 In this video, I wanna summarize the basic techniques we and tips that we've learned from U substitution so far and do some more examples of computing anti-derivatives using this technique of U substitution. So there's two sort of guiding principles one should use, two rules of thumb when deciding whether you should use a U substitution to calculate an anti-derivative or not. And so these two principles are listed in front of you A and B here. First of all, you should be equal to some expression that's inside the integrand, the function you're trying to integrate. If U isn't there, then why that junk are you bringing it up in the first place? I'm not exactly sure. So you just wanna substitute that expression with U. And this substitution generally should be chosen to make the expression become simpler. The integrand should be simpler than it was before. If you've made some substitution that makes it more complicated, then you're probably going in the wrong direction. And typically this function, which is U is inside of some other function as well. So look for that. U should be chosen to be something in the integral to simplify it. And then also the second part, and this is perhaps one of the most important parts, is that whatever we choose to be U, the derivative of that function needs to be inside of the integrand as well. The derivative of U needs to be present. Although if you're only off by a constant multiple, we saw in the previous example how one can correct that. And we'll see some examples of that in this video as well. So if you're off by a constant multiple, no big deal. So let's look at these functions right here. Generally speaking, U is gonna be inside of another function. So use like your parentheses in the function as sort of like as a radar, as a sonar to help us identify who should U be. And this can be difficult at first. Students sometimes struggle to identify who is the best choice of U, right? You don't always know who the best U you are until you practice a little bit more with it. I always love using the variable U because it leads to awkward English grammar when I say things like U is. Sounds like I'm illiterate at the moment. But anyways, we wanna choose functions which are generally inside of functions. So with this one right here, I see these parentheses in front of me, X squared plus six X squared. So that kind of indicates to me that well, because I have these parentheses right here, maybe that's my choice of U. So more specifically, we'll take U to be X squared plus six X. That is inside of the function. And if I were to replace, if I were to replace the denominator with a U, we would end up with a U squared on the bottom. I would say that's arguably simpler. But what about everything else, right? Well, let's look at the inner derivative. Look inside yourself and find your inner derivative who wants to come out and play. If you take the derivative of X squared plus six X, you end up with a two X, a two X plus six DX. And is that what we have in the numerator? What we have in the numerator is an X plus three times DX. But then if you're like, well, and because that's the thing that's like, oh, does that work right? I want a two X plus six. But if you look at the numerator, it's like, well, and then if you look at your DU, it's like, well, I guess I could take out a two X. I mean, that's common to both there, two. Sorry, not take out a two X, take out a two. And that'll leave behind an X plus three, and voila, we found it. It's a miracle. If you take a two over two right here, you can correct that missing coefficient of two you have. And so you're gonna end up with a, that the two times X plus three times DX, this right here becomes your DU. And so you end up with a DU in the numerator and then a one half out in front. And that I would say is a much simpler integral to compute than what we started with. One half the integral of DU over U squared that simplifies the integral and the derivative of the inner function was contained inside the integral here. The calling card of the chain rule was present at the crime scene. So we wanna integrate one half the integral of U to the negative two DU. I'm gonna use negative two for the forthcoming power rule here, right? Because by the power rule, you raise the power by one. So raising negative two by one, you're adding one to it, you're gonna get U to the negative one and then divide by that power and add a constant. And so you're gonna get a coefficient of one half in front. U to the negative one, of course, it's the same thing as one over U. If you prefer to write it using a reciprocal since the original function had a reciprocal it makes sense to do that. But you shouldn't write your final answer with U. U is this like artificial variable we inserted into the problem. What we need to do is go back to the original variable which was the variable X here. And so we're gonna get negative one over two times X squared plus six X and then there's a constant added to that. And so this gives us our anti-derivative of this function right here. Looking at another example, we get two X minus three over X squared minus three X. We wanna find a U in order to make a U substitution right here. And it might not be necessarily obvious here because I don't see any parentheses here but whenever you have a fractional function of some kind, a fraction like this, the denominator is often a good choice to be the U because the denominator, I like to think of that fraction bar there it's sort of like it's a wall like a Berlin wall that separates families on the two sides. Oh no, bring us together. And so the just like parentheses are functions in prison inside the denominator and that's who we're gonna select to be our U typically that function which is in prison inside of some other one take U to be X squared minus three X. Well, it's derivative is two X minus three DX and that is exactly what we have in the numerator, right? Two X minus three DX, oops, sorry about that. You end up with DU. And so then making this U substitution work right here we end up with DU over U or if you prefer one over U, DU. Now the power rule doesn't work in this situation because if you did switch this to a power you get U to the negative one in which case the power rule doesn't work in that situation. And in this case, you just wanna remember that one over U is the derivative of the natural log. So you get the natural log of the absolute value of U plus a constant and then substitute back in your U which is X squared minus three. So we get the natural log of the absolute value of X squared minus three X plus a constant and don't forget the absolute values because X squared minus three is not always a positive expression depends on the choice of X. So you do need to have absolute values that the domain is correct. Without the absolute value you wouldn't have the correct most general anti-derivative. So do consider that one there. As another example we could take the integral of E to the five X and yet there is a chain going on here, right? When the exponential function typically your U is gonna be chosen to be the exponent because the exponent is trapped inside there. Oh no, free me. And so take U to be five X then DU would equal just five DX which I don't see the five there but oh, was that a butterfly going by? And when no one's looking you write the five over five there. So it's like it was there the whole time. JK everyone, no worries. So you get one fifth the integral of E to the U DU. Notice that the five in the numerator times the DX this comes together and becomes DU which we make that substitution right here. And so if you wanna integrate U with respect to U sorry, integrate E to U with respect to U E to the U is its own antiderivative so we get one fifth E to the U plus a constant, right? In which case if you plug back in U as five X you get one fifth E to the five X plus a constant right here. And this is a very simple U substitution that we're gonna see a lot and so we actually might just kinda memorize this formula. If you integrate E to the AX DX the antiderivative will always be one over A E to the AX plus a constant. Nothing too significant about the five right there. Let's take a look at this one. Take the integral of X squared E to the X cubed. If we follow the same idea from the last one the exponent of an exponential is imprisoned and it's usually a good choice for U. So what do we have U equals X cubed? Well then DU would equal three X squared DX and then when everyone's distracted you write your three on the inside and your one third on the outside. And so then we have our DU, the inner derivative three X squared DX. This is our DU. And so then the antiderivative can be simplified to be one third. The integral of E to the X cubed becomes an E to the U and then the three X squared DX becomes a DU. And so just like the last example it doesn't matter how complicated the U is the antiderivative E to the U is still gonna be E to the U plus a constant and plug back in the original expression for U. So one third E to the X cubed plus a constant and this gives us our antiderivative right here. It's pretty awesome how nice this U substitution works here. But don't, I mean, the reason I keep it on working is because I am giving you some integrals for which the inner derivative can be created up to some scalar multiple. That's not always the case but we can explore more of that in calculus two. Now here, let's do one last example in this segment right here. Let's take the integral of X cubed times cosine of X to the fourth plus one. We have a trigonometric function here. We haven't done any of these yet. But trigonometric functions naturally come with these parentheses. This kind of indicates the thing inside the parentheses is probably gonna be our U. Take U to be X to the fourth plus two. Taking the derivative, we see that the inner derivative should be four X cubed DX. We have of course the DX, that's always there. We have the X cubed, we need a four and we can do that of course by times four over four there. So therefore, four X cubed and the DX comes together to give us our DU. And so the simplified version of the integral becomes one fourth. The integral of the four X cubed DX that just becomes a DU. And the cosine of X to the fourth plus two that just becomes cosine of U, cosine of U right here. And remember that although the derivative of cosine of X is equal to negative sine of X, we actually have the opposite true when we do antiderivatives. The antiderivative of cosine of X DX, that's gonna be a positive sign because we're looking for a function whose derivative is cosine. So when we do that here, we end up with one fourth sine of U plus a constant and then substitute back in the original value of U which is X to the fourth plus two. We get our antiderivative one fourth sine of X to the fourth plus two plus our constant here. And so I guess there's some more examples of U substitution. We'll do some more in the next video. Please take a look at that to see it so you can see some more fun examples though. But as always, if you have any questions, feel free to post them in the comments here on YouTube. I'll be glad to answer them if you post any. 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