 Welcome back lecture three. We are wrapping up the review today, kind of an overlap between this course in 141. So if you have questions about any of that overlap, which I kind of picked up on yesterday, that it wasn't complete overlap for everybody in here. So anything you want to ask about that, because we will start with new material on Monday. So here we will use the table of integrals today. So if you don't have yours and you can kind of look off somebody else's, I have one that we can put up here and look at. We're on partial fractions. So let's talk about the two situations that we haven't talked about, at least as far as a review, that address decomposition into partial fractions, and then from that point we can integrate each piece. So these are taken from the appendix in the back of your book, appendix G. This is the first case, we've already addressed that, that if we break up this rational function, this R of X over Q of X, and we break it up into pieces, and each of the pieces is a linear factor in the denominator. Linear factors, regardless of how many there are, this is a linear factor, it gets a constant numerator. Here's a linear factor, it gets a constant numerator, and we'll continue with that. For every linear factor, they'll all get constant numerators. We had an example that dealt with that. The second case is kind of a sub case of this, we have not looked at yet. So we have a product of linear factors in the denominator. Notice there's an X squared in this example, but X squared is really a linear factor, it's X and X, 1X plus 0 and 1X plus 0, but it's a repeated linear factor, as is this. This is a repeated linear factor. Notice every numerator in this decomposition is what? Constant. So if the nature of the factor itself is linear, regardless of to what power that linear factor is being raised, it will command or demand a linear numerator. So here's our first linear factor X, it gets a constant. This is that linear factor X to the second, it's not an irreducible quadratic factor, because it is reducible, it's X times X. It gets a constant numerator. Here's our linear factor to the first, it gets a constant numerator. Our linear factor to the second, it gets a constant numerator. Our linear factor to the third, it gets a constant numerator. So it's kind of the nature of the factor itself, what type of numerator. Now, think of an algebra problem back in ninth or tenth grade, where you were taking this set of fractions, probably one of the first things you'd do would be to decide on a common denominator, what would the common denominator be for all of these fractions that have a constant numerator? Would we need just an X? We'd need an X squared, right? And we'd also need an X minus one to the third. Isn't that the common denominator? Right? So kind of think of that algebra problem that you did in your prior lifetime before you came here. This is exactly the reverse of it. So we're allowing for the fact that there might be a factor of this type. It is possible that A is zero. We allowed for this factor, but in fact, that factor isn't present. And same thing with any of these others. It is a possibility that any of these linear numerators could be zero, which says we allowed for that factor, but it doesn't really end up in the partial fractions decomposition. So we didn't look at an example of this yesterday, but there is a pretty good example of a repeated linear factor. Third case, I think we looked at one of these yesterday. Again, this is from the appendix in the back of the book. Irreducible quadratic factors. So if you've got irreducible quadratic factors, here's one. X squared plus one cannot be factored. Here's another one. X squared plus four cannot be factored. Anytime we have a quadratic, let me rephrase that, irreducible quadratic factor, those kinds of denominators in the decomposition will require a linear numerator. Here's our linear factor. It gets a constant numerator. Here's our irreducible quadratic factor. It gets an arbitrary linear numerator. We don't know what B is. We don't know what C is, but it's kind of generic linear. Another irreducible quadratic, it gets another, potentially another different linear numerator. So we did an example with that in it yesterday. And then the fourth case, we did not do an example with this in the example that we looked at from my 141 exam yesterday. If we have an irreducible quadratic factor in the denominator that is repeated, so I don't know why there's not an example here. Maybe I cut it off too early. So let's say we have an x squared plus four squared and an x squared plus one cubed. And we've got some polynomial of degree smaller than what's in the denominator, in the numerator. So we allow for x squared plus four to the first. It gets a linear numerator. Somebody tell me how to keep going here. What's the nature of the factor that's repeated? It's quadratic. What kind of numerator does a quadratic factor receive? Cx plus d. Cx plus d linear, right? And we would keep going with that till we've got it to its highest power, which were there. So then we'd start with our other factor. This, by the way, I think would be a fair test question to decompose it kind of the first step and then to solve it from there. This would be hideous. So it would not be a fair test question. But this would be, can you decompose it, kind of do the first step? What would it look like in the first step? And then what? x squared plus one squared. And another one, ix plus, I'm going to put another one. Does everybody get the point there? That it's the nature of the factor, and if it's repeated, we allow for that factor appearing. But each time, if the factor is linear, it gets a constant numerator. If the factor is an irreducible quadratic, it gets some arbitrary linear numerator. Questions on that? So all that's in appendix q, if you chose not to write that down. That's where it is in the book. Let's do a couple of table of integrals problems. I've got actually several. Somebody asked me after class, Daniel, was that you? Was that a web assign? That's a trig substitution problem. Actually, let's go to that. I think I have that. I think it's a problem from the book that's also a web assign problem. Does this look like it so far? X cubed in the numerator? So it's a web assign problem. It's also problem 14 in the 5.7 problem set. This is a trig substitution problem. If I were to give you a trig substitution problem on your first test, I would not expect you to come up with the initial substitution. This actually isn't a full-blown treatment of that. But if I give you this kind of statement, let me find it for this one. Well, I didn't write it down. I think it's this. Let x equal 4 sine theta. Is that right, Daniel? Okay. So that would be given to you. Now, it's your job from that point to make the substitution, change this from an x dx problem to a theta d theta problem, integrate it, bring it back to x's, or change your limits of integration so that you can evaluate it over the original x values 0 to 2 squared to 3. Excuse me. So if this is going to be given to you, then you need to be able to take it from there. So we need to know what dx is because we're transitioning this whole problem away from an x comma dx problem into a theta comma d theta problem. So we know how to get rid of the x's. Every time we see an x, we're going to put in 4 sine theta. But we also have to eliminate dx. So if this is our choice for x, what is dx? 4 cosine theta d theta. Okay? So now we can do this transition. What web of sine problem is this? I think number 4. Okay. Web of sine problem 4. I did it without shake, so it's going to work. Okay, how did you do it? I just said u equals 16 minus x squared. And I took x squared out of the top and made it just x. x times x squared? Yeah. Okay, yeah, I can see how that would work. And you'll see when we do this substitution and start knocking out some like terms, your integrand will look a whole lot like this one. Okay. The approach, though, that you used, so you let u or whatever, tell me what you let. I said I'd let u equal 16 minus x squared. And du equals negative 2x. And then I just pulled out a negative 1 half. Okay, all right, I see. So you're kind of stuck with your radical in your problem, right? You didn't eliminate the radical. No, I just meant u to the 1 half. Right. So his approach is not going to be all that much like what we're going to do because we're going to change what's under the radical so that it is something squared and then we're going to take the square root of that something squared which then eliminates the radical. So they're going to not be as similar as I thought originally. So let's see what it looks like when we make the substitution and see why trig substitution and that's common. Different methods are common for problems. The question is, can we get to the same solution from different methods? The numerator is x cubed so it's going to be 4 sine theta cubed. We've got a dx that technically is in the numerator. We've got another 4 cosine theta d theta that's in the numerator and in the denominator which is kind of the real reason for the substitution anyway. We have 16 minus x squared which is 4 sine theta squared and that substitution is going to allow us to eliminate the radical. We're going to have the square root of a perfect square. Questions thus far? Anybody? Should I write in the limits of integration? No, we'd have to change them if we're going to write them in. Aren't these x values? So I'm just going to put a little reminder to myself that we do have limits on this problem. I don't particularly at this moment want to change them to thetas theta 1 and theta 2. I'll kind of bring it back in terms of x before I evaluate. Alright, 4 cubed and then we've got another 4 is that right? 4 cubed is 64 times another 4, 256 we've got a sine cubed we've got a cosine the denominator is going to be 16 minus 16 sine squared theta is that right? Factor out of 16 if you factor out of 16 under the radical you've got 16 times 1 minus sine squared and what is 1 minus sine squared? Cosine squared, so we've got 16 cosine squared kind of getting to why the trig substitution happens to work at least this particular choice for x, why it works in this problem. So the square root of 16 is 4 the square root of cosine squared is cosine what now? Cosine over itself Cosine over itself is 0 what else? Make it just 64 4, get rid of that there divide this by 4 64 so at this point we have I'm going to bring the 64 out in front we still have a definite integral but I'm choosing to not bring that back until we get it back in terms of x Did we do a problem like this yesterday or the day before? No. Odd powers of sine or cosine what was the little procedure that allowed us to integrate it? Pull out one of the sines and throw it out there to the right change what's left to cosines, right? We have a sine squared and a sine sine squared we can change to minus cosine squared I think your method is looking pretty good right now man, kind of got there a little quicker than this but this works and I do want to do an example of trig substitution so we've already made one substitution, right? to get it to this point because we started with x's now we're in thetas I think either we can kind of think through the next substitution or we can actually write it down it's pretty early in the semester let's write it down so I want to let I don't know r the cosine theta and if r is the cosine theta what's dr? So it looks like we need what, negative one here and a negative out in front so this part at the end is all dr and this is 1 minus r squared does that look good? how many of you are doing this kind of stuff in your head? a couple that's a good early goal in this class for you to kind of not have to write this down and say well that's this is sine theta is the derivative of cosine I've got those I'm going to integrate them get back without making the substitution but it's fine so we've got 1 minus r squared dr what's the integral of 1 with respect to r would be r in the integral of r squared with respect to r r cubed over 3 we've got a negative 64 out in front now let's see if we can track our way back everywhere there's an r we should replace it with cosine theta and then to get rid of theta and go back to x we'll probably need a little diagram for that and you'll notice a part of the triangle we're about to label will be similar to a term that appeared in the original integrand I really have a hard time thinking of a time when you use trig substitution that that is not the case so from our original substitution here it is equals 4 sine theta isn't it true that x over 4 equals sine theta is that right so let's label our diagram we'll find an angle theta and then we'll label it as such that when I take the sine of that angle theta we're going to get x over 4 so there's theta so how am I going to label this triangle such that the sine of this angle theta is x over 4 opposite over hypotenuse opposite is x and hypotenuse is 4 does that say the same thing that we wrote out earlier in the problem and now we can solve for the third side wouldn't it be hypotenuse squared minus leg squared square root is that a term that you see in the original problem in fact that was the denominator very common that that's going to be the case so we need to substitute for cosine theta and cosine theta right so what is the cosine of theta this square root over 4 yes and there's another cosine theta so we want to cube that sorry that should be over 4 right yes that whole thing is going to be cubed over 3 now we can bring back our original limits of integration because those are x values kind of a mess right so we substituted a trig substitution to get rid of the radical we're stuck with the sine cubed we pulled one of the sines out we did a trig integral I think is what those are called by making another substitution worked our way back to the original letters which are x and let's see what we can get I don't think the substitution is going to be an issue so let's not take class time to do that but if we distribute the 64 what are we going to get negative 64 now we've got a 4 down here that's technically if you want to put it where it belongs it's in the denominator right it's in the denominator of the numerator so we can put it in the denominator so there's a 4 cubed which is 64 so what's the end result here negative over 3 over 3 64's should that be plus we've got it integrated by using trig substitution and a trig integral and kind of working our way back I don't think that this is going to be an issue you would put in 2 squared to 3 for x and evaluate get a number subtract from that put in 0 a lot of times when you put in 0 you get 0 not the case here when you put in 0 for x you get the square root of 16 which is 4 you're going to get something numerical in both cases when we put in 2 squared to 3 and squared we're going to get 12 just wanted to make sure we didn't get a negative under the radical which we don't everybody feel comfortable from that point trig substitution eliminated the radical little substitution to deal with and I'm pretty sure that's a web of same problem more trig substitution problems this also is in your book it's problem 11 slightly different from what we just did common to these trig substitution problems are that you have a radical that looks like the original problem it's an irreducible quadratic under the radical we're trying to take the square root of it it's just not going to happen square root of the sum of 2 squares we're kind of stuck with that so the recommendation is to let x equal 2 tangent theta I'm going to give you that now you have to be able to take it from there so tell me now that we've done an example of this type tell me kind of what's next dx what is dx d theta right changing the problem from an x dx problem to a theta d theta problem I guess we just go ahead and do that right numerator is 1 and also in the numerator is dx so there's our dx in the numerator denominator is x squared we'll put in 2 tangent theta for x and then square it under the radical which hopefully this is what is going to simplify by the nature of the trig substitution itself x squared does that look right? let's square this stuff in the denominator 4 tangent squared now some ideas I don't know if this happens when you do a math problem but it happens when I do a math problem I've got all these I'm writing stuff down but I'm thinking too much about what I'm writing down I'm thinking about other things that might come in handy later in this problem for example I see a secant squared not only is secant squared the derivative of tangent but what else is secant squared that we worked with yesterday that might come in handy there's a Pythagorean identity that involves tangent squared and secant squared so if I needed to get rid of that secant squared and write it in terms of tangent squared I'd probably do it same thing I'm writing down tangent squared I'm thinking that may or may not be an issue but I know something that I could replace tangent squared with that doesn't look like a tangent squared anymore so maybe you don't like to do that when you work through a problem but those are the kind of ideas that are going through my mind which I've told you is very finite maybe I shouldn't do that 4 tangent squared plus 4 this is kind of why this procedure should help factor out of 4 and what is tangent squared plus 1 secant squared and what's the square root of 4 times secant squared 2 secant so this is transitioned here now we have the square root of a perfect square so where that radical was we have simplified it down to something that doesn't have a radical anymore okay what next 2's are gone and one of the secants in the numerator knocks out with this secant in the denominator can you factor out can you factor out a 2 and get rid of some of the 4 we already packed it out the 2 this 2 and this 2 are gone with a 4 or not we're going to be stuck with a 4 in the numerator we have a 2 in the denominator we have a 8 that's what I would choose to do is kind of get now that we've reduced everything we can reduce let's get all the other numbers out in front so that's not really a 4 it's a 4 in the denominator which is a 1 fourth and at this point we have a secant of theta up here tangent squared of theta because we moved the 4 out in front as a 1 fourth so we did eliminate the radical that appears to have helped but we still got some things to deal with here split them up into sine and cosine that's what I would try I don't really see anything here that's going to tell me that I ought to change tangent squared to what secant squared minus 1 I don't think that's going to be all that helpful to do that but if I had a numerator that I could split up into pieces then I could put the whole problem into pieces but the denominator into pieces isn't going to be helpful which by the way I need to review this this is going to seem ridiculous and you're going to probably think less of me for doing this but let's review something real quickly that you can do and another thing that looks a whole lot like it that you can't do 2 plus 3 over 11 if you have a sum or difference you have a fraction you have a sum or difference in the numerator this is completely legal right? to break that up into pieces and in calculus especially integral calculus doing that kind of thing is sometimes helpful because the original rational function might not be able to be integrated but when you break it into pieces that's the whole premise behind partial fractions here's something that you cannot do let me go ahead and write that there first because you cannot do 5 over 1 plus 5 over 2 I would love to do that with my bank account I would love to do it I would do it repeatedly, daily several times because what's the value of this side right here? that's 5 thirds man, now it's more than 5 in fact it's 5 plus another 2 and a half I want to do that I want that to be a new rule it doesn't work that way you can't say something that's 5 thirds which is clearly a whole lot less than 5 is going to be more than 5 is that okay? we can't do that so as tempting as that is we can do it when it's in the numerator our number system actually operates this way as much as we'd like for it to in the time it does not operate this way so that's why in this problem I don't think it's going to be helpful to break this up into secant squared minus 1 because it doesn't help us to break a denominator up into a sum or difference so I like that recommendation and I need to learn names that is NOAA that made that recommendation let's change everything to signs and cosines and see what that gets us maybe that'll kind of allow us to because secant can be pretty easily written in terms of sine and cosine tangent sine and cosine and then we've got a tangent squared so let's see what happens so secant is in the numerator which is 1 over cosine tangent squared is in the denominator sine over cosine we're going to square that see if this takes us anywhere I have a sneaky feeling that it probably will so the numerator is 1 over cosine we're dividing by this sine squared over cosine squared so we can multiply by its reciprocal is that alright so are we going to get anywhere get rid of the cosine we can get rid of the cosine that's on the bottom with one of these up here let's see if that's going to be helpful before we go that route you can see the next thing cosine over sine squared yes I like it alright so this cosine is gone and one of these is gone so we have a cosine over a sine squared now there's a couple ways actually of approaching this from this point I like that recommendation and that's going to work if you wanted to at this point change this problem into sine of theta to the negative second cosine theta d theta that's going to work isn't it because this is the du stuff and this is the u stuff this would be what u to the negative second du that's going to work or can I multiply the numerator by one would you be angry with me if I did that's legal isn't it so I can do that and then we've got their product what's cosine over sine cotangent and what's one over sine cosecant we don't use this one a lot but isn't cosecant cotangent the derivative of something what's that the derivative of negative so the derivative derivative of the cosecant let me go back another step the derivative of secant is what secant tangent we probably use that more than this one right don't these work in pairs so if the derivative of secant is secant tangent derivative of its co-function which is cosecant isn't it the negative of the product of the same kind of terms but the co-functions instead so for secant we would have negative cosecant for tangent we would have its co-function which is cotangent so the derivative of cosecant is negative cosecant cotangent so if we put a little negative here in a negative here I think we're in business isn't this the derivative of cosecant so what's the anti-derivative of the derivative of the cosecant it ought to be the cosecant so that ought to work this one ought to work also let's show in fact that they get to the same solution here is u to the negative second du what's the integral of u to the negative second du up the power by one divide by the new power u to the negative first over negative one right? our u here is sine theta so that's going to be negative one u to the negative first is one over u and one over u is one over sine theta is that the same answer? same answer two different approaches but they're both going to get you to the same exact solution now we're not done with the problem but the rest of the problem is stuff we have already reviewed this week so I don't know how redundant we need to get when we review the review which reviewed the former review we're going to get a little redundant here but we started with an x problem we need to end up with an x answer right now we don't have that but how could we get that go back to our original substitution make a little picture of it right or you can make a big picture so x over two is the tangent of theta so if this is theta x over two is that correct hypotenuse is what? square root of x squared is that somewhere in the original problem? u square root of x squared plus four is there so it's probably done the right way at least this labeling of the diagram so from that diagram we need to pick off what? the cosecant of theta everybody feel comfortable doing that? picking off the cosecant of theta from that diagram then we've got an answer in terms of x put a little plus c at the end and move on to the next one there are some other good trig substitution problems at the end of 5.7 one we did not do that if you want a little additional practice problem nine is a good one for trig substitution let's look at a couple of table of integral problems today to finish up here's the table of integrals problem that I asked on my 141 exam almost a month ago alright the directions say integrate using the subset of the table of integrals that was provided so the subset that was provided is actually and this is sometimes part of when you're using the whole table of integrals part of something you have to do but you look for an integrand that kind of matches what we have so here's kind of what you look for 4x squared that's the variable part so that's kind of u squared this doesn't have the variable in it but it's going to be something squared that's going to be an a squared so you're in a in a sense looking for problems in the table of integrals that deal with the square root of variable squared minus number squared u squared minus a squared so on the test I provided a subset and those are the forms that involve this particular guy right here 39 through 46 if you're looking at the table of integrals so all of you don't have table of integrals tables of integrals so this is what I provided on the test we can zoom in on that just a little so these are up at the top forms involving this letter squared minus number squared so it's our job then to pick off of this table which one we have so of those which one looks like this original problem the most 40 so it doesn't I mean it is the longest one but that should be an issue it's a matter of getting our problem to match number 40 exactly so here's number 40 it says if we can get an integral of the form u squared that's variable stuff squared u squared minus a squared du everybody agree that's kind of what we have in this problem it's not letter perfect but it doesn't take a whole lot of corrections and compensations to get it to match this exactly so here's our u squared right do we also have that same u squared right here we don't because this u squared is four x squared and this is just x squared so we don't have a literal match right there which we need to have so it looks like our u squared is going to be four x squared that look good okay you could in fact that's good and some people on the exam did that you've got a four here in here which you could factor out in front and then take the square root of that four in front is two so there are other ways to accomplish this but if we're going to leave things where they are our u squared is going to be four x squared so that's not u squared but we can certainly make it u squared by doing what? multiplying by four and dividing by four so now this is a u squared and this is a u squared so although we had this nice looking match it wasn't a literal we're all doing that this doesn't look like it's going to present a problem but sometimes it does because we've got to have d u and you say well this was an x problem we had dx this is a u problem we have du everything's good let's just finish the problem that's not going to work because we kind of like to get it right so if u squared is four x squared then we know that u is two x and if we differentiated this two with respect to x or we could do well let's just kind of leave it as it is what's the derivative of this? two dx I'm sorry this would be derivative of u would be exactly what I just said derivative of u is du and the derivative of two x is two we're deriving with respect to x right so this is kind of derivative of u with respect to x if you want to get real technical about it derivative of two x with respect to x is two multiply both sides by dx so we want to get rid of dx what is dx the same thing as? it's one half du right du is dx so everybody okay that that is now u squared is that alright? this was kind of how we started the problem that was already u squared that's how we started the problem this is a squared that's not going to be an issue a is a squared is sixteen a is four that will be a part in the answer but it's not an issue to correct for and for dx we don't need a dx we need a I want a du there so I'm actually going to use this statement right I have a dx but what else do I need? I need a two so I can multiply by two and multiply by one half and now we're in business so we have a one eighth out in front we've got a four x squared let's make sure that we now have a literal copy of integration formula forty from the table this is u squared this whole mass is u squared minus a squared and what is two dx? du it's du so don't we have it now exactly yes we had a couple of corrections we needed a four we didn't have it we manufactured it compensated for it we needed a two we didn't have it we manufactured it compensated for so now all we need is to put a one eighth out in front of everything go to the right side of the table of integrals now that we have a literal match and everywhere we see a u what are we going to put in? four two x and everywhere we see an a we're going to put in a four might help if I had a table of integrals so that one eighth is because we didn't have a literal match so that's part of the answer there's also another eight right we've got u over eight two x over eight two u squared u squared is four x squared is that right minus a squared that's sixteen this is the easy part it's the part to get the literal match that requires work now it's a matter of you know every time you see a u you put in a two x every time you see an a you put in a four watch me miss it after I say that u squared minus a squared minus a to the fourth what's that 256 that we don't even know what it is it could be hyperbolic signs and cosines anybody dealt with those in a math class before anybody hyperbolic signs and cosines it could be a bunch of those it doesn't matter what it is all we're doing is plugging in every time we see a u we're plugging in a two x every time we see an a we're plugging in a four u two x u squared minus a squared it's got to be because we're out of time have a nice weekend I will see you on Monday