 In this second half of lecture 10 for Math 1220 Calculus II, we're actually gonna start section 7.2 trigonometric integrals from James Stewart's Calculus textbook. This section is gonna focus primarily on integrating techniques applied to functions involving trigonometric functions like sine, cosine, tangent, et cetera, right? Now, the techniques we're gonna use here, we're gonna employ U substitution predominantly in this section. And we will sometimes need to use integration by parts which we saw in section 7.1. And so I wanna show to you how do the U substitution, the integration parts technique combined with trigonometric identities to help us find antiderivatives of trigonometric functions. So in this first video, we're gonna look at examples of functions that involve sines and cosines. So only sines and cosines, we'll deal with the other trig functions later on. If we only have sines and cosines, how can we handle finding the antiderivative? And it turns out our good friend is gonna be the Pythagorean identity sine squared plus cosine squared equals one. So why do we care about this so much? Well, some things to notice here is if you took this equation, solve for sine squared, you see that sine squared is equal to one minus cosine squared. And likewise, if you had solved it for cosine squared, you're gonna get cosine squared X equals one minus sine squared X. So the Pythagorean identity gives us the capacity of switching a statement that involves sine and cosine and can also switch from cosines to sines. So this is why we're converting between sines and cosines in our integral. Now, in order to use this Pythagorean identity, we do wanna have squares, right? Cause the other option is we take sine of X equals plus or minus the square root of one minus cosine squared. And this identity right here is not gonna be a good friend to us. Cause one, we really don't wanna put square roots in the integrand if we don't have to. Plus also, we have to worry about this plus or minus sign depending on which quadrant we're in. And so that adds a ton of complexity that we aren't gonna approach it that way. We'll stick with sine squared becomes one minus cosine squared and cosine squared equals one minus sine squared. And the other thing to keep track of in this situation is that when it comes to sine and cosine, they're their own, they're the derivatives of their complements, right? So the derivative of sine, recall, is cosine. And then the derivative of cosine is equal to negative sine. So there is this extra minus sign in front, but a constant multiple is not a big deal when it comes to integration here. And so because of this, we can play this game of U substitution all the time. So oftentimes when you do with a trigonometric integral, you actually, if you're gonna do a U substitution, you actually choose the DU first instead of the U. So what I mean is take a look at this expression right here, cosine cubed, integral of cosine cubed X DX here. So what I'm gonna do is I'm actually gonna set aside my cosine X DX first, and then we'll leave, well, there's a cosine squared leftover. And so this cosine X DX, I'm kind of setting aside, I want this to be my DU. Now, if my DU is cosine, that means my U should in fact be a sine. Or more specifically, if I take U to be sine of X, then DU would then become cosine X DX, like so. Now, if I want sine to be my U, you'll notice the integral doesn't have any sines in it. But aha, cosine squared can transition to be a one minus sine squared. So if we put that into play here, the cosine squared from above will transition to become a one minus sine squared X. And then this cosine X DX, we're reserving for our U, like so, then notice this statement right here, the one minus sine squared, that just becomes a one minus U squared. And this right here becomes my DU. And so very quickly, I can switch this into a nice U substitution, where this would look like the integral of one minus U squared DU, for which the anti-derivative would be U minus U cubed over three, plus a constant. And so then if you plug back in U, which U is sine, we get our anti-derivative of sine X minus one-third sine cubed X plus a constant, which is our anti-derivative right there. And so with the right U substitution, we can make trigonometric integrals possible, right? And it comes down to identifying what's the right U, what's the right DU? And oftentimes, like I said, when you wanna do a U substitution with trigonometric integrals, you might wanna choose the DU first, all right? Let's look at another example of this. Let's consider this time the integral sine to the fifth X cosine squared X DX. Now, when it comes to dealing with these trigonometric integrals, you're mostly gonna wanna focus on the powers here, right? So we have five sines and we have two cosines. Which one is preferable? Now, like we saw before with those Pythagorean identities, let's take a peek at them one more time right here. With these Pythagorean identities, right, we can transition sine squared into one minus cosine squared and cosine squared. Now, that works really great if you have even powers because if you had like a sine to the fourth X, that's the same thing as sine squared squared. And so if you can transition a sine squared over one minus cosine or a cosine squared into a one minus sine squared, who cares whatever the powers are? So this process right here works really well when you have even powers, right? And then looking at this cosine, we have a three cosine. The thing is we set one cosine aside for the U substitution and then the rest was even here. So what we saw right here, this U substitution looks really great when we have an odd power in the end. So transitioning happens great when you have even powers but for the U substitution, you want an odd power. Let's come back to this one right here. You notice we have an odd power of sine and we have an even power of cosine. So the cosine squareds can easily be turned into sine squareds because we can write those as a one minus sine squared if we wanted to, right? That transition's pretty easy. Let me erase that. Now, since we have an odd number of sines right here, what we're gonna do is we're gonna take the extra sine because in terms of transitions, we want an even power. Take the extra sine since we have an odd number and we're gonna write that sine x dx and this is gonna become our du. That's gonna be forthcoming. That leaves us behind a sine to the fourth and a cosine squared right here. Now, if we want our du, we want our du to be sine x dx. That had to happen because our u was cosine. But jk there, if u is cosine of x, the du actually be a negative sine of x. So we do want a negative sign here and correct the negative there. It's not a big deal. Let's take a double negative. So with that in mind, this negative sine x dx then becomes a du, no problem. This right here, it will become a u squared. What do we do with the sine to the fourth? Well, that's what we were talking about just a moment ago. Sine to the fourth x is the same thing as sine squared of x squared. And since the sine squared becomes a one minus cosine squared and cosine is u, sine to the fourth is to become one minus u squared quantity squared. And so you're gonna make that, you'll make that substitution with your integral. So we have a negative integral. The sine to the fourth became a one minus u squared squared. The cosine squared becomes a u squared. And then finally, the negative sine x dx becomes just our du. And we are able to transition entirely into these us. And now this is a polynomial expression in terms of u. We're gonna have to foil out the one minus u squared squared there. Just be careful as you do that. You'll end up with one minus two u squared plus u to the fourth. Then distribute the u squared throughout this trinomial. Doing that gives us negative the integral. We'll get u squared minus two u to the fourth plus u to the sixth. And now our patient is prepped for surgery. Let's perform the anti-derivative procedure here. We'll get u cubed over three minus two u to the fifth over five plus u to the seventh over seven plus a constant. Distribute that negative sine through. And also now we can replace all of the us with cosine of x. And so we see the anti-derivative will be negative one third cosine cubed x plus, cause it's a double negative, two fifths cosine to the fifth x and then plus one seventh cosine to the seventh x plus an arbitrary constant. Which gives us then our anti-derivative right here in this cute little yellow box. So what we did in these two examples just to highlight what happened there is that if we have some combination of sines and cosines. So we have sine to the A, cosine to the B, right? If either one of these is odd, either the A is odd or the B is odd. The technique we have here is doable, right? So for the odd power, for the odd power what we're gonna do is we're gonna take one, I'm gonna take one for the du. And then everything leftover we can transition to sines and cosines depending on which direction we need to go.