 We are at lecture two, second day of classes in this spring 09 semester. We are in the process of kind of reviewing a little bit, it's a little bit redundant with the end of 141, but sometimes those things are covered rather quickly, so we're trying to pick back up on those things, refresh ourselves, and then get into new material in the end of chapter five, numerical integration and improper integrals. Then we go into chapter six, which is applications of integration. A lot of fun things in chapter six. And chapter seven along with the supplement is first and second order differential equations. That's not all the differential equations that most of you will get because you will get a full course math 341, or possibly two courses 341 and 401 in differential equations, but we get a little taste of it in this course, which again, attributes to the choppiness of this course. And then we'll finish the course with infinite sequences and series, which is probably pretty new for most of you in the classroom. I doubt if many of your high school classes had anything to do with infinite sequences and series. Let's wrap up trigonometric integrals today. This is, I'm going to have to rewrite this or learn how to read upside down. I'll probably rewrite it, but I give me some time and I will learn names. Four of your names I already know because you're repeating and not repeating the course. Going on to the next course in this situation I had you in the previous course, but the rest of you give me a couple of weeks and I'll try to learn names. I did find out that this actually, if you happen to be ill and you miss class, this on a delayed basis, I'm not real sure the delay factor through four days. It runs on Raleigh Cable Channel 18 and it airs at 5 a.m. Doesn't that sound fun? Let's all get up at 5 a.m. and watch 241 Calculus. My wife probably well on occasion. We've already addressed that last semester. And 5 p.m. So if you do miss class and we can figure out when this lecture, which one you missed, you can actually see it a few days later on the Raleigh Cable Channel 18. 5 a.m. and 5 p.m. Nauseating people during the dinner hour all over Raleigh. Okay, that's my goal. Let's wrap up the kind of the basic trig integral problem by looking at one that looks like this. It's got a little twist toward the end that we did not look at yesterday. When we're stuck with even powers of sine and cosine, that's where we have to use that double angle identity or that half angle, whatever you want to refer to it as. So we made a substitution yesterday. You might want to look back at your notes either from yesterday or the notes that you have from the end of 141. What did we substitute for sine squared using the half angle or double angle identity? Okay, so one half. I'll take that over two and bring it out front. One minus cosine of 2x. Is that right? And what if we're again stuck with even powers of sine and cosine, which we are? What can we plug in for cosine squared? Same deal with a plus sign in the middle. So let's see what happens here and then we encounter another situation that we didn't encounter yesterday, but I think we have all the ammunition we need. Let's bring the one fourth out in front. If we multiply the two things that remain, the two binomials, what's the first term of that? Next term. What about the middle term? There isn't any middle term, right? Because it's a minus b and a plus b, right? So we lose that middle term and what's the other term then? So we have one fourth that will stay out in front. We're going to integrate one with respect to x. I could do that. That's not the reason for looking at this problem. This is pretty simple, obviously. Here's the reason for looking at this example. We're stuck with another even power of sine or cosine. What's the derail? What do we do when we're stuck with an even power of sine or cosine? Stick. Here we've got cosine squared x and we made this substitution, right? Well, we don't have cosine squared of x. We now have cosine squared of 2x. Is there a similar substitution? Yes, would be that answer, okay? And the next question is, what is that substitution? See what happens? Actually, I'm thinking of, that's later, okay? That's when we actually begin to integrate it. Cosine squared of x, in that when we came up with this substitution, our answer was in terms of 2x, right? That's why kind of it's a double angle identity. Now our angle is 2x. What do you think is going to be the new one? 4x. 4x, okay? But it's the same concept. So I won't simplify the first part because you know how to do that. But the second piece, because again, we're stuck with an even power of sine or cosine, we can use this same, let's call it a double angle identity. So for that, I can put in, you tell me what to write down, for cosine squared of 2x. One half times cosine 4x. Does that make sense? So whatever the angle is, it gets doubled using this identity. Whatever this angle is, it's getting doubled. So then, which I think is what Nicole was addressing, so we've got a, I'm going to go ahead and put brackets around that too. So I've got brackets within brackets. That's again not the issue. I'm trying to get to the reason why I wanted us to do this problem. We've got a 1 dx and we've got a cosine of 4x dx. This is the kind of problem that if you're not doing the substitution in your head at this point in time, I'd really like for that to be a short-term goal in this class. So I think everybody probably feels reasonably comfortable that if you have cosine of u du, you can work your way back to the antiderivative of cosine of u du. What has cosine of u for its derivative? Sine of u. No substitutions are necessary there. Now, kind of think through this process in your mind. Do we really have cosine of u du in the integrand? Well, we have cosine of u because we can call u anything we want to. But do we have du? What do we need? We need a 4 and a 1 fourth or you can kind of forego the 4 by saying we have 1 fourth of what we need. Does that make sense? So if you like to kind of put the things in there that it's lacking numerically, we need a 4. Can't just manufacture a 4 without manufacturing a 1 fourth as well. So now is this cosine of u du? It is. Now it is. So that's the kind of thing that as we enter into calc 2, those are kind of things that you should be doing in your head more and more. So this part of the problem, I'll just, I don't think there's going to be any problem with that part right now. But if we're integrating cosine of u du and we get sine of u, what is u? 4x. 4x, right? Is that okay for that last integrand? And the other ones, I'm assuming are simplistic enough to kind of battle through the arithmetic and be done. Now we've still got some brackets to deal with and coefficients out in front, but I wanted us to get to that point in the problem. Most specifically, I wanted us to get to this part of the problem. Those double angle identities can be used anytime you're stuck with an even power of sine or cosine. Is that alright? Can I take that away? Any questions on that before we move it? Alright, this is a little different direction, but it is also handled in section 5.7 a little bit, not much, it's not in depth. But there are some other trigonometric integrals that we might encounter and they are not just sine squared, sine cubed, cosine to the fourth, cosine to the fifth, not just powers of sine and cosine, but powers of secant and tangent. So let's at least take a quick look of secants and tangents in the same integrand. Now the thing, there are several things that come to mind when you have sines and cosines in the same integrand. How are sines and cosines interrelated in calculus? Derivative of sine is cosine, right? That's an interrelationship. Derivative of cosine is negative sine. There's another interrelationship and we have that Pythagorean relationship, sine squared plus cosine squared equals one. So we've got some interactions between those two functions. We've got similar interactions between secants and tangents. Give me some calculus something that has secants and tangents in it. What's derivative of secant? Secant times tangent. Right? There's an interaction in calculus between secants and tangents. What's derivative of tangent? Secants squared. There's another interaction between secants and tangents. Derivative of tangent is secant squared. Give me the Pythagorean identity that interrelates secants and tangents. Secants squared plus one equals secant squared. Okay, tangent squared plus one equals secant squared. How many of you knew that right on the tip of your tongue? Okay, well, you did. I didn't, honestly. I always kind of have to think. Here's what I start with. I know this one. Sine squared plus cosine squared is one. Everybody feel reasonably good about that one. Okay? Now, we want this new one. What can I do with this old friend of mine to convert it into my new friend? Divide by cosine squared. So what's sine squared over cosine squared? Tangent squared. What's cosine squared over itself? One. And what's one over cosine squared? Secant squared. So if you don't recall it readily, you've got one that I'm sure everybody does and use that to come up with a new one. So these three things, the Pythagorean identity, and there's different versions if you want to solve it for secant squared, this is the one. If you want to solve it for tangent squared, you can move the one to the other side and make it secant squared minus one. But those will help us solve this problem. Now, in the book, at the end of 5.7, this is one of the problems. And the author gives you this suggestion, which may or may not be all that helpful, really. Let u equal the secant of x. Not necessarily that u, that secant. I'm not going to say, oh, there's u. Now, I've got to deal with the rest of it. There's a tangent cubed. We're going to do this problem as a substitution problem, and we're eventually going to let u equal the secant of x. But keep in mind, if you're going to let u equal the secant of x, we better also have a du in the integrand. You can't integrate u things unless you have a du in the integrand. We agreed on that? You can't integrate u things with respect to x or r things with respect to q and so on. So what else needs to be in that integrand if we're going to let u equal secant of x? What's du? Secant tangent. Can we come up with one of those? We're going to have to move some things around, right? Let's take one of the tangents from here, send it out there to the right, and put it with the secant, and then we've got a du sitting out there to the right. So take one of the tangents. That makes that a tangent squared. Ooh, I got an idea for that too. So that's kind of what we did yesterday with the odd powers of secant, excuse me, odd powers of sine and cosine. We took one of those and we sent it out there to the right. This is the same kind of idea, sending it out there to the right so that this becomes eventually our du. Well, if this is going to be du, keeping in mind that u is secant of x, there's du. Doesn't this have to be something in terms of u? Right. So there we go. We want to get rid of tangent squared and get something equivalent to tangent squared that has a secant in it because we need secants. We don't need tangents. So for tangent squared, what was it? Secant squared minus one? Now, if we just kind of plug in our letters that things are going to represent in the substitution, this becomes, I think, a pretty easy looking problem, easier looking. So with u being secant of x, du being secant of tangent, what does this integrand become? u squared minus one du. Good. u squared minus one and all this stuff is du. That's a much easier looking integrand. So let's go ahead and finish it. What's the antiderivative of what we have at this point? Plus c. Everybody alright with that? And let's substitute back. Every time we see a u, we should replace it with a secant of x. Does that work? Any questions for it gets moved? I'm going to give you another one with a suggestion for a substitution and let you battle it for a few moments. Another secant tangent problem. This time we're going to let u equal tangent of x. Take a few seconds and try to make a decision of how you would slide things around or rename things or how you would do this integral. Keep working but I just want to, we got two chandlers, so I'll try to kind of decide which one I'm referring to. So, okay, hopefully you've decided kind of how to start the problem. What are some things that maybe you thought about or you actually wrote down before you started taking things away or moving things in this problem? Okay, good choice. du is secant squared. We're going to need one of those. If it's not in there, we can't manufacture it because it's not numerical. Can't manufacture variable quantities inside the integrand. What's next? Okay, good. Pull out a secant squared, which leaves another secant squared. So we've got that sitting out there for the purpose of its du, right? And if that's all going to be du, then doesn't this all need to be stuff in terms of u, in terms of tangent? Okay, part of it already is. So what choice needs to be made to get rid of secant squared? Good. How many of you were to that point in this problem you had gotten in? Okay, good. You'll see many, many times in this semester that I am not needed. Okay? I'm just kind of here to, you know, make it look official. But I'm just a figurehead. Okay? You're the ones that actually need to be able to do the work. So does everybody feel good about what we have that it's a solvable problem at this point? So when we convert stuff to u's and du's, this is going to be tangent to the fourth, which is eventually going to be u to the fourth, right? Tangent squared, which is u squared, and then this stuff is du. Everybody all right with that? Okay? Questions before I move it. There's more to it, but we've kind of already done that, and this is review anyway. So, okay, let's switch. I thought actually driving in to work today, I thought I might dig up the 141 final exam that I gave, and ooh, that brought some interesting looks. Looks could kill. I'd at least be severely wounded. We're going to review partial fractions. So let's do the problem that I had on my final exam from 141. We're going to review trig substitution. Let's review by using the problem that was on the 141 exam, and let's review a trig, I'm sorry, a table of integrals problem that was on the final exam, okay? Here's the directions. Decompose into partial fractions and integrate, and then you can also see how ridiculously mean I am that I chose these ridiculously difficult problems. Not really. Okay, so that's a pretty ugly looking integrant. So we want to take kind of off the side before we do the calculus problem. We want to take this rational function that conveniently is factored for us, and we want to decompose it into partial fractions. Tell me how to write that beginning step of this decomposition. Okay, good. I heard that from a couple of you. A, which A is a number. We don't know that number yet, but A is going to be some constant over 2x minus 1. What's the nature of that factor that we chose from the, it's linear. So a linear factor, even if it's a repeated linear factor, right? Always gets a constant numerator in this decomposition. So if this had, well, we'll do some sub cases when we're done with this. But linear factor gets a constant numerator. Next part of the decomposition. Good. So the next factor is an irreducible, right? If it were reducible, factorable, we would do that because it'd be easier. The problem would be easier if it factored. It doesn't factor. So it's an irreducible quadratic, and irreducible quadratic denominators receive a linear, we don't know the numbers involved, but linear numerator. I guess I should have done this before. What's the degree of the denominator? It's x cubed, right? And we've got an x squared in the numerator, so we don't need to do any dividing first before we start this process. But x squared over 2x cubed plus some other stuff, we can go right into the partial fractions decomposition. Everybody had that, I hope, in 141? No? Okay, that's a good reason for us to start this course with it. So if we can convert this ugly integrand into things that are added together, then we can integrate each piece of the sum. Is that right? And this, I can tell you what that's going to be, regardless of what a is. A could be negative 1711th, or some horrific constant. It's still going to be a natural log, isn't it? Isn't this, when we integrate it, going to be a natural log? Where the derivative of the numerator is, for the most part, the, excuse me, the numerator is the derivative of the denominator. Anytime that happens, then that's going to be a natural log. This could be a couple of different things that will actually, probably beneficial for us to review this as well. This, by the way, is in two places in your book. It's not only in this section that we're reviewing, 5.7. It's also at the back of the book, and that is appendix G, appendix G in the back of the book. So if this was something that was not hit extremely hard in your 141, and when we're done with this problem, you need a little bit more. Not only can you look at problems here, 5.7, but also some of the background from appendix G. So what would be the next step? Get a common denominator. So we need an x squared plus 4 here. We can't do that without also multiplying the numerator by x squared plus 4. We need a 2x minus 1 here, which we can multiply the denominator by that as long as we multiply the numerator by that. So x squared plus 4 over itself is really just 1. So we're multiplying by 1. That's legal. 2x minus 1 over itself is 1. So it's another version of 1, but again, that's legal. This is a step that you can probably forego, kind of skip over it. But now we have the denominators the same, right? We've got two fractions with like denominators, so we can add them. How do you add them? You add the numerators, right? And write them over the common denominator. So that's a step that you can probably skip because obviously we were making the denominators the same. That's why we did what we did. So now the denominators are equal. It's a matter of equating the numerators, right? So you could skip this step and go right onto this step. What makes the numerator on the left side equal to this new numerator on the right side? Let's go ahead and multiply things out. So that's what? Ax squared plus 4a 2bx squared plus 2cx minus bx minus c. Does that look right? How many x squared do we have on the right side? There's a term and there's a term. So a plus 2b. So we've taken care of that. Take care of that. How many x do we have on the right side? There's our x terms 2c minus b of them. Taking care of those now. And everything that doesn't have an x squared and doesn't have an x is a constant. So let's put those together for a minus c. So notice there's no calculus going on here. None of this is calculus. This is algebra. It's probably algebra that you never really did in Algebra 2 because what would be the purpose of taking a nice single fraction and decomposing it into two pieces, didn't we do things the other way in Algebra? Didn't we take two pieces and try to put them together into a single piece? Well, the advantage is that we end up with a sum. Each of the pieces of the sum is simpler than this ugly-looking original integrand. So we can work with those pieces and hopefully they'll be easier than the original ugly integrand. So we have this equation. So we want to equate things left side and right side. Tell me what things come to mind from where we are right here. A plus 2b, the coefficient of x squared over here, has to be equal to the coefficient of x squared over here because these two sides are equal. So a plus 2b has to be 1. Tell me another equation. So there's a negative 1x and this is how many x we have here and the third equation. So here's an equation with a and b. Do we have any other equation with just a and b? No. Here's an equation with b and c. Any other with just b and c? Here's one with a and c. Any other with just a and c. So we don't have that easy route. So we're going to have to do some kind of substitution, right? From one equation into the other two. I don't care. Somebody tell me what to substitute. Okay, you want to solve for b? So using the first equation, 2b equals 1 minus a and solving for b, b is, sure you want to do this one? Solve for b in the second one? I think I like that better. I don't want to mess with fractions at this point in the problem. I know I'm going to get them later. I'm going to abandon that one when I realize I've got fractions. It'll work, but I just don't want the fractions. So the next suggestion was what? Solve this for b. So I'll move the b to the other side to make it positive and move the negative one from this side, which makes it positive over here. So that's what? 2c plus 1. And what do I want to do with that new value for b? So we've kind of used this one. Don't we want to plug that into the first one? Is that good? So instead of a plus 2b, it's a plus 2 of these. A plus 2b is supposed to be 1. So now a plus 2 of these things is 1. And the other equation is 4a minus c equals negative 21. So there's a plus what? 4c plus 2 equals 1. So I'm going to move that 2 to the other side. It makes it negative 2, which is what? Negative 1 on the other side. You can accomplish this in fewer steps and kind of writing every step. So we want to get rid of the a's or the c's. What's the quickest and easiest way? Same kind of either way. Since one of these has positive c's and negative c's and the other has negative c's, why don't we multiply the second equation by 4? Does that work? First equation, we'll leave alone. Second equation, everything times 4. The c's will drop out that way. And what is that? Negative 84. Atom, 17a. Those drop out. When you add them, that's the goal, was to get them to drop out. And what do we get here? Divide both sides by 17. And what is that? Negative 5. Now we should be able to get the other coefficients a lot more readily than we got this one. If we know a, what can we now do? Either of these, right? The equations that have a with one other letter. Well, let's see. What's the best, best approach here? Plug it back into this one. Pretty simple one. So a plus 4c is negative 1. A is negative 5. Add 5 to both sides. C is 1. And now we've got a and c, so we can kind of choose whatever equation to find the other. We substituted earlier in the problem. For b, we substituted 2c plus 1. Now we know c. So b is 3. All integers, that's pretty nice. No ugly fractions. That's pretty nice. Right? It could have been a lot worse. If you get a bunch of ugly fractions, that can be bad. So let's go back to our original problem. That was one of our decomposed fractions. And there's the other one. So let's plug in the letters, the numbers for the letters that we now have. So a was, what was a? Negative 5. B was 3. And c was 1. So we had an integral problem to start with. We still have an integral problem. Now we want to reinsert the integral sign. We're going to integrate that with respect to x. And we're going to integrate that with respect to x. Again, this is something that if it's not a quick and easy substitution for you in your head, let's make that a short term goal. Let u equal 2x minus 1. Derivative of u is 2 dx. So this problem, I'm going to slide that negative 5 out in front. Once that negative 5 is out in front, then I've really got that. And how do I need to adapt that so that it's a workable integrand? Need a 2, right? This is u. So I'm going to have 1 over u. I better also have what? Du, which we don't have du. So I'm going to multiply by 2 and divide by 2. So this is now du over u. I don't think we've encountered that yet. We haven't had a lot of class, I guess that's why. This is du over u and what's the integral of du over u? Is that good? Remember that from 141? There'd be a plus c there. So we've got a negative 5 halves. Here's a du over u, which is natural log of u. Does that work? So the first one, we kind of benefited greatly from having this decomposed function. That's a pretty basic integral problem. Now that you've completed successfully 141, that's pretty easy problem. Now the next one we probably want to put into pieces. 3x dx and 1 over that same denominator. That's legal, isn't it? You have a summer difference in the numerator. You can break it up into two pieces over the same denominator. That would be yes. The answer to that would be yes. That's legal. We have nice microphones. We've got to put them into use so you can talk. But these are different approaches. The approach to this problem and the approach to this problem are very different. You might want to tackle one of the two and tell me what the approach is to that problem. We don't have an x on the top. If we did, if this were 1x, oh, this one. Yes, okay. There we go. In this one, we've got x squared down here. We've got the derivative of x squared kind of. We have to doctor it up a little bit. So can't this be u and the numerator be du? Okay, let's do that. I don't want the three there. You don't either. So let's move it out front. Let's not write it down. Let's think through it. What's our u? x squared plus four. What's derivative of u? 2x dx. Do we have that in the numerator? No. No, but it's correctable. So let's multiply by 2 and divide by 2. Now we're in business, right? So Chris, right? Chris said, for this one, it looks like we can tackle this one the same way we tackle this one. This is u down here. The numerator is now du. We've got a du over u, which is integrating back to a natural law. Now I could put absolute value of x squared plus four. You don't really need it, the absolute value, because it's positive anyway. Is that okay? If you're not liking that, because that's kind of too much cognition and not enough writing down, here's what we did. For u, we're calling u x squared plus four. Du would be 2x dx, and that goal was then to put it in this form, du over u. And if you don't like du over u, it's 1 over u du. Same thing. Is that all right? What's the approach on the last one? That's an inverse tangent. And the inverse tangent integrand, kind of the most generic one, is this one. When it's in that form, where this is some number squared, right? And this is some variable quantity squared. And this is the derivative of that variable quantity that's being squared. How's that going to work then? What's that? This is part of 141. Again, maybe didn't get the attention that it should have, but let's give it some attention now. 1 over a, not natural law, inverse tangent of u over a plus c. So if that looks completely new to you, you probably want to look back through earlier sections in chapter five. Where'd we first encounter that? I don't know, chapter five, somewhere before 5.7. That's also, if you want to check that out from the table of integrals, that's one of the formulas from the table of integrals as well. So here's our problem. A little actually simpler version than that one. So our u squared is really x squared, right? And our a squared is 2 squared. So what is that for us? 1 over a would be 1 over 2, inverse tangent, x over 2 plus c. Does that work? The reason I chose that problem for the final exam is it has lots of things in that same problem. So I didn't have to ask a separate natural log integrand because it's here. I didn't have to ask a separate inverse tangent integrand because it's there and the partial fractions that got us starting. So that's really about three or four problems in one. I can see that by facial expressions that this guy right here is one that's not sitting well with many of you. Let's back up before we do our next example. Let's take a look at that because if we review that and you can see what the derivative of that is, then it possibly will make it look a little nicer that the anti-derivative of that ugly thing we just did is in fact an inverse tangent. You might remember that off the top of your head. That's 1 over 1 plus x squared. Now I'm not saying that that's going to be the magical answer to this review that we're conducting. Let's back up from that just a tad. There's the function y equals inverse tangent of x and we want to find y prime. We want to find the derivative. We know what we're going to get. It's going to be this. But let's see where that comes from and maybe we can put all this together and we get this whole concept reviewed. So if we don't really know how to differentiate an inverse tangent, we have to rewrite that. So if y is the inverse tangent of x, another way of saying that is y is the angle whose tangent is x. Does that sound right? That's what inverse tangent means. y is the angle whose tangent is x. How can I rewrite that? If y is the angle that I'm going to take the tangent of to get x. Tangent of y equals x. Those two statements are equivalent. So if you don't know how to take the derivative of an inverse tangent, I feel reasonably confident that everybody in here knows how to take the derivative of a tangent, right? And this is tangent of y, but we can do implicit differentiation. What's derivative of tangent of y with respect to x? What's derivative of tangent of something? Secant squared of that thing. And then we're deriving a y thing with respect to x. Does that look familiar? That's again chapter five. No, that's earlier than five. That's implicit differentiation, which is way earlier. So we're deriving a y thing with respect to x. Every time we do that, we have to kind of acknowledge the fact and the chain rule or implicit differentiation. We're trying to do that. So derivative of tangent of y is secant squared of y dy over dx. Derivative of x, I gave the answer away, with respect to x is 1. And what do we want? We want y prime. That's dy over dx. So it's 1 over secant squared of y. What's 1 over secant squared? What's an easier way of writing 1 over secant squared cosine squared? But that's still, now we've got this answer in terms of y. And what we really want is an answer in terms of x. So let's take this guy right here, get a picture of what that statement says. Y is the angle. And when we take the tangent of y, what do we get? We get x. So how do I label some other parts of this triangle? If the tangent of y is supposed to be x. Opposite is x. And adjacent is 1. Does that diagram say the same thing that this statement says? Is the tangent of the angle y is it equal to x over 1? And if you know two sides of the right triangle, can't we figure out the third side? So these are the two legs. So x squared plus 1 squared. Take the square root. We've got the hypotenuse. x squared plus 1 squared is 1 plus x squared. Square root. So we want the cosine of y and then we want to square it. What's the cosine of y from this diagram? Adjacent over hypotenuse? Is that right? Hey, how about that? What do you get when you square that? There it is. There's the guy. The derivative of inverse cosine is what? The first one. Not squared. This one? Yeah. Derivative of inverse cosine is, now, it's got similar letters and numbers, but they're not in that same position. This is honesty now. I'll be honest with you. At this very second, I don't know what it is, but I know in about 25 seconds I could derive it. So some of those things that I've got limited space up here, probably very limited compared to you. It's what happens when you get older. So I don't want to use up limited space for that. I know I could come up with it in about 20 seconds if I needed it. But it does have similar letters and numbers, but not exactly that same order. But there is kind of the background of why derivative of inverse tangent is something that looks really strange like that. Now, if that's the derivative of inverse tangent, shouldn't we be able to anti-differentiate this back to the inverse tangent? Is that right? So let me wrap it up today with this. If this is derivative of inverse tangent, we ought to be able to anti-differentiate something that looks like this and work our way back to an inverse tangent. Right? We're done for today. I will see you tomorrow.