CHRISTINE BREINER: Welcome
back to recitation.
I want us to work a little more
on finding anti-derivatives.
In particular, in
this video we want
to find an anti-derivative of a
trigonometric function, a power
of the tangent function.
So I would like you to
find an anti-derivative
of tangent theta
quantity to the fourth.
And the hint I will
give you is that you're
going to need some fairly
familiar, hopefully, by now,
trigonometric identities
to get this to work.
And then you will need
some other strategies
that you've also
been developing.
So I'll give you a while to
work on it and then I'll be back
and I'll show you how I did it.
OK.
Welcome back.
We want to, again,
we want to find
an anti-derivative for tangent
theta quantity to the fourth.
And I mentioned that
what we're going
to need is a particular
trigonometric-- well,
I didn't say particular,
sorry-- but we
will need some trigonometric
identities to make this work.
And the one in particular
I'll be exploiting
is a certain one, which
I'll write down here,
which is, 1 plus
tangent squared theta is
equal to secant squared theta.
We've seen that, I
think, a fair amount now,
but just to remind
ourselves, it is important,
and this is the one
we're going to use.
So let me show you how this
works, how this identity will
be very useful in here.
And the idea is
that we can break up
this tangent to the fourth theta
into two, the product of two
tangent squared thetas.
So I can rewrite this
integral above as the integral
of tangent squared theta times
another tangent squared theta.
But instead of that, I'm
going to use this identity.
So I'm going to
write it as secant
squared theta minus 1 d theta.
So just let me make sure
everybody follows what I did.
I had tangent to the fourth
theta as my initial integral,
so then I wrote it as tangent
squared times tangent squared.
And this is actually
equal to tangent squared.
You notice I just
subtracted 1 from both sides
of the starred identity.
So that's my other
tangent squared.
So these two integrals
are actually equal.
So I haven't changed
anything fundamentally at all
in the problem.
All right.
Now let's look at
what we get here.
We get, if I distribute
this, I get integral of tan
squared theta secant
squared theta d
theta minus the integral of
tan squared theta d theta.
Now, you should start
to see that maybe
even powers of tangent
theta are nice to deal with.
Because this kind of
stuff is going to happen,
this is going to happen every
time with, you know, n minus 2,
the power n minus 2, here,
for any n I have up here.
And the reason the
even is-- actually,
I guess the even doesn't
even really matter.
I could just have any n and this
would be n minus 2 down here.
This is an easy
integral to deal with.
Why is that?
Because what's the derivative
of the tangent function?
It's secant squared.
Right?
So this is actually a
straight up u-substitution, or
substitution type problem.
So this, I can finish
and I will later,
but this will be substitution.
So I'll finish that
in a little bit.
But what about this?
Now I have tan
squared theta d theta.
That's, you know,
we don't have any,
we don't have a secant here.
We don't have
secant squared here,
which would make it obviously
nice-- that's what we had here.
So we need to do
something else with this.
Well, what I'm
going to do is, I'm
again going to use the
trigonometric identity.
I'm going to replace this
tangent squared theta
by secant squared theta minus 1.
Let's think about,
why is that good?
Well, that's good
because what I end up
with is, if I have secant
squared theta minus 1--
is what this will
equal-- secant squared
theta is easy to integrate.
Because it's the derivative
of a trig function
we know-- it's the
derivative of tangent.
And 1, I think, is pretty
easy to integrate, too.
So we have two functions we
can integrate very easily.
So I'm going to bring this
back up on the next line,
I'm going to do the
replacement here,
and then we'll
finish the problem.
So let me write this down.
OK.
Now I'm going to do my
replacement and minus
the quantity the integral secant
squared theta minus 1 d theta.
Let's make sure I didn't
make any mistakes.
So I had tan squared theta
secant squared theta d theta.
That looks good.
And then I'm
subtracting tan squared
theta, the integral of
tan squared theta d theta.
And that's that.
So I'm OK.
So this one, again,
I mentioned that this
is going to be a substitution.
If you need to write
it out explicitly,
this is u equals
tan theta, so du
is equal to secant
squared theta d theta.
So this is the
integral of u squared.
Right?
If I substitute in
I get u squared du.
So it's the integral
of u squared,
which is u cubed over 3.
So that first part is going
to be tan cubed theta over 3.
That's my first term.
That's a straight up
substitution pretty
similar to what you've seen.
Now I have two
things to integrate.
I have to integrate
secant squared theta,
and I have to integrate the 1.
Well, derivative of
tangent is secant squared.
So the integral
of secant squared
theta is just tangent theta.
So I have to
subtract, so there's
a minus sign, a tangent theta.
And then I have a minus, minus,
so when I integrate 1 d theta,
I'm going to get a plus theta.
And then, obviously,
because it's
a family of possible
solutions, I
can add a constant
there, plus c.
So again, where did
these come from?
This first one was
a u-substitution
on the first integral.
And then over here I
have another integral
with two terms inside.
The first one is
just the, I just
need to integrate
secant squared.
I get tangent theta.
The second one, I just
need to integrate the 1,
and so I have a
negative, negative.
That makes it a positive theta.
And then I have to
add my constant.
So let's come back and
just remind ourselves
where we started.
We started with
this trig function
that was a power of tangent.
And what we ultimately
did is we took
two of the powers of tangent,
we made a substitution
with the appropriate
trigonometric identity
to make this an easier
problem to solve.
And actually--
yeah-- if you take
two of the powers of tangent
away and replace them by this,
then you're always going to end
up with something of this form,
tangent to the
power 2 less times
secant squared theta d
theta, which you can always
handle by a u-substitution.
And you're going to end
up with an integral--
now here's where it gets
a little tough-- here,
you wouldn't have
tangent squared.
I think I might have said
that incorrectly earlier.
If this was any power, here, you
wouldn't have tangent squared.
You would have had whatever--
if this was power n,
this would be power
n minus 2, so this
would be power n minus 2.
Right?
So if this was power 8,
when we do the substitution,
this'll be power 6, so
this would be power 6,
so this would be power 6.
So you'd have to do
the process again.
This should remind you of the
reduction formulas you've seen.
So it's good of it's even,
because if this is power 6,
you do the problem
again and you end up
with, the second
term has a power 4.
You do the problem again, the
second term has a power 2,
and, oh, we know how
to deal with those.
So we like it when
it's an even power.
So that's kind of how these
even powers of tangent,
you can take, you can
find anti-derivatives
of the even powers of
tangent by this strategy
that winds up, you could
actually get a reduction
formula out of this.
But I think that's where I
should stop with this problem,
and I hope you enjoyed it.
