Evaluating line integral directly - part 2

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.