3.5.2

[STARTS AT 3:53]

Instructor: We want to evaluate this antiderivative. Sine cubed xdx using a CAS. And compare that result to 1/3 cosine cubed x minus cosine x plus C. The result we might have obtained using the technique for integrating odd powers of sine, as we discussed earlier in this chapter. 

Well, first, let's go ahead and rewrite this somewhat. Oh, actually, let me just go and tell you what I got. I used Wolfram alpha, again. And I have the integral of sine cubed. You know what? You already know what the question is, let's write the answer. It is 1/12 parentheses cosine 3x minus 9 cosine x plus C. 

That was the answer that I got. Now, we're going to use some trig identities. This will take a little bit more work than that last one. Use some trig identities to actually show that this is equal to 1/3 cosine cubed x minus cosine x plus C. 

So first, we've got our angle sum identity. So I'm going to write, actually, cosine 3x as cosine x plus 2x so that we can apply that sum identity. So this is going to split into two terms. It is going to be cosine x times cosine 2x minus sine x sine 2x. 

That is just that first term with the angle sum identity. And I will leave the minus 9 cos x. Now, of course, we've got our 1/12. I'm going to leave that out here. We're not really going worry about the constant too much as we go. All right, now, we can take that sine x-- OK, sine 2x and that cosine 2x and we have identities for those. They're double angle identities. 

So cosine 2x becomes-- let me copy this down. That becomes 2 cosine squared x minus 1. So this becomes this. That quantity. And the sine 2x-- so leave the sine x alone. The sine 2x becomes 2 sine x cosine x. 

And if you notice, I like to put parentheses around these. This will get cumbersome as we go. Bringing down the 9 cosine x. And these are all equal. Want to be consistent with that. 

All right, now, let's see. We can distribute some things. Make this a-- this first term a 2 cosine cubed x. I'm going to go ahead and drop the parentheses there. Minus cosine x minus 2 sine squared x times cosine x minus 9 cosine x. And again, we got our plus C hanging out. 

OK, this is starting to take some shape. But that sine squared x is the one thing that's missing and that's not looking like we want it to-- well, what we want it to look like. So we can use our Pythagorean identity and rewrite sine squared x as 1 minus cosine squared x. And I'm going to bring the rest of that down. 

Cosine cubed x minus cosine x. We're getting there. So disturbing that cosine x and the 2 in this term here. That would be plus-- no, that would be a minus 2 cosine x. And that would be a plus-- let's see, you have the negative 2 times negative cosine. So that would be 2 cosine squared. So that would be cosine cubed-- ah, 2 cosine cubed x. 

All right, almost there. Fully, you can see this starting to take shape. We can combine terms at this point. We've got 4 cosine cubed x. And we also have-- let's see, that would be negative 3 cosine x. So negative 12 cosine x. 

Now, for that 1/12. Distributing the 1/12 we get 1/3 cosine cubed x minus cosine x plus C. Now, let's compare that. There we are. So what we might have ended up with 1/3 cosine cubed minus cosine x plus C. 

So the one thing to remember when it comes to integration is that because we have tables, because we have CAS systems available to us, there are a lot of various ways to write an answer like this. So just because there are two different answers that you see does not mean they actually are not in agreement.