Transcript for:
Understanding Trigonometric Substitution Techniques

okay so we're going to talk about trigonometric substitution and where this stems from is all right you can do some integrals great we can do substitutions we can do integration by parts we can do a lot of trigonometric integrals but what if someone was to give you an integral like this one let's one run through our list of things that we can do does this fit in integration table anywhere no okay is a substitution going to work no no cuz if I substitute for this the derivative is something with an X I don't see an X not going to do it uh okay is integration by parts going to work no not really is because well we have a piece of an integral this thing would be the other piece that that' be that'd be bad we take basically do integration by parts when we have a product in there somewhere all right well if we don't have that um is it a trig integral does it have signs or cosines then we got to have something else here's the something else the idea is let's let's kind of change this let's think about this in kind of an out of the box kind of way let me just consider theare < TK of 1 + X2 now somewhere along the way someone said well wait a second you know what this looks a whole lot like Pythagorean theorem whole lot like it in fact this looks like I could do 1^ 2 + X2 same thing yeah mhm well then they said wait a second if that looks like Pythagorean theorem the stuff that we would get by Pythagorean theorem then maybe we can make a triangle out of this thing so they say I can't do it like this well this looks really similar so let's make a triangle and see if we can actually use that here's a crazy idea so we create this triangle and what we're going to do is we're going to fit this piece of junk that we can't deal with as one of the sides of our triangle the whole deal that you're going to develop here this in this whole section um I don't really want you memorizing a lot of the formulas I'm going to give you there's only three of them I don't want you to have to SK sit there and memorize with flashcards I want you to understand the concept what is actually happening here what's actually happening is that we're going to take this and we're going to make it a side of a triangle now there's a very specific way that you do this you have to tell whether it's a leg or whether it's the what's the longest side of a right triangle hypoten whether it's the leg or the hypotenuse it's not hard to tell what would this be a leg or a hyot that has to be the hypotenuse the only way that happens is if I have a leg squar plus a leg squared then I take a square root that gives me the hypotenuse remember this A S Plus b^2 = c^2 mm check it out if it's this hey look at if it's this then this thing stands for a hypotenuse are you with me on that one if it's this what's that it's a leg it's got to be a leg do I know which one well we'll talk about that minute you can choose we we'll we'll pick correctly there's only one correct choice but we get to choose or if it's the other way around that's also oh sorry you know what that was stupid same thing I meant this or if it's that that's also a leg do you guys get the idea here so somehow we can fit this square root of something plus something or some something minus something as part of a triangle in our case this ISS wait a minute the only way that I can make this fit a triangle is if this is the hypotenuse and this is my Angle now here's the deal for the rest of your triangle your idea is going to be to try to make the sides in some relationship that allows you to create tangent s or secant where you have a variable over a constant so I'm I'm going to do this I'm gonna say wait a second if this is < TK of 1^2 + x^2 can you tell me the lengths of these sides it's a piece of cake one of them's x one of them's one that's it now if I do this one and X that's a cotangent cotangent is really hard to deal with okay you know what I kind of skipped a point we're going to use a relationship here that allows us to not work with this one but work with a trigonometric relationship of the other two sides so I don't like the square root that's really hard to work with so I want a trigonometric identity that lets us work with the other two that don't have square roots the easy parts of this does that makes sense so we make it easier if I do this 1 /x well then I'm not really going to get something nice I'm going to get cotangent does don't have some nice that does have an identity but we try to stick with easier identities so I'm not going to do 1 /x I'm going to do X over One X over one now here's the plan the plan was if this is part of our triangle and now we have our other two sides X and one let's pick a trigonometric identity trigonometric function actually that allows us to relate the two other sides in a really nice fashion for instance if I look at tangent tangent is for this specific triangle how much just tangent x x x over one or tan Theta is simply simply X that's great I I I really need you to get this idea before we go any further I want you to really see what's happening what's happening is we don't know how to do this we take this part away we go you know what that looks a whole lot like part of triangle with Pythagorean theorem let's make it be part of a triangle by Pythagorean theorem then we can find the other two sides of that right triangle well it's really easy it's kind of given to you one of them is one one of them one of them is X we put them in this relationship that way when we do tangent of our angle we get simply opposite over adjacent we get X over one or tangent theta equals x so hands feel okay with that idea it will become more clear I promise we do a lot of examples um I'll also give you some the formulas but I'm going to teach this to you in a way that it makes sense not that you just follow the pattern Okie doie it does have to make sense to you now and it'll become easier if it does this is what is awesome about this you agree that t x correct talal x tanal x that allows us to do a trigonometric substitution so we'll have an integral we'll still have a square root notice we're not substituting for the square root we can't do that because there's nothing to to to get the derivative away square root we're still going to have a one are we still gonna have an x no no instead what is our x s so we're gonna have tan wait a minute just tan Theta this was x s hey look at that tan Theta is X this becomes tan^ squ Theta I want to show up hands you feel okay with this so far remember all those times your Calculus one teacher maybe it was me uh told you don't forget your DX don't forget your D Theta don't forget those remember that and how some of you forget the DX and all that junk that you think is irrelevant here's why it's not irrelevant if you just leave that as DX does this match your no no this is an issue these have to match your variables have to match so what in the world do we do with that DX it's really similar to doing a use substitution really similar so we have tan Theta here we have X can you figure out what DX is take a what's the derivative of so equals how much ah now we know so this part becomes our DX I'm going to work over here but um it it is it's not a really complicated idea but it's kind of it's new for you I know it's brand new so I want to make sure that you guys are okay with exactly what's going on here before we go any further you understand we can't do this the way it is correct correct can't do it so we pull it off you understand that this is a hypotenuse of a right triangle it's got the plus there it's a leg squ plus a leg squar and then we take a square root that is the distance of a hypotenuse if we do that it gives us one leg and the other leg upon which we can make some sort of a trigonometric function a relationship amongst those two sides tangent is the appropriate relationship here it's nice we don't want to do cotangent and be one overx that'd be silly I want to do tangent and get X over one that way I have a nice substit tion soal X Tan = x we plug that in for X so 1 + we still have square root DX is an issue we got to change that so just like you did on a u sub you still take a derivative So Tan th = x no no no let's do see s equal DX then we get okay cool integral of s < TK 1 + tan^ 2 what's what what's else squ what that's nice this part that is my DX so instead of DX I've got see squ Theta oh you know what I forgot my uh my D why don't you guys catch me on that we you sleeping my goodness do you all feel okay with that one so far look what it does as soon as you do this that changes into from 7.3 to 72 just what we did you just did stuff like this didn't you in fact this is this is even nicer because it already has an identity built right into it it's already going to match up that's why we pick one of three trigonometric substitutions we pick tangent we pick secant and we pick sign if you got those three the identity is already there it's really really nice you guys ready to do this thing so tell me what is 1 + tan 2 th okay we do need to know that one all right do we have a square root still yeah 1 + tan s is SEC s time SEC s D give me something nice that happens do what now squ yeah that's exactly right this has to happen so this is nice because we have one of these identities which you are going to have look at it's being squared right is squared so you're can to have a constant plus something squared you can factor that out you can make it work so that that with identity your square root gets rid of your Square that's fantastic we love that so this becomes see Theta time 2 D th what are you going to do now com probably combine oh goodness we've actually seen this one before in class do you remember we did this was the this was a horrible one uh actually no no was seeking to the fifth that was really horrible do you remember doing that one we split it off we used an identity we we did a whole bunch of stuff with it we ended up adding the integral back and dividing remember that so I'm not going to do that again this is already in the video if you want to refresh remember on your notes or something it's already back there I don't remember it was 71 or 72 I think 71 I think it was integration by parts is what we did so it's already back there but anyway this integral becomes was it 12 yeah I it yeah 12 sec Theta Tang thet plus Ln absolute value of plus Tang no reduction for form here it's a integration by parts if you really want to prove that that is the way it is you split off the secant squar you take the integral of secant squar it becomes tangent and then you do your integration by parts you guys okay with that one I'm not going to show it it's already on the video we've already done this one uh quick show hands be okay up to at least this point this one we've done before so I don't want to rehash it what I do want to say is okay let's take a look at my integral firstly for I don't know I think this is really cool do you guys see how interesting this actually is that we can do this this we're using a a triangle to do an integral it's it's crazy it's out there it's kind of fun um here's the problem what variable did I start with what variable do I have now that's a problem so you're not done right here don't just put a plus c and go I'm done you're not done but the rest of this is not that hard you see we've got this triangle that we created do you remember when you did a Uub when you did Uub you Ed the substitution to change a part of your integral correct then when you got done you substitute it back in remember that same thing happens here we used our triangle to do a substitution now we're going to use our triangle again to substitute back so if our triangle is the square T of 1 + x^2 and this is X and this is 1 here's what I want to know that's Thea here's what I want to know can you tell me all of these trigonometric relationships just put that anywhere fine uh can you use this triangle to tell me all these trigonometric relationships yeah yeah so let's list them out can you tell me what uh secant Theta is using you get now you got to be good at this you got to know that tangent is opposite over adjacent and S is opposite over hypotenuse and cosine is adjacent over hypotenuse and you have to know what coent and cosine and cose are so tell me what secant ISP very good because cosine is adjacent over hypotenuse therefore secant the reciprocal of cosine is reciprocal of this so what was it again hypotenuse over hypotenuse what's the hypotenuse over let's also do tangent so we're going to do all of our trigonometric functions right here just right by our triangle can you do tangent X OPP brilliant just opposite over adjacent we got X over one do your substitution it's not hard tell me what's going to be here for me please what's going to happen we got a one2 very good what else see what's secant okay then what are we going to have okay plus what changes over here does the 1/2 change no no no does the Ln change no does the absolute value change no what's inside not s x plus X well you don't know what and then 9 plus probably clean it up a little bit I would at least I would do yeah do 12 x you could factor out a 1/2 if you wanted to 1 + x^2 and then plus 12 Ln absolute value < TK 1 + x^2 + x + C oh my goodness let's check is it in terms of X like we wanted it yes so the idea kind of it's a crazy idea it's a cool idea it's use a triangle to make a substitution based on a trigonometric function do the integral like you know how to do it's just a trig integral you guys should know how to do this stuff and then at the very end when you're all done with it use your triangle again so triangle to make a substitution triangle to fix your substitution show hands if that one actually made sense to you do you guys get the idea did you even know that was possible kind of cool though right do you want some more examples of course you do yeah of course now this made it look a little bit easier than it might be for you because I skipped over this I skipped over the secant Cube uh we did that one before which you're fine to use that okay if you've done it before look at okay we've done that uh but you do know how you do know how to do that that is an integration by parts so some of these are going to take you some time do you guys have any questions because these problems take a little bit of of board work so I'm going have to rase this any questions at all before we go on you can ask if you want all right okay so before we go any further I want to give you a rundown about when we can use this and why the way why it looks the way that it does so we can use this when we have one of three situations I'm going to give you the formula again I don't want you to just blatantly memorize the formulas you will okay you will memorize them uh but I want you to understand why they are the way they are because it's really silly to do this stuff just by what's my formula it doesn't make any sense I want you to get why it works and that's important so this idea does work if we have one of these three things if we've got a square OT of a^2 - X2 or sare < TK of a^2 + x^2 or sare < TK of x^2 minus a 2 our first one was of this form now after we do that we'll be using one of our pythagorean identities so then we use either 1 + tan^ 2 = see squ in one way or another and or 1us sin s = cosine s say what oh darn it thank you so here's the big three I'm going to look at these in more detail we're going to understand why they are the way they are why the formulas actually work and then every time we go through one of these trig integrals trigonometric substitutions for integrals I'm going to rehash the same thing until it kind of sticks with you so here's number one let's look at this if we were to create a triangle out of this thing I always draw my triangles in the same orientation that way I have it fixed in my head I don't want to get random triangles up here it's got to be a right triangle of course I always draw it this way maybe you can do the same thing I want you to think of what this actually is it's going to be part of the Pythagorean theorem but think about the pieces up here would this have to be the length of the hypotenuse or the length of a side what do you think it's got to be a side because it's got a minus in there does that make sense pyan theem says the hypotenuse is when you take the sides and add them a side happens when you take the hypotenuse and subtract so what we know right now is that this is going to be the length of a side you guys all understand the idea it's got to be a side now think about which one of these would be the hypotenuse if this is a side one of these pieces has to be the hypotenuse what letter has to be the hypotenuse a has to be X can't be because the hypotenuse is the longest side correct so if this is the longest side x was the hypotenuse I'd get a negative inside my square root and I can't have that you with me so we look at this go wait a minute if this is a side one of these is a hypotenuse the bigger one the first one has to be the hypotenuse that's got to be a now here's your major idea idea for this whole thing for this whole lesson once you figure this out it's not hard I mean shoot if it's got a plus that is the hypotenuse if it's got a minus one of these is the hypotenuse is that clear for you it's the first one the bigger one is the hypotenuse now here's your whole goal your next goal is you are going to make these sides in such a way that you get X over a x a for either tangent or s or secant those are the only three that we're going to use tangent or sign or secant so we figure out where the hypotenuse is then what we do is we make our sides in a relationship so that tangent or s or secant gives us X over a is that clear for you I'm going to write that out and make a little little note here so notice that you get to choose right either this is going to be X or this is going to be X either this is going to be S root of x a square - X2 or this is going to be theun of a s - x^2 you get to choose but only one's the correct choice so put sides in a relationship such that we get X over a for tangent or s or secant these are the three that we use we don't use cosine we don't use cosecant we don't use cotangent we use tangent s or secant and the reason why we do that is because the identities are really nice for those it makes it easy you guys with me on that one so let's try so this is our idea yeah we want this let's give one of them a try if I do this if I put X here the only way I'm getting X over a is if I use what please you got to know your trig cosine would give me X is cosine tangent sign or secant then that's a bad choice so we figure out our hypotenuse first and then you only have two options X either goes here or X goes here or the side goes here or here if x goes here if x goes there can I make x a with either s or tangent or secant which one brilliant just fill up the the rest of the other side with whatever it is a^2 - x^2 that's the idea that's the idea so dep if you understand the idea you see most classes or most books just tells you hey if it looks like this we're going just do sinal x a that doesn't tell you why the reason why we're doing this is because when you create your triangle we're going to do it in such a way that this gives us something nicer something without a square root that's what we want something easy so if I've got a A for my hypotenuse I get my choice of X I'm going to choose it in such a way that tangent s or secant gives me X over a then we write that appropriate trigonometric function so which one are we going to use are you still with me so we go okay cool then s = x a sin = x a and we always solve for x we solve for x because that is going to be our substitution here you guys with me we don't have X over A's generally we have X's what you saying Thea I keep forgetting I'm so excited about doing this I keep forgetting my thetas uh so dorky to forget our forget your data I know it's ironic right but since I'm the teacher ha haa I get to Mark you down does it mean you're right it's true it doesn't that is true that's what I have you guys here for right to catch me on all this stuff so if we solve for x just multiply by a this right here becomes your substitution now you are going to get very good at this and what you're going to understand is that when you see something like this all you have to have is X = a sin thet you're going to have to draw your triangle every single time I still do it for myself especially because at the very end we're going to have to substitute back in so you're going to need that triangle so draw it get used to the idea of drawing your triangle don't skimp out on this stuff uh but this is where that's that's actually coming from you show F feel okay with that one good all right the next one number two let's talk about a^2 + x^2 now we actually already had an example of that one Michael uh so we just skipped the other trick functions because they're just difficult to use in this way you don't need them uh we use the ones that that that that work all the time uh those ones are the tangent the sign and the secant could you do it with the other ones probably um but we don't because these ones work these are the the big three this one takes care of tangent and cotangent this one takes care of s and cosine this one takes care of cosecant course so otherwise we'd have the different identities these ones work really nice you guys ready for the next one yep first thing we do every time we see one of these that we can't fit on our table that we can't do a substitution for that we can't do integration by parts for and that's not trigonometric we do a trig sub that means you're going to draw a triangle so so we draw our triangle here you've got to be very good at telling me whether this piece is a hypotenuse or a side what is that one hypoten very good so this makes it actually pretty easy our hypotenuse is a^2 + x s now we get to put our sides in a relationship we we are either going to have by the way what are our two sides here A and A good it's not a squ it's a so if this was a nine you'd be putting a three does that make sense sense okay so we have a and X where's the a going to go opposite or ajacent good because I'm trying to make I'm not trying to make S or see here because those have to deal with the hypotenuse I don't want that I want tangent tangent is going to be this relationship I don't want a overx that would be a bad idea what I want is X over a so in our case our substitution would be tan Thea = x a you okay with that one solve it for x x = a * Tang thet if you look back at the ex example which I I left on the board left part on the board when we did this when we did our substitution that was it X was 1 * Tang th that's what we had here then we squared that still okay with it let's talk about the last one then I'm going have to erace all this so if you have any questions you let me know but again I don't really care if you memorize these ones the way I'm hoping to teach you is that the relationship of a triangle actually should make sense okay it's then you don't have to memorize the formulas you can just do it whenever without having to look at it as a formula okay number three okay let's think through squ < TK of x^2 - A S would you say that X2 - A 2 is a hypotenuse or a side what do you think side it's got to be a side so it's either going to go here or it's going to go here I don't know yet until I determine where the hypot what the hypotenuse is so let's talk about what the hypotenuse is what's the hypotenuse how do you know it's X yeah if it wasn't if that was not a hypotenuse this would be bigger when I Square those numbers I'd get a negative and that's a problem so X is the hypotenuse now we got to put our sides which are a a is a side here x is already taken care of a is a side and this whole piece is also a side all right where's the a go in this triangle opposite okay let's try if I do opposite oh no sign would give us ax do I want that I want X over a and I can only use tangent s and secant so let's let's see which one of those work would tangent give me X over a Would S give me X over a okay so if that doesn't happen what's it going to be it's got to be secant and the only way to do secant is to move a from here to there is hypotenuse over adjacent so if I say hey the x is the hypotenuse guess what I'm using see hello that's the only way it works so X is hypotenuse I'm using secant does that make sense then this is x^2 - A 2 in this case we go okay well let's make our trigonometric substitution let's make something so that our trig function gives us x a the only one that works here is secant so see Theta is X a if we solve for x our substitution becomes x = [Music] a in each of these cases we've used a triangle we've created a trigonometric substitution after this we take a derivative of both sides that lets us change our DX into a d Theta DX d D DX into D Theta we do our substitutions and then we have a trigonometric integral which we just covered that section four get the idea kind of cool right what can't you forget at the very end of your integral to do use triangle yeah so we use a triangle to do this we're going to use a triangle to undo this as well it's time for some practice this is it this is the whole idea if you get this idea you're done all right it's practice it's not it's easy to made it sound all right uh right now it's like oh this piece of cake they do get a little tough um mostly it's just working with the trig and not the substitution you guys should be almost pros at that just after looking at this it's typically the trig integral that gets like oh man because you're going to be doing the last section stuff again lots of practice here you guys ready for some practice exciting little bit I'm so excited it I don't know it's fun for me will we be able to do the whole section knowing this will you able to do the the whole section just knowing this yeah is this pretty much like the bread and butter the whole thing this is trigonometric substitution about yeah sorry 73 this pretty much 73 pretty much it okay yeah which I think unless there's anything else that uh that happen yep that's it so we can do our homework yeah won't we just end right now then huh no examples whatsoever fantastic you know one more thing that I want to put up here for you that I didn't put up um if you're going to do this substitution you're going to be getting a s squared see it if you get a s squared then you're going to be using this one you follow me you're going to have something sorry something minus sin squ you'd be using this identity if you're going to have a tangent up there you're going to have something plus tangent you'd be using this one does that make sense so let me write this for you here so once we have this the substitution we would have is 1 -^ 2 thet = cosine 2et this would be the substitute the identity we'd be using once we have a 1 + tangent 2 th you'll be using 1 + tan^ 2 th = see squ this one if you have a look at this if you made the substitution you'd have see s minus something you have s- 1 we'd use SEC 2 - 1 = tangent and those would be the follow-up identities that you're going to use here now let's practice do you have any qu do you have this written down everybody yes no yes yes okay let's practice again I don't care if you memorize the formulas it's going to be obvious what you do if you me if you understand the idea so let's do some how about this one go through these integrals not just that we're in in a certain section for trigonometric substitution go through them and see if you can actually do them without it okay like if I give it to you on a test I'm not going to tell you what to do so go through it see if you can do it with a substitution if if you can cool do it see if you can do it integration by parts of it if it's really quick you go okay I can do that see if you can do it with a tri just basic trigonometry if it has signs and cosiness up there do it go to trigonometric substitution of course you can have something like this but try some of those things like if this was just X and that was just X I'd go for it I'd go for a substitution I would start a trig sub because it would be easier than that does that make sense okay in our case we have well nothing but a trig sub here what piece are we looking at to substitute yeah so let's take a closer look at the square root of 9 - x but you know what I'm not going to think about it as 9 - x^2 I'm going to think about it as almost begs the question wait a minute what if it wasn't nine what if it was seven could you still do it yeah you got to think outside the box though if this was a seven you could do something like the square < TK of s square that would give you seven does that make sense it's harder but generally we get those perfect squares which is really nice but you don't have to okay here's where the rubber meets the road all you basically all you have to do is know how to draw your triangle I mean not just a triangle but know how to put the sides on the triangle that's the key thing if you can do this correctly and make the appropriate substitution everything else falls into place really nicely so let's walk through it I don't want you to just look at your notes and memorize the formula I want you to think through this would you say that this is a side or a hypotenuse side it's a side for sure can you tell me what the hypotenuse is here so create a triangle that's three tell me what uh one other side is X now we want to put this in a relationship so that we get X over 3 with what three functions are we looking looking at I heard a lot at once hopefully you said s tangent and secant that's right s tangent secant let's see which one would work for us is is tangent going to put X over a for us no tangent would put whatever you're going to have a square root somewhere that's not good uh would secant work for us no secant could put X over sorry 3 overx but not X over3 I don't want to do that so what's the only other option for us sign let's make this work with s x would have to go there to work with s does that make sense to you the other side now we change it back to 9 < TK 9 - x^2 let's go for this let's do our substitution so the first thing we got to do is well all right once we have our triangle what's our relationship again one more time what do we [Music] have good so s of theta = x 3 that makes sense doesn't it sin of = x 3 opposite over hypotenuse therefore x = 3 sin th quick show hands if you're okay with that one cool now what else are we going to do right now that we've done our substitution yeah that's great idea so DX would equal 3 cosine [Music] so hey we got our substitution are you guys okay with this so far we've got x = 3 sin thet no problem take a derivative DX = 3 cosine D Theta now we get to substitute with all the stuff so let's see how it works out for this in this type of process everywhere you see X you're going to have to substitute there's going to be no elimination of x's because you're getting rid of all the X's with your thetas so tell me what's going to be on the numerator please we got to move a little little quicker here three what okay so three is the three going to get squared here yes yes it's part of your X so this whole thing get squared if you put three sin^2 thet you missed out on your constant you're going to miss that denominator n's still there very good what's 3 sin Theta and then what all squared very good I like it and then oh man we got a D Theta as well oh sorry DX as well what's the DX become okay honestly if I did this at the start of class you'd probably be like what in the crap is going on here uh but let's see if you it makes sense now are you guys okay with a triangle it's just a triangle you got to be okay with triangles they're fun make up the relationship such that this fits okay and that you get a trigonometric relationship you want s tangent secant where you get x a that's the whole deal once you get that everything else works out real nice so we got well most of the time we got DX = 3 cosine D Theta that's where this comes from we got x = 3 sin Theta s 9 cool- 3 sin th s now it's just a matter of simplifying don't try to work with it now simplify what's going on here so let's do it this becomes 9 * sin 2 Theta this becomes the sare < TK of 9 - 9 sin s Theta and this is still 3 cosine Theta D Theta we're just doing some algebra right [Music] now now here's the issue this is almost almost an identity but it's not why is it not an identity the nines are there well let's Factor the nines yeah let's factor that out so 9 sin squ th over Square < TK of 9 if we Factor those nines I get 1 oh beautiful look at that love math cool that's the same this we factor out the nines we got 1 - sin^2 th we got 3 cosine Theta D Theta still you guys still okay with it tell me what happens when I have a product within a square root what can I do with that pull yeah you can separate products within square root so let's see if we can do we've got 9 sin s we've got what's a square of 9 and then the square root of 1 - sin 2 and then we got up here a 3 cosine thet D threes are gone three here three here also check it 1 - sin s th this is why we did that substitution cuz this will work for us so 1 - sin what is 1 - sin sare that's why we use sign okay because that works nicely for us also what else is going to happen Okay so n is here I've got s Square th I have square < TK 1 minus oh sorry I messed up I've got cosine 2 time cosine thet D Theta we still okay so far it's getting nicer noce have we done any calcul this right now no no not really at all all we've done is manipulate it so we've got nine out front threes are gone this is cine s it's still in a square root tell me what happens with that square root [Music] so we have cosine th time cosine D Theta those are gone we have N9 integral of sin 2et D thet can we do the integral of 9 squ sorry 9 integral of s^ s how would you do it reduction you could do reduction formula you could how how could you do it if you didn't want to half angle yeah very good sorry I didn't hear you so if I did half angle we could do 12 1us cosine 2 th D Theta 12us 12 cosine 2 th D Theta stop me if you see a mistake I know I'm working kind of quickly here uh what else pull half out you could you P pull the 1/2 out then we're going to have 9 half if I factor the 1/2 out 1 - cosine 2et D thet don't just pull this one half out though you'd have to factor to make that happen why didn't do it on the earlier stuff you could have you totally could have don't know why I didn't oh you're about right there yeah you could have done it right here just pulled the one I see that right there can you take the integral of one what is that it's not x no keep in mind this is why it's importanta what's the integral of cosine 2 Theta good so Theta minus 12 sin 2 Theta are we still okay okay we're going a little long today I'm sorry for that um did you guys want to finish it right now un bet you do we have no class coming in so we're good wo let's go another hour okay no no no show we're not doing that I'm hungry I'm wasting away lost three pounds in a week I know I've been moving it's great this anyway it's not not important SOA I don't you feel sorry for me after to do math and I'm moving 12 sin 2 thet we still all right with that one M now here's the issue this is done calcul is done but we started with we have use your triangle you're always going to use your triangle so here we go let's start with it what do we know about this thing we know what sign is oh goodness gracious well we know that sin Theta let's start with this okay cuz you you can't have any thetas left you got to have all X's we know that sin thet = x 3 we know that don't we Okay cool so keep it keep it simple use this substitution again s this is why we write this one we don't go directly to this one because often times we'll have to use this sin th = x 3 if that's the case the only way we can solve this is say well let's take it a left inverse of sign on both sides if we if we do this sin th = x 3 if only if Thea = sin inverse x 3 that's one little piece here's the problem our our main problem our main problem is that sin 2 thet that's a problem if I look at sin 2 th is that 2 thet no no you don't just get to pull out a two or multiply something by two you actually have to have the actual Theta so when we look at S 2 we need to know that that is the same thing as 2 sin thet cosine Theta hey look at that do you recognize that identity sin 2 Theta yep is a double angle so sin 2 is the same thing as 2 sin now now it becomes easy can you do sin Theta yes we already know that one sin Theta is X over 3 Let's do cosine Theta let's look back at our triangle for cosine Theta how much is cosine so adjacent adjacent over hypotenuse you like our back there so good toy guys yes okay I'm gonna have to erase this we're g to move over here we're almost done I promise uh but I want to make sure that this actually makes sense to you are you okay with our triangle substitution for this that was pretty this the easiest part actually is this it's easy now that I've taught you how it actually makes sense you don't have to follow formula hopefully you understand hypotenuse sides put them in the right relationship get whatever trig substitution is necessary everything's simplification all of it is simplification until right here then we have a little bit of calculus you know how to do that already that's your last section we get all the way down here and then you go ah junk now we got to substitute back in for any thetas that we see some of them are easy some of them are are take some some time okay so check it out Theta whenever you have just a Theta you are going to have an inverse trig function so if sin th = X 3A is sin inverse of x 3 you got it sometimes you'll need an identity sin 2 thet no no can't have that get 2 sin Theta cosine Theta well sin Theta would be X 3 that's from more triangle s is X 3 cosine thet is 9 - x 3 and now we're going to put everything back together so 9 Hales we've got Theta - 12 sin 2 thet I'm going to draw it out to make it hopefully very clear for you you this is 9 - 12 * 2 sin thet cosine thet by the identity that we just used so the 1/2 is here this is 2 sin thet cosine thet understood 1/2 and two are gone so we get nine halves Theta is sin inverse x over3 minus s was x 3 cosine was < TK 9 - x^2 over three we'll make it just a little bit nicer looking here you could factor out um a a 1 nth but don't do it with this three because that's part of your Angle now you can't change that so oh not sorry not angle that's uh side relationship so 9 halfes we have S inverse X over 3 - x * < TK 9 - x^2 over 9 and then finally plus C oh my goodness yay is the triangle the hard part no it's all of this junk that hopefully afterwards I need to show fans you feel okay with what we've just done fantastic all right three more examples and we're done with Section 7.3 again we're going through some of these these integrals that we can't otherwise do without this thing called trigonometric substitution so the idea is look for a square root in your problem sometimes you will have to create a square root in your problem if you look for a square root in your problem that fits the Pythagorean theorem of a right triangle we can do the substitution so let's take a look at this one this one's not going to take us too long just want you see a couple other techniques that we're going to use here so integral from 1 to the3 of 1/ 1 + x^2 to 3 do we have a square root in this problem yeah might be a good idea to write it like that so instead of just like this just take a second and rewrite it so that we have that square root so instead of three halfes what does that three halves actually mean to the square okay so I'm going to have a square root like this and then what I'm going to do is I'm going to have that three outside you should know from your algebra that you can have the the power three inside or outside we're going to choose the outside that way our substitution is actually possible you guys okay with that so far now we should be to the point where you can make up your own triangle here so give me the idea let's walk through this one I'll have you do the next one on your own what's the idea what are we trying to do hypot so we want to find the leg okay so we want the legs and the hypotenuse let's look here we get the square root and what I'll do is I'll have uh square root of 1 plus + X2 that's going to give me something here so we're trying to make the < TK of 1 + x^2 be either one of the sides one of the legs or the hypotenuse now one thing I'll always do is I'll try to write whatever this number is as a square that way it gives me the side of one of the lengths of the the legs so instead of one I know it seems a little bit trivial but I'll put one squared if it was like four I'd put 2 squared 16 I'd put four squared that way I'm not accidentally putting the wrong number on one of my my legs does that make sense to y'all yes okay so let's think about it we got the < TK of 1^2 + x^2 you guys should know right now what that is whether that's a hypotenuse or the length of a leg what do you think that's going to be how do you know all the out of like sorry how do you know automatically what it is if it's got a plus there it's got to be the hypotenuse not because it's a formula but because we understand what the Pythagorean theorem does Pythagorean theorem says that the length of hypotenuse is one leg squared plus another leg squared and then you take a square root of it so there's no way that can be a leg this has to be the length of the hypotenuse does that make sense to you if we have a minus well that's the other case Pythagorean theorem says for the length of a leg we taking the hypotenuse squar minus another leg so what we know is if we had a minus right here this would give us the length of hypotenuse and this would give us the length of the other leg this would be the length of uh the the last leg itself so let's go through this how much sorry how much do you know about what Trigon trigonometric identity to make which three do we always use good those are only three because those are the identities that work well for this now what we're trying to get out of this is X over a constant X over a is what we say so we got to use either tangent or s or secant to make this happen in that case we're going to pick X for here and one for here so we look at the length of our sides we got X and we got one X and one where's the X going to go here option one option two option option two try put X here oh wait no option one if I put X well think about why though don't just fall into the pattern look at the formulas I hate that I don't want you to memorize all the formulas and okay CU what if you forget a formula that sucks if you understand the triangle it will make this easier let's look at this uh firstly do you all understand why this has to be the hypotenuse yes or no okay if that's the case either this is one and this is X or this is one and this is X if I have this is there any way in the world I'm going to want to use sign or secant on this triangle right now sign I'd have this I'm trying to get rid of that that that's the cool idea I don't want I don't want that how about uh how about seant I'd have this I don't I don't want to use that either I want to relate the easiest two sides the X and the constant but I want to do it in such a way that I get X over the number not the number over X so is this going to work for us no tangent would give us 1X but well let's make it nicer let's put X here and one here now if I take tan Theta I get X over 1 which means that X = tan Theta typically what we'll do is remove our constant over here since it's one no big deal okay keep on going what now we got xal tan Theta what's the next idea before we plug it in because we have two things that we're going to be doing here what else should we do say what you take the let's do that if we take the derivative you see when we do this we're going to be changing this into some sort of a function involving Theta but we also need to change our DX into some D Theta so in order to do that we take our derivative we get DX equals oh everybody tangent d d Theta perfect guys so far we so good go ahead and make the substitution let's see you guys do it Go From Here plug in what you need to plug in I'll do the same thing on the board I just want to make sure you can do it on your own okay so go I'll give you about 10 seconds and I'll put it on the board make your substitution notice when you're doing trigonometric substitution the square root it's never going to change the idea is that we're changing just this variable X because when we Square it when we square whatever that variable is of our substitution we're going to get some sort of One Plus or constant plus that we're going to be able to make an identity out of like 1 plus sin squ or one plus Tang squ or one one of those actually 1us s Square sorry or 1 plus Tang squ that automatically fits one of the identities that we have in hand so one does the one change no no and the plus doesn't change what's the X become s good so if the x is tangent then this is tangent Square thet very good and we still have that power three let's make sure we're not losing anything now the reason why we did our our derivative right here is so we also substitute for our DX what's our DX become let's make sure do we have all thetas and a d Theta looks like it we're pretty good to go there's one thing I haven't done yet and this is kind of up to you I like doing this uh some people don't some people like to wait till the very end I like changing bounds I like to because sometimes I'll forget to replug in my substitution that kind of sucks so right now I'm going to go ahead and change bounds we do that at this step we don't want to wait till later do it right now when you're doing your substitutions so right up here we'd have X = < TK 3 and 1 write that down so if that's the case if xals a < TK of 3 then what we would get is is tan th = < tk3 and tan th equals 1 please notice I'm doing this on purpose I want to make sure that you guys see how to actually change the bounds in this scenario if you have x = 3 look at where the x is X is Nota what we're trying to do is solve for Theta does that make sense to you so we'd have the S < TK of 3 equals tan Theta we would have 1 equals tan Theta you guys follow this idea well now what we're going to try to do is put our integral in terms of theta not tan Theta we don't want one and 3 here again what we're going to have to do is use an inverse tangent to figure out what that is so basically said okay cool I'm glad you got this now go one step further figure out how much Thea is so recall I'll do this on one of them have I lost you I don't want to lose you yet recall that this is true if and only if this is true do you remember doing that this is true if only if and only if this is true so basically it's asking hey um what angle does it take for tangent of that angle to equal one and same thing here what angle does it take for tangent of that angle to equal s of three do you guys know what it is or can you think of the the inverse what's tan inverse of the three what's tan inverse of of one and pi over 3 and four yeah it's it's a nice order for us too so this becomes pi over 4 this becomes pi over 3 let's make sure that you guys all have this um are you okay on writing this as a square root to the third power yes no you guys are dead today huh my goodness was Monday but geez more activity I'm I'm excited about this stuff you should be too goodness gracious to this is fun I guess come on people we talking about math Jedi already jeez okay so are you okay with writing that as a square root to the third power still wasn't any better was it lame people in the video think you guys are lame right right now right on anyway uh so this is now that hotus real we talked about why we're trying to make our sides in a relationship that work with Tangent or s or secant that give us X over constant we've done that with Tangent So Tan the = X no problem now if you have a definite integral change your bounds here if we change our bounds x = 3 okay tan theta = < 3 that's what we're getting that means that you're going to have to do tan inverse of 3 to figure out your angle in this case we get 3 in this case where tan = 1/ 4 tan of Pi 4 gives you one so we've changed our bounds from PI 4 to Pi 3 and still in order which is nice for us okay next step come on tell me what we're going to do the bottom from the bottom so we're going to use our identity get rid of that thing this is always going to happen if you do your trigonometric substitution appropriately you're going to have an identity that you'll use so we'll have pi over 3 Pi 4 to pi over 3 got SEC s th over tell me what this is going to be exactly what's still there okay square root is still going to be there I like that what else this this is an identity this is equal to see square and what else don't forget any little pieces here cool almost done what's the next thing that we're going to do quickly come on you you end up one over Square you squa to the second and then the square roots would cancel out you end up with one okay so let's simplify this part let's do that square root of something squared what happens so we got p 4 we got Pi 3 we got s we got the square root of SEC squ is just but then I have it to the thir power so this gives me secant Theta to the third power are you still okay so far notice all the simplifying you're doing really we haven't done any calculus this is a calculus idea sure trigonometric substitution but have we actually done any maybe a derivative here huh maybe a little bit that's pretty easy we're just really simplifying a lot of us to make it nicer for ourselves so Pi 4 Pi 3 we're going to have 1 over seea D Theta don't get hung up on this stuff think back to what I told you about trigonometric integrals how trigonometric integrals work is this if you have signs and cosiness I told you how to deal with those right yeah if you got tangents and secants I told you how to deal with those if you have cotangent and cosecant I told you how to deal with those if it's not one of those things try changing it into SS and cosin try changing that if it doesn't man what's one over secant well let's think about one over secant is cosine so pi over 4 to pi over 3 no problem this is just cosine Theta and all of a sudden we have a very very easy integral to do everybody what is the integral of cosine Theta what do you think perfect oh you got everyone now we woke up sin Theta now here's the question do I have to go back and change everything into X's again no this is why you'll want to change bounds here because if you don't you're going to have to use your triangle you have to go back and change this back into X's before you plug in 1 to the of 3 did you just catch what I said yes okay so if you change balance this is kind of nice here because you don't have to go back to your triangle if you didn't do this right now you'd be looking at sin Theta you would have X look at this please you'd have X over this 1 + x^2 then you plug in the squ of three and you plug in one and you subtract those two fractions does that make sense to you if we change balance no big deal we change these from X's into thetas hey it matches now so now these are in terms of theta no need to plug it back in we're just going to go Pi 4 to pi over 3 what that means is we'll have s ofun / 3us s ofun / 4 that should be pretty easy for us at this level s of piun 3 is what it's over two that's good isn't thegal of cos sign no it's not but think about why okay sare 3 over2 very good andare < tk2 over 2 make them one fraction 32 over two that's the idea took us a little bit longer than I thought but show hands if that move made sense for you good deal okay let's work on the next one so definite integrals you might definitely want to try to change your balance here that way you don't have to use your triangle to make your resubstitution just a good idea are you guys okay down to this far show fans if you are very good that's fantastic that's what I'm looking for let's do the next one questions comments concerns is it more understandable than the first time we went over yes it's getting easier is it getting easier for you I like definite integrals better these ones yes why there's no res substituting that's why that's why you change your balance well of course Rob now that you've said that I'm not going to give you any more different intervals we're going to practice these ones now right on you're welcome okay here's what I want to do I want you guys to try to do this problem basically completely on your own but we're going to do step by step so the first thing I want to do I want to see if you understand that that's going to be a trigonometric substitution can you see it just by looking at the problem we're looking for well substitution firstly substitution ain't going to work we're looking is it uh by Parts well not really we don't have a product that's going to work for us very nicely is it a trigonometric integral well not yet but we're going to make our substitution so I want you to pick the piece that's going to be the trigonometric substitution go ahead and do that now pick the piece that you're going to use it's always a square root always if there's no square root in the problem you need to make a square root in the problem you can do that algebraically so pick a piece of that that's going to be your substitution make a triangle from it make a triangle think about think about the piece that you you just picked is that the hypotenuse or is that a leg a leg leg okay so you have it as a leg pick the appropriate piece of this that's the hypot new first so label that out do it individually do it on your own let's see if we all get the right Tri do you guys know that making the triangle is like the most important Point part here after that once you translate it to a trigonometric integral it's it's pretty straightforward I taught you how to deal with it the triangle is important here I want to make sure y all got this it's a right triangle the piece you should have picked is X - I'm not going to write x^2 - 4 what I'm going to think of it as is x^2 - 2^ 2 okay now let's think about it did you find out that this is actually the length of one of our legs yes okay now how we start this we always start with the hypotenuse first always because then we can determine what our legs are after that so if this is a leg then one of these pieces is going to be my hypotenuse everybody what is my hypotenuse in this case good yeah it's it's always the one that comes first otherwise you have a negative inside of your radical that's not possible so X is our hypotenuse the reason why we write this as 2^2 rather than 4 is so that we don't accidentally put four here or four here we put two here or two there depend on which side that we need to use for this does that make sense to you okay now we'll think a little bit harder which trigonometric relationship would we want to use here would we want to use tangent at all no tangent is going to have a square root in it and that sucks I don't want that are we going to use sign no sign would have X in it but it' be something over the X this is the way you go through you actually think about these don't just follow the pattern think about what's going on here tangent sucks got a square root No Deal s sign's pretty close because we could make it constant overx but I don't want it overx I want X over something so the only other option that we have is so set up this triangle so that secant is going to work nicely for you where does a square root go bottom leg right leg that way we get it out of the secret relationship so the square root is x^2 - 4 I'll put the four back because I don't need to worry about that anymore and this one what's this going to be two two not four two do you understand why it's two this is why we do that let's make sure this works for us we have hypotenuse got it we have one leg here this is going to set up our secant relation ship and then we have this as another side show feel okay cool okay go do the next step if you haven't done it already write the trigonometric relationship we just talked about as equal to those two sides find your derivative and make your substitution I'll give you about 30 seconds to do that if you can do that fact okay which one works again yeah secant is going to give us hypotenuse over adjacent so secant Theta makes that relationship that we want we want X over some sort of constant uh then what we do after we get X over some constant solve for x yeah solve for x cuz that's going to be our substitution I don't see any X over 2s over here neither do you I see just X's so let's make this x = 2 Theta okay everybody what's the next thing that we're going to do take yeah because not only do we have to substitute for our X's we also need to substitute for our DX so by taking our derivative by the way just question here should you use a product rule here no it's a constant don't use don't make things harder than they are don't do something silly okay I know that a lot of people think that this is going to be a really really hard class and some parts it is but don't make it harder than it is if it's a constant you wouldn't do this what's the derivative of 2x product rule let's do it no that's silly right yeah it's a product but that's a constant in front it's a factor that's happens to be a constant so we do two and then what's this become oh that's good and then d Thea very good okay if you haven't done so already make your substitution have you all substitute on your integral not yes or no have you made it that far okay go go do that next do the substitution let's make sure that we're all 100% correct there's a lot of pieces here lots of stuff going on so make sure we're all right where you have X where you have x what are we putting in for X OH Close don't forget that too if you forget the constant there is your integral going to work out no it will but your constants are all going to be wrong so make sure your constants are correct if if it's 2 see Theta make sure that goes in every place for x and your DX make sure we have the correct thing for DX so let me make sure you get the same thing I'm about to get let me have your eyes on the board here real quick please so X S I still have a square root but x s no no no for X2 I'm going to have this whole thing squared notice it's not this do you see why not yeah when I have X this whole thing is the X so when I Square it it's got to be like that did you catch that yes that's going to affect this because what happens here is when you square the two it allows you something to factor out you can take that four out of there and that's what we want to do also to the 4th well it's now going to be 2 SEC theet all to 4th so we got 2 see Theta no problem we got it squared minus minus 4 4 got it we got a square root around it we got a 2 secant Theta now it's to the 4th power and lastly what else do I need up there so 2 SEC Theta Thea D Theta what I want to know is show up hands if you got exactly that on your paper I know you do now but like before I did it how many people got had that that's good that's great that's a first step if you didn't does it make sense how to get that are you sure it's going to take some practice right we got to get this stuff down what's the next thing that we should do should we start doing substitutions and all that stuff right now what we going do simp always simplify these things don't do substitutions or crazy stuff until you have this simplify so if we simplify well it's just a whole bunch of algebra I'm going to go through kind of quickly if you have questions of course let me know when we Square this this will be square < TK of 4^ 2us 4 all over 2 4th is 16 SEC SAR sorry sec to 4th Thea and then of course we still have a 2 see Theta Tang thet we can't afford to lose any of this stuff interal won't work out next thing what would you do if this was your problem on a test because it very well might be Factor the inside of the yeah what we know is that identities work with see squ minus 1 not SEC squ - 4 so when we have the same number here that's what we have to have that's that's important we got to have that so if you miss the two here you're going to miss the four here that's a problem so once we factor that four pull it out sure I'm going to actually do that step right now okay so watch carefully there four factors correct so what we get is a^ 2us 1 if I factor the four and take a square root it's going to give me two does that make sense this is this to me in my head is it to you in your head yes okay so see squ - 1 no problem the four is factored I take a square root because we can separate square roots like that with product 16 4 time you'll notice that when I'm doing this I'm not doing a whole lot of math in my head there's a few things but not a lot because I don't want to lose any constants that's the big deal not con no losing constants no losing negatives tell me some things I can do now pull one turn it into a 10 okay let's be careful about that that 1/ 18 oh yeah yeah it's going to be a44 oh yeah 2 * 2 is 4 4 16 is 1/4 so 2 this is 4 over 16 1/4 comes out integral of so this is 1/4 all the constants are gone that's what we want how much is see Square - 1 everybody come on how much is SEC Square - 1 obvious identity because that's what we were going for that's why we did our substition in the first place once we get hey look at that something s minus 1 this will become a one something squ minus one that's great we want secant because we know see squ minus 1 is tangent squ a square root of something squared goes away gives us just tangent this is the whole idea folks that's why we're using S and secant and tangent because those identities are so nice for us so we got uh let's see square root of tangent [Music] over SEC 4 * Theta tangent Theta D Theta are you all still okay you sure what's the next thing you do we're still simplifying how much does this become what now canc canel let's do that so this becomes to the third power that's gone so we'll have the 1/4 is still there yep tan squ over see to the thira D what I want to know on a show of hands if you can make it down to that far on your own does the triangle make sense to you is it starting to make a lot of sense can you do that this is the idea here folks this is trigonometric substitution that's all it is after you do this and you simplify it that is your last section now this is kind of cool because we can do all of these problems you won't be facing anything that you haven't at least seen before somewhere I've given you some pretty tricky ones in this class so nothing should be above you okay nothing's Beyond you at this point now just follow the pattern this does have tangents and secants like I told you about but are they being multiplied no okay so it's not sign and cosines it's not secant tangent how we like it it's not cotangent cosecant if that's the case then what are we doing with it we're changing it to signs and cosin so we're going to be changing this to signs and cosines if we can does that make sense if it doesn't fit change it to signs and cosin and see what you get right now there's no way I'm doing that that integral I'm looking have no idea there's no way but if I change it signs and cosiness and simplify it after that maybe we got something so let's try so instead of 1/4 integral of this bunch of crap we're going to have 1/4 this should be I'll do this in two steps this is tangent s time cine 3 D you're all should be okay that 1/ SEC Cub is the same thing as cosine Cub yes no yes keep going I don't know anything that deals with Tangent and cosine to you was that in your trigonometric sub integrals from last section no we didn't have tangent cosine we had SS and cosin but if we do this if we know the tent is sin s over cosine s you okay with that one yeah do you see how changing things in the signs and cosiness can really help you sometimes otherwise you man you get stuck what am I going to do here substitution ain't going to work try changing it signs and cos signs once you do that maybe you get lucky and get something that that you can really work with here this is really nice those are gone all but one of those is gone we have integral sorry forget the 1/4 1/4 integral of sin^2 cosine th now take a look at it what would you do if you were doing this problem I would do you sub for sure just so remember listen this is a perfect example of go through the patterns that I taught you do not immediately jump to one technique or another honestly go through the easiest thing you can do in integrals is it well firstly it fits the table so we check it does this fit an integration table no it doesn't can you do a substitution yes yeah then for heaven's sakes don't start doing uh identities and stuff if you can do a simple substitution does that make sense to you so what I'm trying to get you to do is really think through these don't jump to a certain section because we're in it really think it through we start here we go hey can we do it like it is no can we do a substitution no is it trigonometric no integration by Parts no trig sub yeah do a trig sub simplify it down to whatever your trigonometric functions are after that does it fit a substitution no does it fit any of the trig models that we have no change it size and cosign can you do a substitution notice how I'm always going back to can I do a substitution are you listening we're always doing that always checking for the easiest not the hardest things to do do your substitution go for it aren't you glad we did a substitution didn't use an identity or something that would have been silly right if we use an identity we'd have 1us cosine squ wouldn't we then you distribute you have cosine that's that's don't do that what we want to have is s with one one power of cosine off to the side if we have that already this is the whole idea from the beginning right make a substitution substitute for sign cosine th D Theta is right there so we know this is going to be 1 14 integral of this piece is our DU so this is going to be what was it how much easier can we get just don't forget about the 1/4 just follow us all the way down this would be 13 U to the 3 still okay so far okay now are we done almost done almost simplify okay we're going to simplify sure we're going get 112 do we keep the U what's the U become okay we're going to have a couple things to do here so this becomes 112 s Cub th very good sin Cub Theta now we've already substituted once because we had one substitution here we're done with with that we got it back in terms of theta look back at our integral did we start with thetas x's so now that you have sin Theta can you make that back to X's where are you looking right now triangle use your triangle everyone should be able to do this go ahead and do it right now make sin Theta equal to something and then use that substitution so I want something right here and then I want you to substitute we don't need that big what relationship is sign whatever what do you have an opposite yes do you have a hypotenuse notice this based on that Theta this is exactly what we're talking about here so we hopefully you should have the < TK of x^2 - 4 over X did you all get that one how people did get that one that's great that's almost everybody that's fantastic so one last little step that we have we've got a 112th we've got a sign to the 3 so basically all this stuff is being raised to the third power so we get a 112 we get aunk x^2 - 4 to the 3 and we get an x to the 3 are you guys okay that both these are going to be raised to thir power It's s to the 3 so this is sin Theta to the 3 both numerator denominator ra to thir power if you want to make it just a little bit nicer is really kind of up to you um I might choose to write this as this because that's how we were given do you see that how we were oh sorry that was a previous problem my bad um whatever you want you can leave it like this you can write it like this I really don't care see so I know I'm done I never stop an integral until I either have the number if it's definite or I have a plus c I need to show pans if that made sense to you guys good deal all right fantastic question any questions we're going to start our last problem of the section we're going to get through the first little piece of it and then I'm going to ask you to do most of it on your own think you can do it on your own so we're going to do the first part though cuz the first part is a little tricky and I want to make sure you see this at least once last chance for any questions at all like days now I don't know just for simplification purposes before you erase it uh that that one right there under the triangle can you turn that into like a two square and then got rid of the Ral B let's down one more down one more step that step right there to get ready the radical can you turn that into a two squ this one yeah you can turn that into a two squar but there's no way that's going to simplify that square root no don't make up your own math it's never a good idea unless you're Isaac Newton who invented this are you Isaac Newton I'm notti no I'm Valentine making my own math awesome he's Cupid it's C back to the work okay I think we can agree that looks nastier than some of the other things that we have done so far so first things to do first thing I know it's fun right first things to do uh go through your whole process does that fit the integration table heck no no not even close uh can you do a substitution no do you understand that every time I see an integral and I've been doing this for years that's what I do I'm not just fing you down a rabbit hole this is literally what I do so I'm trying to tell you this is the best way to do it um I don't think I've got an integral wrong in a really long time really long time but this is what I I always stick with that pattern I think doesn't fit the table if not okay go to substitution if not all right integration by parts that's not going to work here I don't have a product all right if that doesn't work uh or I don't have anything to work with an Ln here uh that doesn't work does it have trig in there well no not yet can I fit it to something that has trig in there maybe maybe does it have a square root that's a good indication I might be able to do that so I'm going to write this just for my own mental well-being as a square root to the fifth power don't forget that fifth power now here's the issue this part sucks cuz that part right now does not look like a right triangle does that make sense to you it's not something squared plus or minus something squar and that's what we need for a right triangle for Pythagorean theorem so the idea is if this listen carefully please is the whole whether you're going to get this right or wrong if if this does not fit something squar minus something squared you make it fit something squar minus something squ or plus something squ so we have let's take this off to the side here little asteris 3 - 2x - x^2 take a look a little bit more closely at that one you ever heard of completing the square yes we're going to complete the square we're going to make this thing be a perfect square so how in the world are we going to do that well uh first thing you might want to do is have this thing in order because that's how completing the square works so we'd probably write this as x^2 - 2x + three you with me on that one by the way these are special cases these this is not going to be true for all the ones that look like this at all these are special cases that you can actually complete the square and it'll work so not all of them you can do this universally giving you special ones here does that make sense to you okay so put this one we're going to see it as well next thing you do you make sure that first term is positive that's got to be the case so what I'm going to do is I'm going to factor out this negative here I'm going to have x^2 + 2x I'm going to leave some space here I'm going to leave that + three you'll see y it's complet the square you guys okay with that one so notice this isx2 this is - 2x but the plus three I'm leaving that alone you follow now how you complete the square you have your first term positive you don't worry about this right now you look just in here we take the middle number the middle coefficient which is your b a A plus b sorry ax S Plus BX plus C take half of it Square add it now whatever you do here you have to also undo it otherwise you're changing the value of this thing does that make sense but watch Real Care First are you guys all okay with where the one came from that's kind of a big deal it's complet in the Square so we go hey take half of this number and square it add it over here now you have to undo what we just did I know it says plus one but watch carefully we really see how the negatives can distribute we we really subtracted one to undo that we're going to add one so are we adding one twice no we're not we're adding one here because really when you distribute you subtracted one notice how this going to be a minus one - one plus one gives you zero these black letters should be gone if you were to distribute this so if if you want to check your work if you distribute this and combine like terms do you understand that it's going to give you this back again it's going to be the same thing what we've just done right here is we've completed the square so this becomes x + 1 squar this becomes plus 4 you guys all right with that one completing the square if you do this you automatically have a perfect square uh trinomial it factors negative we kept that up front this is x +1 2 this is plus 4 show hands feel okay with that one write it in the appropriate order we want something squared minus or plus something else squared this doesn't cut it right now for that negative so I'm going to rewrite this as 4 - x + 1 squared now working back to our asteris we got a one we got a square root we got a nasty fifth power we got a DX in there watch careful first are you guys okay with completing the square idea look at what happens to us when we plug this back in we've got 4 - x + 1 squared do we have something squared plus or minus something else squared yes do notice why it's a special case because if this was like a five it'd be kind of difficult to do you could still do it all right could but it'd be a little bit harder um I think You' do it I remember now pretty sure but this is nice for us this will be really good here's the next idea I'll give it to you right now because I want you doing this on your own at home okay this is this is for you to do you need this completed by the time you come back to class CU I'm going to rip through this thing next time in about 5 minutes do a substitution right now get rid of the X+ one so if you did u = x +1 du = DX this just becomes a u so we get 1 4 - u^ 2 under < TK to the 5th power du U is this so 4 - u^ 2 5th power fth power square root square root DX becomes du show hands if you can follow that one could you do could you have done that on your own probably not before this can you do it now of course you can can you please finish that up for me this is not going to be super super difficult I think if you can all do that uh can you make a triangle with this yes okay so of course don't forget to substitute back in for you at the very end we work on this next time all right so let's finish up our problem um as you know what we're trying to do is we're trying to fit part of our integral into a triangle to use a trigonometric substitution on either tangent or s or secant because we have some nice identities with those so we're working on our problem now and right off the bat we understand it's got a square root but it's not something nice what something nice is a number or a variable squared plus or minus another number or variable squared we want something squared plus or minus something squared that's the idea if we don't have that immediately don't give up on it you might still be able to work with it so square root is the first indication that I might be using a trig sub here but substitution wouldn't work of course nothing else will works so we're down to just a trig sub square root says yeah you know what you might want to give this a try with a with trig sub but we got to change it a little bit so whenever you have a case where man it's not obvious what this is going to be take it up to the side see if completing the square is going to work so that's what we did here we organized it we factored out the negative we added half of this number squared notice that when we add it here it's really like subtracting it so I undo that so I know it's both plus one plus one but because of that negative this is really like a minus one so I do it with that plus one we organize it again we have a quantity minus a quantity here we have two squ minus something else squar this is our our first little manipulation are you guys okay with that first step it's just completing the square I know that many of you haven't done that in a long time but that's what it is maybe a review about completing the square on how to do that now at this point we could go straight to our triangle if we wanted but to make things easy to make it nice for me what I'm going to do is just a very basic substitution I want to get rid of that x + one you guys follow that way yeah I don't have to look at X+ one I'm going to look at a u instead so whenever this is the case when you get something squared you don't really want to deal with it right then feel free to do a substitution so off to the side I'll do u = x +1 this is really easy because du is just equal to DX I mean it's a straight sub it's nice if you're getting like uh after your derivative here something in terms of X yeah don't do that just the constant is what we're trying to get out of that does that make sense to you so here we go all right cool well we got an integral then we got one we still got this bracket with the square root we got a four still but here we're just going to do minus u^2 to the 5th and then du straight substitution for DX now this looks a little bit more manageable so right now what do we do yeah we're going to do that triangle it's a square root it has something squar minus something squar that's fantastic so right here I'm going to take my square root I'm not going to think about it as four I'm going to think about it as yeah that gives me the length of a side or a hypotenuse or I'm sorry a leg or a hypotenuse depending on what we're given and which order we're given and I'm going to draw my right triangle from this what I definitely don't want to do is leave that a four and mistakenly put four here or here or here then my integrals will be all screwed up I'm not going to get this thing right so I really want to make sure that we understand that this number is the length of a hypotenuse or the length of a side a leg uh square is what that is so right triangle always drawn in the same orientation that we always have thing right uh by the way what is this thing is this whole thing the length of a leg or the length of hypotenuse yeah how do you know yeah the minus right there this says what this is which one of these pieces is the hypotenuse then the two the two not the four but the two so the two is the hypotenuse the first one is the bigger one otherwise we'd get something negative inside of the square root we can't have that so if the two is the hypotenuse the only way this comes about is if I take the hypotenuse squ minus another leg squar and took a square root of it that says this piece is one of the legs now we just have to be really good about fitting the the the legs in the right place we want a trigonometric substitution that works it's got to be S or tangent or secant so uh which one's going to work for us would tangent be a good choice no heck no I want to deal with the two at someplace right so it's either going to be sign or it's going to be secant however I want the two to be on the denominator I want the constant on the the denominator whatever fraction I'm going to create here whatever relationship so it's going to be sign which means X goes opposite or X goes adjacent make it so this stuff works for you don't make it it is me yes I know for away it's not me it's you anyway and we have theare < TK of 4 - U ^2 that's right I did the substitution so I have U now two feel okay with that one so we go hey hypotenuse uh no problems two this stands for a leg just put it in the right spot so that we get U over a constant for whatever trig sub we're going to do right now what's the appropriate one again yeah s is going to give us U over 2 s Theta so sin Theta = u 2 and then we're going to do what solve for U okay so u = 2 sin th and Next Step what is that sure because we're going to do a couple things here we're going to substitute for our U we're also going to substitute for our DU make hopefully we're all okay with the derivative of 2 sin Theta being 2 cosine Theta no product rule nothing like that now we get to put this back into our integral and make it a lot nicer look well not nicer looking but easier for us ultimately it's going to look nasty at first you oh my gosh there stuff everywhere but it'll all simplify for the most part so we got a one what's going to be inside of our square root right now can you tell me please 4- 2 sin 2 and if you do two sin Theta s that should all be in parentheses and then this whole thing is raised to the fifth power and then finally I do got to remember about my du some of you guys are forgetting some of this I can't have you forget this du has to be substituted for so instead of du instead of du we're going to put two perfect okay let's make sure we got this one stays four stays minus stays you no no no no U is 2 sin Theta we got it we have it squared we have 2 The 5ifth Power du no no no not du 2 cosine Theta D Theta D theta's got to match everything we had we can't have two different variables up there otherwise that doesn't work for us so far so good now what do we do yeah we're going to simplify as much as we can we don't start any sort of substitution right now that'd be crazy right I mean that how we going to do that what we do want to do is well simplify it as much as we can change it into an integral that we know how to do so one over if we do this this is going to happen a lot for you where you have four minus the same number whatever number is and then sin power are you still okay okay awesome what now out we're going to do the four we got to have this as one minus something for for this otherwise you know we got 4 4 - 4 sin s thet isn't an identity 1 - sin thet is so we're going to factor that if we factor out the four just be careful with your algebra don't do all of this stuff in your head give yourself some time to really think about it I don't want you just pulling out a two and have one2 you really need to see that what's going on here is well we're pulling out the four yeah the square root of four is two so we'd factor out the four and take a square root it's two we still have a square root of 1 -^ 2 are we all okay with that one factor the four we can separate square roots by product square of four is two no problem it's outside the square root this is inside the square root but it's still being all raised to the fifth power so don't do something silly right now and cross out this two with this two order operation says you do exponents before you start simplifying things like that well what that means is then we got an integral we got two cosine Theta oh do that all we got two cosine thet what's that two going to become 32 yeah because it's is 2 5th power so we got 32 and then we got 1 - sin 2 Theta yeah 2 the 5th D we're going to do a couple steps right now what's the first thing we're going to do yeah 116th goes up front di identity so this is 116th we got it we got a cosine Theta this is why we did that substitution for sign because we knew it was going to be 1 minus something squared well that something squar has got to be sign for this to work for us so 1 -^ 2 well it gives us the square root and that's all raised to the 5th power uh notice how I'm writing lots of steps I'm not trying to do everything in my head right now I know you guys are I Know It uh but I don't want you doing everything in your head cuz what if you forget a little constant is that going to be a good thing it's a bad thing if you forget a little square or fifth power that's a bad thing I don't want that to happen square root of something squared uh itself awesome so I'm going to have 116 integral of cosine Theta let's do this at once okay this gives you cosine thet 116 1 cosine 4A D Theta now before you start working with this integral the way it is 1/ cosine to the 4th it's probably a good idea to check if it fits one of our integration models first which means that when I'm looking at that I'm not thinking one over cosine cuz I don't know what to do with one over something do I I know what to do if this was cosine of the 4 that'd be great I'd split it up that's cosine 2^ s i' do 12 * 1 + cosine 2 thet squ i' distribute i' do that again but it's not that it's 1 over cosine 4th so let's change that into a trig function that we know how to do it so 116 integral of yeah let's talk about this asant to 4 Theta D Theta and fortunately for us we know what to do when we have seant to the fourth power when secant is even think back to uh to your tra seven to your last section what do you do when see is even break it up to a squ that's right so we're going to take a sec squ out there so 116 2 * s Theta D Theta the reason why we do that is because we're trying to get a tangent uh if we get a tangent by some sort of a Pythagorean identity here then derivative of tangent is secant squ so reason why we break off the secant Square it's CU we're looking for a tangent from the rest that's why we did that with even powers of secant because with even Powers we can strip off the secant squ it's still even we'll be able to get an identity of one plus tangent squ somewhere in there if I'm if I'm right hopefully identity now I'm going to have to erase part of this do you have any questions on this before we we continue so you got all this written down right this was basically just an algebra completing the square and then doing one basic substitution one thing I do want to remember is this okay I'm going to leave this on the board because this just kind of big for us at the very end and I'm I'm sure going to leave this on the board so 116th no problem integral uh looks like we're going to have uh what do I change that secant squ into what's that going to be plus or 1 + T theet whatever you want to write that as it time a see s Theta D Theta look how every what this does it changes it back into some sort of a trig model that we know how to deal with that's what our trig sub does change something we can we know how to do it's it's that's what substitutions always do change it into something we know how to do yeah now it's it's kind of basic after this we know a u sub for Tan Theta we've already dealt with this stuff du = 2 th D Theta that's right here that's exactly what we want that's why this step worked oh what I forget so we can no nothing think I just wondering if we can oh you know what yeah we have a u you might want to change it to a different variable so uh I already used a u so that's a good point forgot about that one let's make this into like a w just so we don't get because we will be res substituting back in for you and I don't want any confusion there so good point thanks for catching that I saw the look on your faces and wait a minute wait Mr Lanford you got yeah I made a mistake I mean I was testing you you passed making sure we were awake that's right I'm making sure you're awake that's the ageold excuse for mistakes from Math teachers so integral uh tell me what's inside my integral right now please what is it not U w^ 2 + 1 D good this whole thing is D DW whole thing is DW integrate yeah let's do it let integration can you all do that yes so when we integrate here we've got okay that looks pretty painless but we have some things that we're going to do we've done two substitutions here so first thing we're going to do we had W's here we started with some thetas originally we started with exit so we're going to have to translate this back twice now first I want to make sure you're all okay with this so far show hands if you are at after this point it's pretty easy I me it's stuff you know how to do so just do it like we've been doing okay so we're going to have 116th we'll have 13 well what's that W become so sure 10 Cub plus how much we still okay you got to put your X's back in okay so we are gonna we are gonna put X's back in but it's kind of weird we've done three substitutions uh we substitute first with a u then we made U's into thetas we made thetas into W's so now we've got our our W's we got back into thetas we're going to use this to translate our thetas back into U's then we use this to translate our use back into xes does that make sense to you all right so as long as you have your breadcrumbs to come back with you'll be fine all right just don't forget this stuff uh we don't want to get too involved in this to forget what we actually started with so with here we got 116 going back up top 116 we got 1/3 we oh let's think about tangent can you figure out what tangent is tangent Theta is that's why we leave our triangle here this is So Tan is over cubed the whole thing plus can you do tan Theta same same thing but not you know I'm going change that to a parentheses just so it looks a little bit better by the way is it the 1/3 that's also getting cubed or not no okay that wasn't inside there it's just the tangent so tangent of theta hey look at that tangent of theta is U over this guy of 4us u s no problem we got it to the Third Power Plus another tangent same idea opposite over ja it's right there are we done no one more thing to do now at this point if you are going to choose to simplify this this is where you would do it okay you try to make one fraction out of this right now instead of plugging in X's first all right so we get all the way down to as simple as we can with 's then we plug in the X's does that make sense to you so if you choose to do that so this this may or may not look like what it is in the back of your book it might look different they might find a common denominator here it wouldn't be super hard to do you'd multiply this by uh three and another uh 4 - ^2 factor and you can find a common denominator get one fraction out of it and then you plug in your X's you follow that's the idea now uh for the sake of time I'm not going to do that I usually do on this problem to show you how insane it gets with your algebra but I'm not going to uh that's going to be on on you what I'm going to do is do one more thing what am I going to do us we still keeping the 16 outside you can distribute it you can keep it outside I'm going to keep it outside just for the sake of seeing where everything's coming from okay um again if you were to simplify this you would do it here before doing what I'm doing right now because otherwise you have a lot of different Factor not factors there a lot of different terms of X plus one you're changing every U into two terms that gets kind of crazy so we'd have an x + 1 we'd have 4 - x + 1 squared and then that's under a square root and this whole thing is raised to the thir power then we'd have another x + 1 we'd have 4 minus x + 1 2 that's too small and that's what we would deal withy blessy let me know I'm that did you get it yeah yeah could you follow it yes do you want to simplify this one and see what you get no sir you sure we can do it if you really want to do you know how to yes no for some of you sadly that's true oh oh good how makes me want to right now go for it go for it yeah of course you say no I know and then I just completely said I'm going to anyway forget you guys let's see if I were to do this uh here's how I would do it I would think of this as U cubed over 3 < TK 4 - u^2 to 3r I would multiply this by three and 4 - u^2 squar I'm both the numerator denominator that would change it to a third if you simplify you can see that I'm right this would cross out all but one of those this would cross out you'd have just the U so this is this is the same after that you do a little bit of simplification we'd get 116th we get U to the 3 you'd get three bunch of junk 2 the 3 this right here square and square root have gone we get + 3 U * 4 - u^ 2 are you still with me you catch me if I've made a mistake sometimes this happens it happened to be once ear today after that 116 we'd have a u Cub + 12 u - 3 U 3r over 3 < TK 4 - u^2 to the 3 after that combine your like terms if you can I got a 116th doesn't really matter how you do this but I'm going to have a 12 uus 2 U 3r over same stuff that's as simple as we can make it right there uh the only not as simple we could factor out a two if we really wanted to that might be a good idea so factor out of two then what we would get is 1/8 6 u - 2 U 3 say what wrote it right wrong what' I say well obviously it's the camera was in the way I didn't see you told us to catch a mistakes you did yeah okay that's as simple as we can make it well not really we can factor out the three too EAS one over 24 let's just do that here okay we'll get rid of this three we'll make that a 24 are we still okay so far I hope is all my algebra right I certainly hope it is let's see not yet do the thank you at this point when you've had it simplified here's what you plug in X+ one so 6 x + 1 because you is x + 1 - x + 1 3 power all over 4 - x + 1 2 sare < TK 2 the 3r we can get rid of the square root on the bottom we can't no we can't we have we still have the square root uh what I'm going to I'm going to leave this I'm I'm going to leave this as three half power I am going to simplify this one because I want to show you that what you're going to get back is exactly what we started with uh 3 - 3 - 2x + x^2 or - x^ 2 it's going to be the same so if I do this this would be 124 we got a 6 x + 6 minus if you're going to distribute that it's X Cub + 3x + sorry 3x2 + 3x + 1 but that's inside parentheses there we got Square < TK of 4 - x^2 + 2x + 1 in parentheses 2 3 moving on distribute your negative and then we'll combine some like terms almost done so let's see here we've got a five we've got plus 3x we've got a - 3x2 and a - x Cub can you double check my combine like terms for me please yes no now it's thank goodness so uh this is a test that so I want this all right no all right I just want you want you to see what potentially you can get out of this thing uh we stopped right here you need to see that this and this are equivalent by algebra they're the same thing so can your answer look a little different but be the same yeah yeah this is simplified don't actually it's not you can actually factor out an X plus one here so if you factored that out then it would be then it would be simplified but that's the idea she feel okay with the idea all right take a deep breath we're do twak sub for like 10 minutes well here's the point you now have a whole bunch of integration techniques right yeah we know substitution we know integration by parts we know how to do trigonometric integrals we know how to do a trig sub all those play of oh we know um we know integrals of inverse trick functions right hyperbolic and all that stuff all those things play a part in the following sections all of them we're going to use all of them that's why we did integration techniques so that when I give you an integral you can do it there's only one more integration I think one more one more integration te technique that I'm going to teach you it's this this next section 7.4 it's much more algebraic than the stuff we've been doing there's not a whole lot of Tri trigonometry in unless you have to use a trig Seer part of it you got me and that will rarely happen because what we're going to deal with in this case are rational functions integrals of rational functions any questions at all before we continue you sure this made perfect sense to you perfect sense perfect sense up to here i' love to take a picture and show