particular,
sorry-- but we

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will need some trigonometric
identities to make this work.

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And the one in particular
I'll be exploiting

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is a certain one, which
I'll write down here,

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which is, 1 plus
tangent squared theta is

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equal to secant squared theta.

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

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but just to remind
ourselves, it is important,

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and this is the one
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

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this tangent to the 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

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of 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

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squared 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 at all

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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 of tan

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squared theta secant
squared theta d

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theta minus the integral of
tan squared theta d theta.

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Now, you should start
to see that maybe

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even powers of tangent
theta are nice to deal with.

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Because this kind of
stuff is going to happen,

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this is going to happen every
time with, you know, n minus 2,

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

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And the reason the
even is-- actually,

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I guess the even doesn't
even really matter.

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I could just have any n and this
would be n 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,

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but this will be 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,

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we don't have a secant here.

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We don't have
secant squared here,

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which would make it 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

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again going to use the
trigonometric identity.

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I'm going to replace this
tangent squared theta

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by 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

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with is, if I have secant
squared theta minus 1--

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is what this will
equal-- secant squared

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theta is easy to integrate.

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Because it's the derivative
of a trig function

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we know-- it's 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,

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I'm going to do the
replacement here,

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and then we'll
finish 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

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the 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

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theta, the integral of
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

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is going to be a substitution.

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If you need to write
it out explicitly,

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this is u equals
tan theta, so du

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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,

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which is 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

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similar to 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,

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and I have to integrate the 1.

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Well, derivative of
tangent is secant squared.

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So the integral
of secant squared

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theta is just tangent theta.

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So I have to
subtract, so there's

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a minus sign, a tangent theta.

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

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I'm going to get a plus theta.

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And then, obviously,
because it's

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a family of possible
solutions, I

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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

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with two terms inside.

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The first one is
just the, I just

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need to 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,

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and so I have 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 ourselves

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where we started.

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We started with
this trig function

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that was a power of tangent.

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And what we ultimately
did is we took

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two of the powers of tangent,
we made a substitution

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with the appropriate
trigonometric identity

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to make this an easier
problem to solve.

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And actually--
yeah-- if you take

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two of the powers of tangent
away and replace them by this,

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then you're always going to end
up with something of this form,

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tangent to the
power 2 less times

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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,

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you 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 have tangent squared.

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You would have had whatever--
if this was power n,

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this would be power
n minus 2, so this

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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,

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so this 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 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

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with, the second
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,

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you can take, you can
find anti-derivatives

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of the even powers of
tangent by this strategy

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that winds up, you could
actually get a reduction

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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.