 It should have finished the second WebAssign homework because the paper homework fits up. It's on the class schedule webpage, so that's where they will all be. If you've had trouble getting into WebAssign, please first try and contact them and then let me know and I will yell at them. I know some people have trouble. I don't know. Any other questions? Also, if you're having, if when you look at your WebAssign, I think it's problem 9 on homework 2 in particular has some issues with grading. I don't remember the answer. Am I going to write something wrong? I think it's number 9. I don't think it's a sign. But if the correct answer is something like, I don't know, 1 half sine cubed to the x, it takes an answer written this way and it does not take an answer written this way for some reason. I complain to WebAssign about it. They're going to fix it but it's already out there. So, if you got burnt by this and lost points, click on the ask your teacher link and get it fixed. It's number 9. Does anyone know? The answer is really something like a co-changer to proceed in, to proceed in. Does anyone know? Yeah. Writing something just like that, which you can see around the sign and the x, which is the bigger one, and then you can click on it. Yeah, well, so all of them should take this, or this, or this. One of them, it's either number 5 or number 9, and I think it's number 9, doesn't take this. And if you put this in, it gets marked wrong. If you change it to this, it gets right. So if you put it in like this and it gets marked wrong, if you change to this and it gets marked right, click on the ask your teacher and say, WebAssign was dumb on this problem. Please give me my half a point down. So we don't. The point is not to rob you for notation. The point is to know whether you can do the robbery. So the only problem that I know of is that one. All the other ones will take any of those three for now. Okay, so where we were, as we were doing integration by parts, which is useful when you have, it's a way of creating derivatives for integrals. We have something. Let me do one more example. Let's say the integral from, where should we have to get integral this time? No? Okay. I feel like Obama when the guy shouts out, you lie. Let's do something like this, 0 power 4. It's not a good question. Do this integral. There's actually several ways I can do this integral. Should I use parts on this? What should I take for the parts? Take one part for sine x and take the other part for sine x? Okay. That means that du is the cosine and v is minus the cosine. So that gives me minus, did I do that right? Yes. Minus sine x cosine x from 0 to pi over 4 minus the integral of du. So I think I did that right so far. This is from 0 to pi over 4. And this is, so the sine pi over 4 is root 2. Yeah. Well, that's the other way. So let's do it by parts or you can do it by using a half angle angle. Either way is fine. So let's do it by parts. Because they wanted to do it by parts. So the sine pi over 4 is root 2 over 2. Cosine pi over 4 is root 2 over 2. So this is minus, minus. The sine of 0 is 0, cosine of 0 is 1. So that part's gone. And now I have to do the integral of cosine squared. Now if I do the integral of cosine squared, I can do it by parts again. But in fact, it's easier to remember a different identity. So I actually, if I do it by parts again, I get stuck. Pretty sure I get back to the original integral. If I do it by parts again, I get 0 equals 0 or something like that. So that's not going to work. Feel free to try it but I'm pretty sure it doesn't work. So I have to use the fact to do this. What is another way I can write cosine squared? Yeah? 1 minus the sine squared. Right. So I remember that sine squared plus cosine squared is 1. And so this becomes the integral from 0 over pi over 4 of 1 minus the sine squared. And now again we're in one of these tricks. Yes. We're in one of these tricks where I'm going to get the integral sort of related to itself. So this is 2 over 4. So this is a half from here. This is the integral from 0 to pi over 4 of dx minus the integral from 0 to pi over 4 of the sine squared. And that equals the integral from 0 to pi over 4 of the sine squared. So I have a thing that I don't know equals to a number plus something I can do easily minus the same thing. So that means... So that means I can bring this and add it to this to get 2 times the integral from 0 to pi over 4 of the sine squared equal to negative the half plus... Well, the integral from 0 to pi over 4 of dx is 0 is x from 0 to pi over 4. So that means... So this is pi over 4 minus a half equals twice the integral I want. So the integral I want is pi over 8 minus 4 or pi minus 2 over which is a little bigger than I want. Okay? So there's 2 things going on here in this example. One, I'm doing an integration by parts which is a derivative integral and one of the things that you can do if you don't have to, you can go all the way to the end and then plug in is that you can plug in as you go the derivative sum of the parts and that's what I get here. The other thing is if I had tried to do this cosine squared the same way I did the sine squared I would actually wind up with 0 equals 0 so that doesn't work. So sometimes when you do integration by parts you wind up with something true, 0 does equal 0 but you wind up with something that doesn't really get you anywhere. So this is true of integration in general. Going backwards is often harder than going back forward. Trying to impart and figuring out how to do it takes perseverance. Okay. So I'm going to stop pursuing integration by parts right now because I'm tired of it and we have to do the long line from the hurricane. So but as somebody else noted I can also do this integral by using well, let me do the e. So suppose instead of that I had something like I can't control the sine cubed of s dx and let's just make it an indefinite integral just fine? Let me throw a cosine in there. So say I have an integral like this what would I do? So I hear people mumbling substitution factor the sine off so why would I do that? Just because I can. I mean is it true? I don't know that it helps so what these other people are shouting out is true, let's make a substitution. I mean I can do this but I don't want to do it. So if I make a substitution what should I substitute? I hear a bunch of mumbles. Okay so I should substitute the sine. Why should I substitute the sine? Because the derivative is sitting right there. So I make the substitution just so you don't call it u substitution. I'm going to make the substitution w equals sine of x but because right there it might take w. So when I make that substitution that makes this a really easy integral and so then this is just one quarter w is the fourth and w was the sine. So let me point out something that students always do because they think it's quicker. It's really important that we write things down that it's true. Especially when you do something like this notice that everything I wrote equals the previous thing and I wrote equals every time and I didn't leave off bits until the end. It's important even though it makes you sound like I'm trying to be an evil accountant or something that you keep track of all of the bits because you need them later and so rather than just writing stuff all over the place try and write it in order and make sure every stage is equal. When you write equals it means the same. It doesn't mean it becomes. Equals means equals. It means the same. If something becomes something else don't write an equals sign. Also don't write arrow when you need equal. I see a lot of students that would write this arrow this arrow this. Now in this case it doesn't matter whether you think of equal as a process that gives an outcome through we just say this equals that and this equals that. You don't write equals if it's not equal because then you get in trouble. You don't get in trouble for me you get in trouble because you wrote garbage and eventually you get confused. Sorry that was a little off to the side tirade. I see it all the time in the calculus that students just write junk all over the page and then magically the answer appears and that's fine because that has no mistakes. But if you make a mistake we can't find it so we can't give impartial credit and also you're less likely to make mistakes if you proceed in orderly fashion. If you're going to write junk all over the page write it on another page. I lost track of where I was going because I had to write it for a minute. There we go. This is actually a general trip. When we see an integral in one cosine laying around it's easy. When you see an integral with a pile of cosine in one sine laying around it's easy. So if we had an integral like sine to the fifth of x cosine squared of x dx let's make it sine of q. It's going to be easy. Suppose I have something like sine cubed of x cosine squared of x dx I'm going to make a substitution. So I'm going to make this a quicker question. So choices are make some manipulations and then make the substitution equal to sine. Make some manipulations and then make the substitution equal to cosine. You can do either one and get the answer or no this is not a useful way. I started with this. Use a trig identity. Looks like we have if you need more time you're supposed to go like, yeah I do. Anyone need more time? So most of the people think we should manipulate it and then substitute the sine. So since 57% of you think that's what I should do maybe at least one of you knows what I should do. U equals sine cubed of sine squared of x sine of x dx Wait a minute, what am I doing? Sorry. I'm going to lose pain. Three sine squared of x Well he said cubed. So cubed isn't going to work because now I have an extra business. It was a bad idea. Square So then du is 2 sine x cosine x No it doesn't. So if I do that so then that integral becomes I have a sine x cosine squared of x dx One of those is du So this is u x cosine x du I don't know what to do with that. I guess this could be a square root of u but this guy is still there. So this doesn't work so well either. Yeah. U in the back. I think the cosine x squared equals 1 minus sine x squared. Okay. I guess I'll go over here. I'm going to let I'm going to use the fact that cosine squared is 1 minus sine squared. Just go on. What? Then it should be sine x2 star minus sine x to the face. Okay I can do that. Then it should be the sine x2 star integration minus integration of sine. I still can't do this. Oops. I'm still stuck. I don't know how to do either of those integrals. Use u equals sine x. Let's just take this one. Yeah sine x. I'm not trying to make fun of you. I'm trying to point out there is something going on here and if I just tell you, you're going to forget. So anything else? So this is a great idea except it's a little bit wrong. Yeah. There are two of the sines internally in the cosines. So this is a bad idea too. So this is close and as he said instead of focusing on trying to turn the cosine of the sines let's turn most of the sines into cosines. So write this integral as sine x. I'm just going to save it for later because I might get hungry. And then I have cosine squared of x dx. So this is the same but this is one minus cosine squared. One, or how about cosine squared of x minus cosine of the fourth times the sine of x dx. And now I'm happy because when I eat these cosines I have the sine left over for dessert. I can let u be the cosine and so du is minus the sine that I need. This becomes negative du and this becomes u squared minus u to the fourth which I can do which is easy. So let me just finish. So now I get the integral of u squared minus u to the fourth which is negative 3 minus u to the fifth u to the fifth which is I'll write it in the other word u to the cosine. Cosine to the fifth minus cosine. So that means that this choice for sure works. I don't know how to make this one work. And so if a doesn't work either to c OK. So this is a general procedure Whenever I have an odd power of a trig function laying around so in general if I have the integral of sine to the m cosine to the n then if either one of these are odd let me just say if n is odd then I convert sine into cosines by sine squared is one minus cosine squared and I do the same thing but I convert sines into cosine by the other piece and if they're both odd I can do either way. So this answer is sort of right in spirit but it depends on which power is odd it doesn't matter if n is greater than n. It doesn't go care. It would be negative too. So the trick is that I want to have laying around so what I want is the integral of stuff in sine let me just call it a function of sine times the cosine x dx or the other function of cosines times sine x dx. I want everything except one of them to be the other guy and then this becomes easy so straight forward substitution away we go so the thing to remember is not to memorize this formula but just remember that you know the most useful trig identity is not even trig with a diagonal theorem on the same thing this identity is not even trig it should be that I hope you've learned more than one thing from trig but that's the one thing that is most important from trig which is just the diagonal theorem so using that trick we can always we can always get rid of all but one of an odd power okay um running a little low on time let me not do another one of those he gets to do some other homework what's the time is here can I stop this if they're both even if they're both even like the integral that I did at the beginning or oh let's just do that so here the power of the cosine is zero that's an even number the power of the sine is two that's an even number okay because if I take one sine away and if I turn both sines into cosine I'm in the same place that I was in before and if I don't do it well then I have no hope so I can't use that trick and I always forget what the leeches plus and this is minus so I have to use a different trig identity so I want to use either this identity or this identity which maybe you remember and maybe you don't I always tend to forget them so I have to cheat but I get to cheat and so here I'm going to use this one and then this integral becomes pretty easy so this is so I integrate one and I get one half x and I integrate one half cosine two x so when I integrate cosine two x I get one half sine x sine two x and another half so this is minus one quarter check that you can check it by taking the derivative and then using this formula so if both are even so if you have an even number and notice what this does this takes an even power and it turns it into half of that power it changes the angle by a factor of two but it turns everything into half of that power and so ultimately if you take an even number and you keep dividing by two unless it is zero eventually it becomes odd and then you can use this trick so if I wanted to do say sine to the fourth do it twice and get down to the power of one so these things can be a little tedious but so I want to focus on this business for a while but these are easy once you get a handle down other than sometimes there is a little bit of a tedium with manipulating trick identities over and over and over again but I'll leave that to a presentation another thing that we can do is go through techniques that work for various types of integrals so related to trick identities we have this one sine squared plus cosine squared equals one and we do one that doesn't work so the problem with these sometimes they don't you choose one that doesn't work and then there are three I didn't want to do that one one like this is not going to be so good but there's a right to it the trick to notice here is actually there's some trigonometry going on and this doesn't look like it somebody said something so if so the problem here is this student squared root of x squared plus four and I'd like to get rid of that and I'd like it to be like a square root that I can do so I remember maybe this should be a one to make it easier maybe it would be a one to make it easier to come out maybe I can do it when it's four or two but let's do it when it's one so here I remember some trick identities so this guy I remember this guy which is not useful here I remember this guy and this one is useful here tangent squared plus one is secant squared which you can just get by pushing this one around and here I have something squared plus one becomes a square so that means that if this x were only a tangent then this would become just a secant and then I could do this integral so let's say well I wish the x were a tangent so I'm going to let x be the tangent of some angle I don't know let me call it theta and so that means that dx is secant squared d theta fine and so here by doing this I can now rewrite the integral to get so dx becomes secant squared x squared becomes a tangent squared the square root becomes secant squared and now I can do a little trig manipulation so let's reduce this the square root of secant squared is secant so this is let me just rewrite it slowly so I have a secant squared on the top pick up another secant on the bottom and I have a tangent square here and so that I can keep track of everything I mean this is really one over cosine squared this cancels this and we can make a cosine on the top and we can write everything in terms of sines and cosines so that I know where I am so the secant squared is one over the cosine so I'm going to put that on the bottom and actually first I'm going to cancel this guy with that that can be one over cosine that's from this bit and then I've left with the tangent is sine over cosine so that gives me a cosine on the bottom and a sine on the top so I think I did that right is everyone clear on where I went from here? yeah so this x squared plus one becomes the secant squared taking the square root of the secant squared square root of something squared is just the thing and then this secant cancels one of those secants and then a secant on the top becomes a cosine on the bottom this is upside down see this is a cosine squared on the top and a sine squared on the bottom so again I can cancel this with that and so this one is this one but after all of that u-ha-fufuru you wind up with that integral of cosine over sine squared well that's one we can do it's on the bottom so this is sine over cosine but it's on the bottom so I need cosine over sine so so after all of this mess u equals sine well yeah I didn't use u yet so this guy becomes just the integral of one over u squared u which is one over u negative I need a two yeah that's right well u this is the cosecant negative cosecant if you want but my question was not about theta my question was about x so if x is the tangent then theta is the arc tangent so this is really the cosecant of the arc tangent but the cosecant of the arc tangent has another name so we should figure out what that is I'm going to draw a picture I'm going to draw a right triangle so you have to remember some trigonometry now and I'm going to let this angle be theta now if this angle is theta x is the tangent of theta because that was the choice I made the tangent is the opposite and I want the tangent of this angle to be x so if this is x and this is 1 then the tangent of theta is x that means that this hypotenuse I use the thethagorean theorem is this squared for this squared take the square root now what I want is not the tangent I want the cosecant one over the sine so the sine here is this over this one over the sine is this over this so this is minus x no that's the sine is minus squared plus 1 or x which seems like magic but it's reasonable I'm integrating some algebraic function some bunch of stuff involving trigonometry and eventually I get back to another algebraic function so the short of this so what went on here is when we see the square root or actually just something squared plus a constant this trig identity can be helpful and you turn it into a trigonometric integral which then is not so bad you do just the way I said how to do that and then you go back if I see something involving for example 1 minus x squared I would use this identity and if I see something involving x squared minus 1 I would use the other ok look at this so I won't go through the whole business so I'm going to do exactly the same integral but when it's x squared plus 4 I'm actually going to do that so this squared plus a thing tells me that I want to use where'd you go? ok thanks it helps people look at something this tells me that I want to use something like an octan but if I use this one it's not going to help me so what I want to do is divide everything by 4 and multiply that by 4 so I rewrite this which I don't have to rewrite explicitly but I'll rewrite it in my mind but I'll write it explicitly so you can see my mind as well I'm going to factor this 4 out and so when I factor 4 out here I get x squared over 4 but I can actually write x squared over 4 over 2 squared like that I'm going to let x over 2 be the tan I want to get rid of this square root if I get the same well it isn't quite the same I'll pick up some factor of 4 but it's almost the same ok let me not finish this now since it's 5 then I need a square root of 5 you can take the square root of 8 so if this had been how about a 41 this would be a 41 and then here this would have to be 41 over 2 but I want to take the square root of 41 here so this would be square root of 41 and this would be square root of 41 and this would be square root of 41 it's the same now you need that square root square root of 41 what's the same I could have done 2 tan we're here square root of 41 tan of 8 equals x same I don't see maybe I'm misunderstanding what you're suggesting 2 tan of theta so what do you mean by so let's put it back to 4 which is 41 2 is a lot easier right ok so it's a 4 again what are you proposing I do now instead of doing this much because x squared plus 4 would be no no I'm converting it into a secant so the trick here that I want I mean it is I have to divide by a square root I don't know how to get around it so I mean if I let x be 2 tan of theta that's ok but that's exactly what I did here I said let x over 2 tan of theta so that I could um so