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# Evaluating line integral directly - part 2

Video transcript

All we have left to do is
evaluate this integral, which is really an
exercise in kind of evaluating trigonometric
integrals more than anything else. But it never hurts
to get that practice, so let's just do
it step by step. So we see sine cubed theta. No obvious way to directly
take the antiderivative of sine cubed theta. But if we had some mixtures
of sines and cosines there, then we could start essentially
doing U-substitution, which at this point you
probably can do in your head. So what we could do is we
could write this is a product. So we can write this
as the sine of theta-- so I'll do this part
right over here. This is sine of theta
times sine squared theta. And sine squared
theta can be rewritten as 1 minus cosine squared theta. So this is the same thing
as sine of theta times 1 minus cosine squared theta. And if we multiply
this out, this gives us sine theta minus sine
theta cosine squared theta. And this is much easier
for us to integrate, although it looks like a
more complicated expression because it's easy take the
antiderivative of sine theta. And now it's easy take
the antiderivative of this because we have the derivative
of cosine theta sitting right over here. So this is going to be
cosine cubed theta over 3. So, essentially, we're doing
U-substitution right over here. But I'll save that for second. Let's rewrite all of
these in a way that's easy to take the
antiderivative of it. Cosine squared theta, we know
this is a common trig identity. That's the same thing as 1/2
of 1 plus cosine of 2 theta. And once again, this
is a much, much easier to take the antiderivative of. So I'll write plus 1/2
plus 1/2 cosine of 2 theta. And now, all of this
is actually quite easy to take the antiderivative
of, and so I'll just rewrite it again. So minus 4 cosine theta plus
4 cosine theta sine theta minus cosine theta sine
squared theta d theta. I just was able to sneak it in. And so that's our integral
between 0 and 2 pi. So let's just take
the antiderivative in every one of these steps. It's starting to get
a little bit messy. I'll try to write a
little bit neater. The antiderivative
of sine of theta is negative cosine of theta. If you take the derivative
of cosine theta, you get negative sine theta. Then the negatives
cancel out, and you get that right over there. Then over here, we
have the derivative of cosine theta, which
is negative sine theta. So we can essentially
kind of treat this-- we could kind of
make the substitution that u is cosine theta. That's essentially what
we're doing in our head. So the antiderivative
of this is going to be equal to plus
cosine cubed theta over 3. And then the antiderivative
of 1/2 with respect to theta is just going to
be plus 1/2 theta. The antiderivative of
cosine 2 theta-- well, we want the derivative of
this thing sitting someplace. The derivative of this
thing over here is 2. So if we put a 2 here, we can't
just multiply by 2 arbitrarily. We'd have to multiply
and divide by 2. So we could put a 2 here,
and then we could also-- but we'd also have
to divide by 2. So then that would become a 4. And we haven't changed this. Notice, this is now
2/4 cosine of 2 theta, the exact same thing as
1/2 half cosine 2 theta. But this is useful because now,
the way I've written it here, we have the derivative of
2 theta right over here. And so we can just
say, well, we'll just take the antiderivative
to this whole thing, which is going to be sine of 2 theta. But we still have the 1 so that
the antiderivative of this part right over here is
the sine of 2 theta. And then we have
the 1/4 out there. So plus 1/4 sine of 2 theta. And then the antiderivative
of cosine theta is just sine theta,
so minus 4 sine theta. Antiderivative of
this right over here, we can kind of pick whichever
way we want to do it. But we could say, well, the
derivative of sine theta is cosine theta. So this is going to be the same
thing as 4 sine squared theta over 2. Or instead of saying over 2,
instead of writing that 4, I'll just divide the 4 by
the 2, and I will get a 2. So let me erase that and
put a 2 right over here. And you can work
it out yourself. If you were to take the
derivative of this thing right over here, it'd be the
derivative of sine theta. If you'd just use
the chain rule, which would be cosine theta,
and then times 4 sine of theta. So that's exactly what
we have right over here. And then we have this last part. The derivative of sine
theta is cosine theta. And so once again, just
like we've done before, the antiderivative
of this whole thing is going to be negative
sine cubed of theta over 3. And we need to evaluate
this entire expression between 0 and 2 pi. So let's see how it evaluates. So first, let's evaluate
everything at 2 pi. So this evaluated at
2 pi is negative 1. This evaluated at 2 pi is 1/3. This evaluated at 2 pi
is just going to be pi. This evaluated at 2 pi is
0, because sine of 4 pi is going to be 0. This evaluated at 2
pi is going to be 0. This evaluated at 2
pi is going to be 0. And this evaluated at
2 pi is going to be 0. So that's a nice simplification. So that's everything
evaluated at 2 pi. And from that,
we're going to have to subtract everything
evaluated at 0. So cosine of 0, well, that's
going to be once again 1. And we have a negative
sign, so it's negative 1. Then you're going
to have plus 1/3. And then you're going to have 0. And then all these other
things are going to be 0. And so if you
simplify it, you get-- this is going to be equal to
negative 1 plus 1/3 plus pi. And then we have plus
1 plus 1 minus 1/3. Well, that cancels with that. That cancels with that. And we deserve a drum roll now. It all simplified
just like when we use Stokes' Theorem in
like the four videos. Actually, I think it
was a little bit simpler to just directly evaluate
the line integral over here. We got it simplifying
to just pi.