 Welcome back with today's lecture 13. The goal today is to kind of gather together everything that we have covered for the test material. I think it was Kelly yesterday made a good suggestion that I locate an appropriate old test that I've given over this section. It may not be exactly the same material, but I'll try to do that and that'll be what we do tomorrow. Kind of look at some problems, how they might be asked on the test. Sometimes it's appropriate, I know most of you have never had a test from me, but sometimes it's appropriate on a test to say, you know, set up this volume problem. Do not integrate and do not evaluate. So that's a way that I could actually ask more questions without having to get to a place where we do that in class. Sometimes when we get it to a point where it's set up, we've got the limits of integration, we've simplified everything we're going to do. A lot of times we integrate it and don't evaluate it. Sometimes we don't even integrate it and evaluate it. That's a potential test type question also. So that would be primarily to save time. Get more questions on the test and you may think that, well, I don't want that. I don't want more questions on the test. I think that you do because if we had five questions and you miss one, they're 20 points each. But you know, if we have seven questions and you miss one, they're 14 points each. So the more questions, the less value for each question. So if somebody probably never happens to you but you might forget how to do something, then it's a little more forgiving when there are more questions. So let's really web assign questions today. We had some after class yesterday. Any web assigned question that you have is fair game today in class. Not probably to the point where we're going to do it all the way through, but we'll at least get it to the point where it's jump started for you and you can take it from there. But let's start to kind of gather where this course started and what you should be prepared for for test one. Even though 5.7 and 5.8 are technically review, I know that we've got people from not only other 141 classes on campus that are coming to this 241 class. Probably a couple of you in here did not take 141 on this campus and you had calculus at a community college or a high school. So it's kind of a starting point. We obviously expect you to know it, but because it's where we started in 241, we could have questions from 5.7 and 5.8 on this test and we will. So from 5.7, we had three types of problems. I don't know that today... Are we going to have web assigned questions today? Anybody? Yes? You have some web assigned questions? I mean you have them with you? Anybody else have web assigned questions? All right, so what we may not be able to do at each topic has come up with a kind of a sample problem, but if you have a web assigned question that's related to, let's say, trig integrals, now would be the time for us to do that. Now for trig integrals, what are we talking about here? We're talking about powers of sines and cosines. So if you have a sine cubed and a cosine to the fourth in the integrand, we've got to be able to deal with that. Also powers of secant and tangent. We did some problems with secant and tangent. Since this is kind of gather up day and do those kinds of things, refresh and review, sines and cosines, now if we see them in an integrand, what are the things that you want to think about when you see sines and cosines? How are sines and cosines related to one another? In calculus. Okay, derivative of sine is cosine, derivative of cosine is negative sine, so they're kind of derivatives of one another as long as you correct for a minus sign on one of them. How else are sines and cosines related that might come in handy when you are handed an integrand that has cosine to a power of an angle, sine to a power of the same angle? Sine squared plus cosine squared is one. So you can get a couple of different versions of that. Sine squared is one minus cosine squared, cosine squared is one minus sine squared. What's the bad case here? One that kind of isn't as smooth as the others with powers of sine and cosine. The double angle identity and when do we need to use the double angle identity? Even powers of sine and cosine is kind of the stubborn one. So if we have a sine squared and we're stuck with that, or a sine to the fourth, or even powers of both of them, cosine squared, sine to the fourth, then we're in a position where we need to use the double angle identities. So this kind of arsenal of weapons then allows us to deal with these trig integrals. So I don't know if I mentioned in here that I had an outside possibility of going to the Super Bowl that fell through over the weekend, last weekend. I kind of thought it was kind of lame originally because the tickets are going for like $1,600. I just don't have a spare $1,600 to bolt down to Tampa and see the Super Bowl. So we had a connection on the coaching staff, the receivers coach of the Steelers. A friend of mine who lives in Cary coached him in high school baseball in Pennsylvania. So that was our connection but it fell apart pretty rapidly. But I've never been and I'd really like to go to a Super Bowl sometime. So if we're stuck with sine squared or we're stuck with cosine squared and we have to use that double angle identity, what are those? What's sine squared? And this is one plus and if that kind of is gone from your memory when you need it you can kind of develop the cosine of 2 theta as cosine squared minus sine squared that kind of double angle identity and come up with both of these if you need to in a matter of a few seconds I hope. So all that kind of is in the arsenal of weapons to attack this kind of problem. Now does anybody have a secant tangent type integrand or a sine cosine type integrand and a web of sine question? Okay how about let's just go ahead and verbalize but not write down how secant and tangent are related in the same kind of line of thinking if you have a secant squared and a tangent squared of the same angle in the integrand and we have their product what are some of the same thought process? One plus tangent squared equals secant squared. Okay one plus tangent squared is secant squared which will give you the what tangent squared equals secant squared minus one right? How else are the tangent and secant related that might come in handy? Tangent being sine over cosine. Well we could change them to sines and cosines that if something's not coming to mind and you kind of draw a blank on the interrelationship of secant and tangent change it to sines and cosines and maybe go with this. That's possibility. Derivative of tangent is secant squared that was kind of stereo I like that and derivative of secant is secant tangent. So those kinds of interrelationships allow us to simplify the integrand and write it in terms of u to a power du and then hopefully once we get it to that point it's fairly easily integrated. So I'll bring in an old test tomorrow and if I don't have one of these on there I'll make one up and we'll go through it. What else in 5.7? Trig substitution. And I guess really it'd probably be fair to say in 5.7 the appendix in the back of the book that covers partial fractions is that g? Yes. So appendix g especially when we get to partial fractions if there's anything back there that's going to be helpful to you I guess that's kind of the algebraic breakdown or decomposition into partial fractions. Trig substitution they do not have trig in them but they have a variety of radicals there's one of the types a is a number we may not like what a is a might be the square root of 7 but a squared is okay that's 7 so don't ever worry about that that shouldn't be in the way of getting to a solution in a problem u squared is the variable thing that's being squared so if it's just x squared u squared is x squared and that's fine but also it might be 9x squared so u would be 3x so that might come into play we could have number squared minus variable squared under the radical or variable squared minus number squared now you're going to be given just like in the book the author gives values that you're going to substitute for u and I'm going to provide this on the test if it's the sum of things squared we just had a Pythagorean identity was what 1 plus tangent squared is secant squared right so if we've got the plus thing going on we could let u equal a tangent and that ought to get us somewhere because eventually we could have a 1 plus tangent squared under the radical what is 1 plus tangent squared it's secant squared so you can simplify that this will be given to you so you don't have to come up with that yourself but what to do from that point on you need to know how about the difference here so if we could come up with a about a 1 minus something squared that itself would be a perfect square what's a candidate there? sine again that's going to be given to you so it'll say let u or whatever the variable quantity in the problem is equal 4 sine theta or just sine theta or 9 sine theta whatever how about variable quantity squared minus number squared under the radical Pythagorean identity has is going to secant squared minus 1 right so these things will be given but you'll need to be able to take it from there so one of the key steps does anybody have a website question with a trig substitution one of the key steps then is once you are handed what you want u to represent is you need to find what du is based on that right so if u as in this case is a tangent theta what's derivative of u? a secant squared theta d theta because we're going to change the problem to a u sorry to a theta d theta problem so we're going to have to get d theta in the integrand somehow and that's how we're going to do it so where there's a du in the integrand we will replace it with a secant squared theta d theta same thing here find du and same thing here find du questions or issues with trig substitution and no web assigned questions right that anybody has the other part of 5.7 is integration after decomposing into partial fractions so if you felt comfortable with this stuff at the end of 141 and what we did earlier in this course refreshed you on this stuff then this is not going to take a lot of your study time to go back and refresh yourself again with partial fractions and there's really technically I guess there's four categories but really two main categories that if you have let's say a linear factor and an irreducible quadratic factor now we've got to have the degree of the numerator smaller so the degree of the denominator is x cubed right so if we have something of smaller degree then we can go right to the partial fractions decomposition so let's say we have 2x squared minus x plus 7 so we've been handed an integral problem so this and kind of product and or quotient form we don't like it if we can decompose it into pieces that are added together we can integrate each part of the sum so here's one of our rules let me backtrack because I'm afraid I'm going to forget this I've already forgotten it once today just like 30 seconds ago if the degree of the numerator is larger than the denominator what is the first step before you decompose? Long division or if it's equal right x cubed and this is x cubed do the long division first but as long as it's smaller than the degree of the denominator then you're okay to just start this process so we have a linear factor what kind of numerator will a linear factor get no matter how many times that factor is itself used as a factor constant so even if we had an x plus 2 to the third each time we use that right get a linear factor for the linear factor and the denominator gets a constant numerator linear factors constant numerator here's an irreducible quadratic and if it's reducible then reduce it and you don't have to use this so if it's x squared minus 9 break that up into x minus 3x plus 3 well because when you're done you've got a number over x plus 3 that's a pretty quick integral problem isn't it natural log or you've got a number over the other one x minus 3 another natural log so the more you can break it down ahead of time the easier it's going to be once it's broken down can't reduce or simplify x squared plus 9 what kind of numerator does it get linear some kind of new letter other than a it might be the same letter but we have to allow for the fact that it's different and then we would what from this point get a common denominator equate coefficients of x squared on the left side of the equal sign take coefficients of x squared on the right side of the equal sign and go from there yeah question do you have one of those? I don't know if it's done the same way because I have no idea how to start it but it was x squared over x minus 3 times x plus 2 squared x plus 2 the quantity squared okay let's look at that one because that's got the other thing that this problem didn't have so you're integrating x squared is the numerator what's the rest of it please times x plus 2 squared it's an indefinite integral and that's a web assign question do you have what section or whatever that is and the problems might differ a little bit they have that's what? that's number 7 appendix g number 7 thank you so we don't want to do the first things first can we go right to the decomposition no what's the degree of the numerator x squared what's the degree of the denominator x cubed right here's an x and this x is squared so we've got an x squared times an x so we've got x squared over x cubed we can go right to the algebra piece where we're going to try to break this down okay tell me what to write down slightly different but has some of the features of the problem which is kind of a generic problem we just looked at A over x minus 3 good B over x plus 2 squared and C over x plus 2 good everybody alright with that all linear factors even though this is a repeated linear factor it still gets a constant numerator why do you have that x plus 2 on the end let me try to answer that question with an algebra I'm going to reverse your question if you had this problem right here in an Algebra 2 class in high school at Middle Creek High School the Mustangs which is where she went to high school you would say okay I'm going to add these fractions what is the common denominator kind of what's the least common multiple of these three denominators and what would you want to have when you added these fractions together you want to get a common denominator of what don't you need an x minus 3 right because it's in this denominator and you certainly need an x plus 2 but that's not quite enough because that wouldn't accommodate this denominator so you're going to need an x plus 2 squared is that right so if you were doing an Algebra 2 problem and that problem is to put these three things together you would then multiply this one by x plus 2 squared over x plus 2 squared this one by x minus 3 over x minus 3 and this one by what x minus 3 x plus 2 x minus 3 x plus 2 so you're kind of reversing that thought and you're saying if I have something that has that already in its denominator is there a chance that there could be a term like this yeah there's a chance that so we have to allow for it so if it was x plus 2 cubed we'd have another one you would have x plus 2 squared and then x plus 2 you'd have x plus 2 have a constant over x plus 2 constant over x plus 2 the quantity squared and another constant over x plus 2 the quantity cubed so for repeated linear factors you allow for that to the first to the second to the third or to whatever degree it's repeated I don't know if that answered your question but it's the your thought process was in algebra you're now kind of allowing for the possibility of that kind of term being present everybody okay with that get a common denominator equate the numerators well let's go a little further since that's a web assigned question we need a x plus 2 squared here and here we need an x minus 3 here and here we need an x plus 2 because we only had one of them and we need it squared and an x minus 3 now I know a lot of you consolidate steps and that's fine to do on the test on Friday don't consolidate steps to the point where you're making too big of a leap and you're going to make sine errors and coefficient errors and squaring and cubing errors but you don't have to show every step necessarily that we're showing right here let's see if we can do this part in our head how many x squared are we going to have in all these numerators right we're going to have an a x squared here when this gets expanded and a gets distributed no x squareds here here we're going to have an x squared and then that's going to be multiplied by c so we're going to have a plus c is that good so what's a plus c equal to one right because we've got one over here and if they're equal denominators are equal so therefore the numerators have to be equal one x squared better be equal to a plus c x squared alright how many x do we have well middle term here is going to be 4x so there's going to be a 4a is that right there's a b here in the middle term here is what minus one right so minus c does that work it doesn't have an x squared or doesn't have an x is a consta so we're going to have 4a is that correct in what 9 I'm sorry negative 3b and negative 6c so a plus c better be one 4a plus b minus c right equals zero because we don't have any x over here we have 0x and then 4a minus 3b minus 6c is also zero what method would you want to use to solve this from here set the 2 at the bottom equal to each other okay and what are we going to accomplish by that you'll get rid of a couple things and it's easier with a and c you can knock out the 4a so we'd knock out the 4a's that way so we'd have b's and c's and you could then take then you'd have one equation that had b's and c's in it so what would you want to do with that right if you equated these are both equal to zero so the suggestion is that we set them equal to each other strike the 4a and 4a on both sides and you'd have one equation with b's and c's in it what are we going to do with that one equation with b's and c's in it c in the first equation I'm not opposed to it but I want to know what we're going to do with it when we get to that point I like the 4a's are gone solve for c and then plug it into another equation and then solve for c solve for c alright so if we did this here's where we would be those are gone so we have one equation with two variables in it so let's say we put it all in the same side so we add 3b so there's 4b and we add 6c which would be 5c is that right now where do we go I mean solve for c solve for b solve for b so b equals negative 5c divided by 4 does that work and then solve for c in the first one okay c equals what am I saying and then put all of that in the second one and change the c to an a here so you want to use this in here okay we're kind of working our way around the corner here so if c is 1-a can I recommend a different path okay if we say I like this and you can either solve for a or for c and if we substitute let's say c equals 1-a if we plug that in here and we plug that in here don't we still have two equations right now what about those two equations are we rid of c and we have two equations with two variables right and then we can kind of mess with them and get rid of the other variable so the problem I think with something like this is as tempting as that is that was actually my first thought too is I like the 4a they're both equal to 0 let's set them equal to each other 4as are gone now we have two letters in that equation so why don't we use this substitution to help this equation to reduce it from three letters to two and this equation from three letters to two and see what we come up with that way so the second equation 4a plus b minus c well what is c equal to 1-a that's 0 this equation 4a minus 3b minus 6c would be minus 6 times c is now 1-a now we lose the luxury of having 4a and 4a but at least now we have two equations with two variables so the first equation is what 5a because we've got minus a minus a 5a plus b equals 0 and what's this second equation 10a minus 3b minus 6 oh we've got a minus 1 thank you and here we've got 10a minus 3b minus 6 minus 6 now we've got two equations and two variables so what would be the plan any time you have two equations and two variables you could solve one let's say the top one for b plug that into the other one or some type of linear combination where you multiply one equation by a constant add the two equations together top by three that'll work so 15a plus 3b minus 3 second equation leave it alone and add 25a those are gone minus 9 so a, boy that's a nice number 9 over 25 does it really matter that a is an ugly number is it going to make the integration that much worse here's the decomposed piece that has a in it if a is 925 is that going to radically disturb our world this morning it's going to natural log isn't it just bring the 925 out in front question do you have to do x plus 2 squared for b or can that go under c let's say that when you broke this down you had b over x plus 2 and c over x plus 2 the quantity squared and we did it this way your answer for your b would be our c and vice versa still going to be a number and you're still going to be integrating that term your b and c would just be the reverse of our b and c alright so we can get to what we need to get to today now that we know a c how do we find b c is 1625 ok well we do know a nice clean interrelationship of a and c right there so if we know a c is just 1 minus a which is 1 minus 925 so c is 1625 another delightful number if we know a and c then we can plug back into an appropriate equation not the first one because it doesn't have any b's in it but any of the equations and we can solve for b b's going to be a number so let's look at our original decomposition there's our original decomposition a's a number so when we integrate what was it 925 over x minus 3 what's the answer think about this more than 15 seconds you really need to do some digging in before friday there we go 925 natural log of the absolute value of x minus 3 that's a du over u with a little extra baggage alright c isn't that the same kind of animal do we have c what was it 1625 what's the answer to this 1625 natural log absolute value 1625 natural log absolute value of x plus 2 now this isn't going to be a natural log inverse tangent not going to be an inverse tangent but we do we need to talk about what constitutes an inverse tangent before we leave this section you would say u is equal to x plus 2 then you're going to have a u squared or a u to the negative second did anybody run through and get a value for b negative 20 over 25 oh great I like that I'm sorry that we missed that doing that together what is it negative 20 over 25 I'll go right back to my office and do that when we're done so here's our middle one so whether you actually write this down or you think about it we're going to let u equal x plus 2 is everybody convinced that this is not a natural log what has to be the case for it to be a natural log next to the first in the denominator right or u to the first or r to the first here we've got it to the second so that throws that out and we don't need that special case because this is just a power rule can you pull out the negative 20 over 25 yes and I think that would be wise to do that especially if it's ugly like this just get it out of the way just you know you're ugly would you sit in the other room please if you go out there where I can't see you anymore not really she's a lot nicer than that I do get to come in the house so if u is equal to x plus 2 du is 1 dx so this problem really becomes dx is really du and this x plus 2 the quantity squared is really now u squared so I think that's kind of a goal I think I said that earlier in this class that the goal is for you to kind of think through this stuff now that you've seen it several times at the end of calc 1 and now the beginning of calc 2 that you recognize first of all it's not a natural log that's u to the negative second and what's the integral of u to the negative second negative 1 over u negative 1 over u u to the power rule 1 degree larger over that new exponent and then you sub back in what u is so the first one our first one and our last one were natural logs this one in the middle with the denominator x plus 2 the quantity squared uses the power rule so u was representing x plus 2 temporarily so we plug that back in is that okay how many of you are doing most of this in your head I mean I can see several of you wasn't a shock to me those that raised their hands because I can see your engines working your way ahead of where I am writing things down but that would be a nice goal is to do more and more of that in your head and then the stuff that does require us to write down more steps we can write down more steps inverse tangent came up so let's before we leave partial fractions does that work for that problem I mean at least we got it integrated we think if you have and I had a problem up here a minute ago that see if I can write it down again I had x squared plus 9 right in a x plus 2 and I'm just going to kind of do this in the numerator we have a term up here that is less than third degree I think I had a 2x squared I made it up so I know it's not going to work out very nicely we just had one that had some ugly coefficients but let's say we take this and we decompose it and we get 3 over x plus 2 and here is our irreducible quadratic denominator and we allow for a linear numerator some bx plus c we have b is negative 5 and c is 1 so I'm getting to the inverse tangent thing but I'm trying to relate it to partial fractions so this is our original problem we did the decomposition alright first one we just did that stuff that is going to be natural log absolute value x plus 2 what are we going to do with this one ok not yet eventually yes split it up we've got negative 5x over this now also I think be a wonderful cognitive experience that if we know why we're doing that why would this problem why is it to our advantage to break it up into two pieces x squared plus a squared ok there's one of them to have a number it doesn't matter really that it's one but some number if it's not one you just bring it out in front this is an inverse tangent integrand because the derivative of inverse tangent is 1 1 plus x squared we don't have exactly that but we have another version of it this is not an inverse tangent but this also works out nicely to have the x term in the numerator and the x squared term in the denominator what is this guy with and what's the result going to be if u is x squared plus 9 don't we have kind of du in the numerator correctable version of du natural log this is a natural log so if the bottom is x squared and the top is x that's a very strong candidate for natural log and in fact this one's going to be a natural log sometimes I can read your faces and I know it's time to move on to the next problem I didn't see that look on this problem so let's take both of these pieces and finish them so here that quantity in the bottom x squared plus 9 what's derivative of u 2x dx negative 5 is in the way let's move it out front so that's gone what do we need in the integrand we need a 2 I can't just manufacture a 2 so I have to divide by 2 also so now what's the numerator that's du so we've got du over u and the integral of du over u as we said just a minute and a half ago is natural log so we've got negative 5 halves natural log and you can use absolute value but you really don't need it here alright with that one now generically and when we covered this we actually derived this in class and when we covered that I said it's probably a good thing to remember because we're going to see it a lot and we have seen it a lot so far so I think it's probably valid to say that when we see something in this form where this is variable squared plus number squared 1 or some other number that you can farm out front that integrates to 1 over a 1 over a tangent inverse u over a u over a plus c so if you have that committed to memory then you don't have to do the battle each time with this hopefully you recognize that as kind of looking like the derivative of an inverse tangent but this lead coefficient in the u over a now that we've kind of done the battle and seen that that's the case all the time makes this a little quicker so the answer would be 1 over 3 of x over 3 plus c let's move on to a tougher problem one that's actually going to challenge us a little bit right so that all falls under the category kind of decomposition into partial fractions alright real quickly 5.8 I realize we got kind of bogged down there but I still think that's time well spent because I could tell by facial expressions that wasn't where it needs to be by Friday using the table of integrals and if you have a 5.8 web assign question we can start class with that tomorrow table of integrals remember is I mean you think this is the easiest thing we've ever done in any math class and it can be but remember you've got to take what you're given and you've got to make it match exactly so if it doesn't match exactly I remember a problem we did in class that had an x squared here and a 4x squared I think minus 7 there well if that's what we're going to let be u squared then that can't be u squared also right when you look in the table of integrals you're going to see something that looks like this but if we have it exactly and we want to use this formula from the table of integrals what we're calling u squared better be here exactly not we need to kind of doctor it up so that it is you can doctor up numerical stuff you can't doctor up variable quantities so what could we do if we want to call that u squared what could we do to this multiply by 4 in the denominator what would compensate for multiplication by 4 in the denominator multiply by 4 in the numerator but I don't really want it I just want to take it and move it outside so that becomes a u squared just exactly like that's a u squared so we do have to make our problem match what we see in the table of integrals and once we do that then we don't have to integrate it it's already been done for us but if somebody has a table of integrals problem we can start class with that tomorrow I'll bring an old test tomorrow and either hand you out a copy of it you basically have it and any web assigned questions that you want to ask tomorrow those are fair game too