 Welcome back to our lecture series, Math 1220, Calculus II for students at Southern Utah University. As usual, I will be your professor today, Dr. Andrew Missildine. In this lecture, Lecture 11, we're going to continue our discussion of trigonometric integrals as can be found in Section 7.2 of James Stewart's Calculus Textbook. In the last lecture, we saw lots of examples of calculated integrals of functions involving sine and cosine, and our technique of calculating those integrals was to employ appropriate trigonometric identities to help us better utilize the techniques of U-substitution and integration by parts. So what I wanted to do is do one more example of finding an integral using sine and cosine in this lecture, but one that involves some different trigonometric identities. Now, I don't want you to get overwhelmed by the number of trigonometric identities we use in this section. There are a lot, but be aware that despite the large number, the ones that you're actually going to need are provided to you in these lectures here, right? Now, there might be occasionally one that uses a more exotic identity that you haven't necessarily seen before, but again, for like 99.9% of the trigonometric integrals you might be faced with, I have presented to you the identities that you would want to know. Those are ones you'd want to remember are record in your notes for future usage. We use things like the Pythagorean identities, the Haphingal identities, plus derivations of those identities. So what happens if you are looking at an integral that looks something like the following here? You have the integral of sine of 4x and cosine of 5x. You'll notice that unlike the previous examples we've seen here, the period of sine has been adapted and so has cosine. Now, for the most part, that's not such a big deal, right? We've seen things like take the integral of sine of 4x, and there's things more complicated than that. The issue here is that the periods between the sine and cosine are different. We have a 4x and we have a 5x. And because of this mismatch between the frequencies of these sine and soda waves here, the Pythagorean identity, the Haphingal identities we've seen in the past are insufficient, right? Those all are dependent upon having the same frequency, but we have different ones in this situation. So what do we do? Well, it turns out we have this Deus Ex Moxina set of identities, the so-called product-to-sum identities. These are identities one often sees in the trigonometric course at SUU at Smith 1060. And these identities almost have one purpose in mind, and the purpose is really to help us out exactly in this setting. So I want you to kind of remember these identities for future usage here. So these product-to-sum identities are useful when you have a product of a sine and a cosine, or a sine and sine, or a cosine and cosine. And here A is the inside of A and B are the inside of the trigonometric functions. So it could be things like 4x and 5x, like we see here. A is going to be 4x, B is going to be 5x. And the insides can be as exotic as you want. So if you have a product of sines and cosines, the product-to-sum identities can convert the product into a sum, a linear combination of sines and cosines, where you're going to add together the quantities A and B. And the right-hand sides are going to be much more preferable to integrate than the left-hand side. So these are what we want to use in this situation. Now, for this example, evaluate the integral of sine, 4x, cosine, 5x, dx, that's going to be the first type that we're going to use. Other types, if we had to like a sine, 4x, cosine, sorry, sine, 4x, sine, 5x, the second type would be useful here. So you can come up with similar examples for the third one there. So applying this identity here with sine, with A being 4x and cosine being 5x, just use the appropriate identity right here. We're going to give one-half, it's a constant multiple, so I'm going to take it out of the integral. We're going to give one-half sine of 4x plus 5x. I'm putting out all the details here. You could try to simplify this step right now. If you want to or wait to do it later, you don't have to be a hero. You can do a couple steps here. So we have sine of 4x plus 5x. Then we're going to have sine of 4x minus 5x right here and then squeeze a new dx. Now, do notice that with this first identity, there is a negative sign. And for all of these ones, there are some negative signs that do show up here. So look out for those negative signs. For this one, you get negative because of the placement of the cosine and the sine. For the other ones, we have sine and sine, cosine and cosine. You might be worried, what if I pick the wrong ones? But there are some symmetry properties about sine and cosine that will actually save the day. Remember, sine of negative x is the same thing as negative sine of x. And likewise, cosine of negative x equals just cosine of x. So for all of these ones, these symmetry principles can be useful, simplified, and shows that it doesn't actually matter which one is A or B for the second two cases. It does matter for the first one, when you have a sine and a cosine. But simplifying, we get 1 half the integral. We're going to have sine of 9x plus a sine of negative x. And using that symmetry principle that I was talking about just a moment ago, I'm actually going to make the integral a little bit easier. I'm going to switch from sine of negative x, 1 half sine of 9x. That's a poorly written sine, oh well. So the sine of negative x becomes the negative sine of x. I prefer the angle of sine just to be x if possible. And so now we're ready to integrate this thing. We're going to get the 1 half still out in front. The antiderivative of sine of 9x, because of the 9x that's on the inside, the antiderivative is going to have a 1 ninth in front of it because there's a very basic u substitution there. The antiderivative of sine is actually a negative cosine. So we get negative 1 ninth cosine of 9x. And then the other ones can become positive cosine of x plus a constant. Feel free to distribute that 1 half through if it helps you there. So in pretty form, we're going to write this as 1 half cosine of x minus 1 18th cosine x plus a constant. Sorry, not cosine x. That should be a cosine of 9x. There we go. And so then we found the antiderivative. So if you ever have to integrate a product of sine and cosines and they have different angles inside, use these product to some identities to help you out. And they worked out really nicely. Now I do want to mention that this exercise here could be solved using the technique of integration by cycles. Although that technique a little doable is extremely more taxing. It's much more lengthy process. And that's the only I would recommend that only in the case of like emergency like Desert Island Calculus, where nothing survived the plane crash except for yourself and a volleyball, but you have to integrate this thing. Sure, use integration by cycles if you don't remember the identity, but it's better to use the identity to simplify this calculation.