 So let's look at the integration of trigonometric functions. This is actually a fairly broad topic, so this is just an introduction to some of the basic ideas. In general, if I'm trying to evaluate an integral involving a trigonometric function, I'm going to do one of three things. First, I might try to see if I can simplify or expand the trigonometric expression into something a little bit more useful. This is really in the nature of an algebraic manipulation of the integrand, and it's probably the first thing we really want to try, because everything else is a little bit more work. And by a little bit more work, that means we'll apply an integration technique, we'll do a u-substitution, or we might do integration by parts. And in this particular video, we'll take a look at integrals, which revolve around the angle sum identity. So we have the sine of a sum and the cosine of a sum. And the important observation to make here is that if I look at the sine of a sum or difference of two angles, what I get is an expression that involves products of sine and cosine. While if I take a look at the cosine of an angle sum, then what I get is an expression that involves products of cosine and also products of sine. And what this does is it means I might be able to go backwards if I have a product of sines or if I have a product of cosines, I might be able to rewrite that in terms of a single cosine. Or if I have a product of sine and cosine, I might be able to rewrite that in terms of a single sine value. Now if I wanted to, I can do the algebra beforehand and figure out an expression that tells me the sine of something times cosine, something what that is, what cosine cosine is, what sine sine is. I could get these other identities, but it's generally not worth memorizing them as long as you're willing to do a little bit of algebra. So let's take a look at how that might work. So here's a typical integral, find cosine 3x, sine 7x. And the first thing we note is that what we're looking at is a product of sine and cosine, and these products appear in the angle addition formula for sine. So we might, because this is the way we tend to write the identity, we'll rewrite cosine 3x, sine 7x, so that the sine is first sine 7 times cosine, and then I'm going to write down the angle addition formulas that produce a term like sine cosine. And so what is going to produce sine cosine? Well that's part of the formula for the sine of the sum of 7x plus 3x. So our angle addition formula, sine of a sum, sine of the first, cosine second, plus cosine of the first, sine of the second. So there's our sine 7x, cosine 3x, which is what we're looking for, and there is some other stuff. Fortunately for us, there's actually a second identity that produces the sine 7x, cosine 3x, and again that's going to be the sine of again, 7x, 3x, but this time we'll be looking at the difference between those two values. And so sine of a difference, sine of the first, cosine of the second, minus cosine first, sine second. And so there I have my two equate two identities that involve sine 7x, cosine 3x. Now if I add these two expressions, a remarkably useful thing happens. First off, I keep sine 7x, cosine 3x, that's actually what I'm interested in, but then this term here drops out, and then on the left hand side I have a sum of two different sine values. So over the left hand side, sine of 7 plus 3x, sine of 10x, sine 7 minus 3x, sine 4x, this twice two sine 7x, cosine 3x, and then these terms drop out entirely. And well now I have a problem because I didn't want to find two sine x, cosine x, I wanted to find sine x, cosine x. Well actually that's not too bad, I have a way of solving this through the expedient of dividing everything by two. I can find an expression for sine of 7x, cosine 3x is one half sine 10x plus one half sine 4x. And what that means is that this integrand, which I don't really know how to approach, I could rewrite in terms of a sum of sine values, and I know how to integrate sine. So let's go ahead and do that. So my original integrand, I'm going to replace that with what it is identically equal to, one half sine 10x plus one half sine 4x, and each of these two parts individually I could integrate without any difficulty. So that's going to be minus 120th cosine 10x, minus 180th cosine 4x, and don't forget our constant of integration. Well let's take a look at another example, how about sine squared of x. So here I'm going to try to integrate sine squared of x, and again if you have a lot of free time on your hand you can go around memorizing trigonometric identities. For the rest of us we'll try something easier and do a little bit of algebra. So sine squared of x, well here's the useful thing to remember, that's just sine of x times sine of x. And here we have a product of sines, and this appears in our angle sum identity for cosine. So if I want to find sine x sine x, one place that's going to appear is when I take a look at the cosine of x plus x, cosine of a sum, cosine first, cosine second, minus sine first, sine second. And there's our angle sum identity for cosine. I also get that product sine x sine x when I find the cosine of a difference of x minus x. And again if I expand that identity out, cosine of a difference, cosine first, cosine second, plus sine first, sine second. Now I want to solve this for sine x sine x, so I can do that by subtracting the first equation from the second equation. And what that will leave behind is the sine x's and then these cosine products will drop out. So I'll subtract the first from the second, and if I do that this minus this is going to be two sine x sine x. I'll do a little bit of simplification. This is just cosine of zero, this is cosine two x. And I do know what cosine of zero is, I don't have to carry that around. And finally I'll solve for sine x sine x by dividing everything by two. And that gives me sine x sine x, one half minus one half cosine two x. And again what I've done is I've taken this integral, which I don't have a good antiderivative for the integrand, but I can replace it with something that I do know how to antideferentiate. So here I'm going to have my sine square root of x, and instead of writing sine square root of x, I'm going to replace it to something that it is absolutely always equal to, and I can do the integration at that point. Integral of one half gives me one half x, integral one half cosine two x, cosine goes to sine, there's a factor of two that shows up, and I end up with this as my antiderivative. One more example just to bear in mind that anything we can do once we can do as many times as needed. So what that means is we already know how to handle the product of a sine times a cosine, this is a product of a sine times a cosine times a cosine, and in principle this is something that is a product of sines and cosines, and we already know how to do that. And since we can do it once, we can do it as many times as we have to. Now that doesn't mean that what we're about to do isn't going to be painful and tedious and may take a lot of time, but it does mean that it is not actually difficult, it is just time consuming. So let's take a look at that. Something we might start with is something we've already figured out, we know that sine seven x cosine three x, we've already figured out that this product of sine and cosine can be rewritten as the one half sine of the sum plus one half sine of the difference. So I don't have to carry around this product here, I can replace it with a different value, and so that allows me to do that a little bit of algebra. I'll expand that out and guess what? I now have a product of sine and cosine, sine and cosine, and once again I know what to do with a product of sine and cosine, and again I can apply my angle sum identities. Sine times cosine appears in the angle addition formula for sine, so if we do that we find that sine cosine, sine ten x cosine five x is a half sine of the sum plus a half sine of the difference, and likewise this sine four x cosine five x, again half sine of the sum plus half sine of the difference, and note that that difference is actually going to be minus x. And what that means is that this integrand here can be replaced with a whole bunch of sine functions. This sine ten x five x, one half sine fifteen, one half sine five, this sine four x five x is going to be one half sine nine x, one half sine negative x, and again this is now an integral of sines, I know how to do that, not a big problem, and I have this thing here, this horrible mess here, and we could probably do some simplification, so we'll take this great and enormous set up of simplifying this. Cosine of negative x is the same as cosine of x, so I'll change that, and there's our final answer.