 Welcome back to our lecture series Math 1220, Calculus II for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misseldine. In lecture 16 in our series, we're going to take an official speed bump in our curriculum here, particularly in section 7.5 of Stuart's Calculus. We want to review the different techniques of integration that we've learned about. In Calculus I, we learned about the substitution method, particularly how it pertains to u-substitution. We've also learned an adaptation, a modification of the u-substitution technique, which we call trigonometric substitution. We've learned about that in section 7.3, excuse me. 7.2 helped us with trigonometric integrals, which are necessary to do trig subs. Also in this lecture, we're going to learn about something called a rationalizing substitution or a ratio sub. We'll see that a little bit later. It's a third variant of this substitution method. Also in chapter 7, we've learned about integration by parts, which is a very useful technique. It helps us undo the product rule, much in the same way that u-substitution helps us undo the chain rule. We saw some variations of integration by parts, our integration by cycles, our integration by hope. Then we also saw that algebra itself can be helpful to help us with trigonometric, sorry, just for integration, right? Like that in 7.4, we learned about partial fraction decompositions, which are purely algebraic techniques to help us integrate. We've seen how completely the square can help us do a trig substitution. And just basic factoring and simplifying algebraic expressions is essential to help us rewrite an integral in a way that is much easier to use, that is to compute the anti-derivative. So in section 7.5, which we'll call street-fighting integrals, I use this name here because I borrowed this actually from a textbook, from a book that's entitled Street-Fighting Mathematics. The idea is when one does real mathematics outside of a classroom, there's no one there to tell you what the rules are. It's like, I can tell you, you must complete the square, or you must use partial fraction decomposition. That may work well in a test, where it's like you have to do it this way, but in real life, no one's telling you. There's no rules, the only rule is if you can do it, you can do it, right? Is it wrong to scratch someone's eyes out or kick them in the foot, or whatever, if you bring a gun to a knife fight, guess what, you get to use a gun, right? And I hate to use such violent imagery here, but when it comes to computing integrals, there's no one telling you what you have to do. And so it behooves us to determine what technique is best to use. And so I wanna look at a few examples to illustrate this type of principle here, right? So if you look at this integral right here, the integral of cosine of x times one plus sine, sine squared x dx, how does one calculate the antiderrid of such a thing, right? Well, there's basically three things we wanna use. We either wanna use algebraic identities, integration by parts, or use substitution. Some are some variation of these things right here. And so my first inclination is, well, if I have a cosine, I should distribute it. And if you distribute it, you're gonna get the integral of cosine x dx plus the integral of sine squared x cosine x dx. And we spent a long time talking about antiderivatives of trig functions in 7.2. Antiderivative of cosine is pretty nice. It's just gonna be sine of x. For the second one, a U substitution is appropriate here. Take U to be sine of x, and then take du to be cosine of x dx. This then would look like U squared du for which the antiderivative should then be one third sine cubed x plus a constant. So the key on this one was just distribute it, right? If you distribute it, you're good to go. And that's really what we wanna do in this situation here. Cause we kinda treat the two things differently. And you could actually done a U substitution from the very beginning. You could have taken U to be sine of x du to be cosine x dx. In which case this will look like then one plus U squared times du as your form. And so you could have handled that as well. So there's more than one way one could approach these things. For the second one, B, which also involves some trigometric functions. Again, I would try to use some algebra with our basic trig identities to help us out here. Take sine of x over tangent dx, and then take secant x over tangent x. So some things to note here that tangent is the same thing as sine over cosine. So if you are dividing by tangent, that means you're timesing by cosine and you're dividing by sine. So the first one actually becomes cosine of x dx. The second one you end up with, let's see here, this one I'll have to try a little bit harder on. You're gonna get a cosine on top, a cosine on bottom that will cancel, and you'll get a sine on the bottom. So this is just a cosecant dx. In which case you find the antiderivative of cosine. That's a sine, we've seen that already. Look up the antiderivative of cosecant if you don't remember it, and then go from there. This one right here, it's not about solving them. This is about just getting started, knowing what the right technique to use. On this one, number C here, x plus one over x squared minus four, x minus five. This is a rational function. It makes me think maybe I need to do some type of partial fraction decomposition. If I'm gonna do that at the first factor, the denominator, which factors of negative five that out to negative four, you're gonna get x minus five and x plus one. So I want you to stop there immediately, right? Notice x plus one shows up in top and bottom. You can simplify the fraction. This would then just become the integral of dx over x minus five, for which the antiderivative is just a standard logarithm you get from the U substitution. The natural logarithm value of x minus five plus a constant. So don't underestimate the ability of algebra, right? Algebra and other type of trigonometric identities can be useful as we try to simplify these things because sometimes simplifying it dramatically changes what we have to do. Because we recognize the cancellation of the x plus one here, we didn't have to do a partial fraction decomposition, even though that's how we were getting started with it. Let's take a look at another example here, right? Let's take tangent cubed over cosine cubed. I'll go through this one a little bit slower. We want to use algebraic simplifications to help us out here, right? If you have cosine cubed on the bottom, tangent and cosine don't work really well together, but I could churn the cosines cubed on the bottom into tangent cubes like so, in which case we get a tangent cubed x times a secant cubed x dx. So tangent and secant get along very nicely here. And since we have a tangent cubed and a secant cubed, what I wanna do is I'm gonna actually pull aside a tangent x and a secant x dx. I wanna set this aside because this is gonna be my du right here for a substitution. That leaves behind a tangent squared x and a secant squared x. Now secant is what I want to be my u because it's derivative is tangent secant. In which case if you're gonna do that, you don't want a tangent squared. Instead, you wanna replace tangent squared with secant squared minus one. Or in terms of u substitution, this would look like u squared minus one. That is more specifically as a u substitution, this becomes u squared minus one times u squared times du. So we do a nice substitution right there in which case then distribute the u squared. We get u to the fourth minus u squared du. Anti-derivative is pretty nice. We get u to the fifth over five minus u cubed over three plus a constant right here. And then replacing u with secant, we get the final answer is one fifth secant to the fifth x minus one third secant cubed x plus a constant. So in terms of a trigonometric identity, a trigonometric integral, it's fairly standard form, nothing too complicated there. The key thing was really, you had cosine and tangent interacting, use the identity to switch cosine to secant and it becomes much more natural what to do. All right, let me give you one more example to emphasize how awesome algebra can be here. And so this one right here, let's find the integral of e to the x plus e to the x. And now I'm gonna give you a challenge. Pause the video and try to do this one on your own. Try to do that and see how well it goes for you. Now, some of you might be able to see it pretty quickly, but many will struggle with this one. And so now if the video is on pause, hopefully you gave it the old college try on this one. This one sounds really hard when you look at it in this present form, but if you take e to the x plus e to the x squared, remember when you add x once together, that means multiplication. Adding x once would be the same thing as e to the x times e to the e to the x. And then of course, if we rewrite that, you get the integral of e to the e to the x times e to the x dx. If we were to do a u substitution where u equals e to the x, then du is itself e to the x dx. You have your du right here, you have your u right here. This thing looks like the integral of e to the u du, which we know that one, that's just e to the u plus a constant. And then so substituting back in x, you end up with e to the e to the x plus a constant. And that's all there is to it. Super simple, super slick, but it only is super easy when you see this, algebra. This algebraic identity about exponents is what saves us here. And that's what we need in order to help us with this one. So by all means, use integration by parts, use substitution to help you with these integrals, but don't underestimate how useful algebra could be. It could be that the reason you don't see the anti-derivatives, because you're looking at the wrong form. You might need to do some type of simplification or modification of the integral to make the substitution or the integration by parts much more obvious on what to do. And also be patient with yourself. With these ones, practice, try, try again. If you make a guess, go with it. If it doesn't work, try something else. This section is all about practice, practice and practice. You'll never get good at this until you practice. And you have to also be willing to make mistakes. If you get something wrong, that's okay. We learn more from our mistakes than we learn from our triumphs. If you get stuck, don't be afraid to ask questions. The hardest part of calculus is not actually the mathematics itself. It is fear. If you're afraid to try, if you're afraid to make a mistake, you will fail. If you're afraid to ask questions, you're gonna be stuck for a long time and be wasting a lot of time. Get the help that you need. Work with your study groups or whatever you have. You can do this. I really know you can. Just be patient with yourself.