| Start | So, Let's see
We want to talk about the trigonometric substitutions
I want to start to have kind of a way of ...
When I see something, right?
|
| 0:30 | So, I want to say like If...
I want to say, If, If I see
If I see, and then I want to say something like a)
and then a usual formula
If I for example have b^2+a^2x^2 squared
Then, what is that trigonometric change that I have to use?
|
| 1:04 | So, my trigonometric change is going to be
x equals to, I need to get rid of this "a" to be able to factor out the "b", right?
So, I am going to take x=b/a And, then, I need a formula that gives me after I factor this term Im going to get one plus, ok, what happen with the "b"? |
| 1:33 | So, If I have b squared plus a squared x squared and I have, and I need to put more stuff in here, but I want to say what I need here So, If I plug in b over a instead of my x I am going to get b squared plus a squared times b over a and then Im going to put something in here squared |
| 2:03 | Ok, so all this is inside the square root and then, when I square this things, then I am going to have b squared plus a squared b squared over a squared and something here now this a's I am going to cancel out I am going to be able to factor out the "b" and I am going to have b squared times 1 plus I dont know something here squared |
| 2:35 | and then these b's are going to go out
I am going to have b times square root of 1 plus something squared
and I want an identity here that lets me get rid of this miserable square root, right?
So, what is that? I know that 1 + tangent squared is equal to... |
| 3:00 | It equals to? secant squared!
So, when I see b^2 plus a^2 times x^2 my change of coordinates should be "b" over "a" secant of theta right? because then... ohh 1 plus tangent secant, right? erase that! tangent of theta, right? because when I put it over there |
| 3:31 | If I put my tangent over here
tangent square of theta
one plus tangent square of theta equals to... secant squared!
so I can replace all this one plus by secant squared and then the square root of that is just going to be b times secant of theta, right? yes? |
| 4:02 | so, you see what I mean? Why are you looking me like that?
you are supposed to know this! now, this is not the only possibility. This is one possibility. now, there is another possibility b) What if instead of a plus I have a minus What if I have a minus here? |
| 4:32 | now, I am going to have this
"b" squared minus "a" squared times "x" squared
What is my change of coordinates?
So, ... I have to have "b" over "a" "x" has to be equal to "b" over "a" of course, I am not thinking that "a" is zero, right? because then it is not going to make sense if "a" is zero, then what am I doing? just get the square root of "b" square and that is it |
| 5:01 | So, what should be that thing
that I have to put here instead of tangent?
because tangent is not going to work what will work? Exactly! sine or cosine For example, I am going to say sine of theta Why is he saying that? because now all will I have to go is change instead of a plus, a minus |
| 5:30 | instead of a plus I have a minus
now he told me I have to use sine of theta
then this becomes a minus
I have a sine of theta over here
this is a minus
I have a sine squared of theta over here
and what do I know? 1 minus sine squared of theta is equal to??
cosine of theta! |
| 6:01 | so Im gonna have here a beautiful cosine squared of theta
and square root of that is going to be "b cos theta"
which is something that Ive been talking.
correct? yes? so, this is very easy to very easy to deal with now, If I have another case oh, I should not move this all right, so what if I have now |
| 6:32 | uhm Im going to change, now Im going to go like this a squared x squared minus b squared I just flipped the signs I need a different, a different trigonometric identity so, I see that here what am I using here because I know that one plus tangent squared of theta is secant |
| 7:06 | can you repeat?
ok, good, and then I am using here one minus sine squared of theta is cosine and then what do I need here? "b" over "a" and now, what do I need? now I dont have this case any more |
| 7:31 | now what I have is the other way around so I need another trigonometric identity like this that gives me something so what I have now is this I have a squared x squared minus b squared and I said, I wanna do my x equals b over a a squared b over a but I need to put something there and then minus b squared |
| 8:03 | and then, so, what do I need?
secant? Ok, I'll go for secant, let's try secant. If I try secant, then they are trying to tell me that the formula they are using is now secant squared minus one is tangent squared right? |
| 8:31 | come on! I need some more support here ok, so, that is what they told me, let see they told me to put here secant so If go over here, I am going to put here secant so If I cancel out, this is gonna look like this a squared b squared a squared secant squared of theta minus b squared |
| 9:00 | cancel this thing, factor out the "b"
and you get secant squared of theta minus one
which is gonna give tangent squared, right?
ok, so these are all the possible trigonometric substitution |
| 9:30 | any questions?
any questions? no questions? ok, so lets do an (example?) What is this for? Ah! to solve an integral that I am going to give you! right now these are formulas blah blah blah It is that I have a reason. And the reason is |
| 10:02 | if I wanted to solve the following integral
Im going to give you the first this is gonna be a bigger question
so you tell me what is the change of coordinates
and then I finish the integral, ok?
so |
| 10:37 | so, I am going to, we are going to solve right now uhm, problem 11 in section 5.7 so, this is section 5.7 and it says like this it says find the integral of the square root of nine minus x squared over x squared dx |
| 11:07 | ok?
so you dont need the book, you just need to see the problem and then the question is going to be what is the substitution that I have to do and then Im going to give you something like this say a), change of coordinates should be 3 |
| 11:30 | x equals to 3 and then I use what I have over there 3 tangent of theta b) x equals to 3 sine of theta and then c) x equals to equals to 3 secant of theta |
| 12:03 | I am opening the webpage now I am going to open the clickers ok, so, choose your change of coordinates |
| 12:32 | I dont want you to solve the whole integral
are we done?
are we done? what is the problem? I have the formulas here, what is the problem? |
| 13:04 | are we done?
yes! ok, are we done? I'm gonna close it, OK? If I don't close this I get in trouble. Ok, so what was the change? a), b) or c)? b) of course!, |
| 13:32 | b) and it is very clear because you have a beautiful minus
and I told you whenever you see a beautiful minus there
you have to use this change, yes?
Your b=3, your a=1... What else do you need? Nothing! Ok, so that is your change, so lets do it! so, if "x" is equal If "x" is equal to three sine of theta |
| 14:02 | then I just to plug everything in there
that has to substitute all these values in there to solve this integral
so this is gonna give me
nine minus nine times sine squared of theta
over nine sine squared of theta
and now I need to figure out what is "dx"
this is three times cosine of theta d-theta, all right?
|
| 14:34 | Ok, so now
what do I know now?
ok, so what I know is that I can factor out the nine as I promised you that's why I put, that's why I made that change of coordinates because I want to be able to factor out the nine |
| 15:02 | so, this is gonna be square root of nine one minus sine squared of theta over nine sine squared of theta and over here I have to put three times cosine of theta d-theta this nine is gonna go out like this three times three is nine divided by nine is one |
| 15:30 | so, Im going to get
cosine of theta, Im gonna simplify like this and then we know what to do
Is this, Is this clear?
or, did I go too fast? too fast? ok, so is it clear why I got rid of the nines? |
| 16:05 | three times, one minus sine squared of theta is cosine squared of theta and I have a nine here and I have 3 times and cosine of theta here so, three times three is nine divided by nine gives me one so, this is cosine of theta times cosine of theta over sine squared of theta |
| 16:33 | d-theta
yes?
so, this became cosine squared of theta over sine squared of theta d-theta ok, now I need help how do I solve that problem? we have a problem I have to solve the integral |
| 17:01 | cosine square of theta
all of my integral that look like this now look like this
how do I solve this problem?
cotangent squared of theta and then? |
| 17:31 | and then!?
the square root here cosine squared of theta because you have square root of cosine squared of theta so when you take the square root when you have the square, they cancel each other so the guy inside of here is cosine of theta yes? because I have to multiply by dx |
| 18:03 | and dx happens to be three cosine of theta d-theta
that's why I write like this
ok, my friends?
what is the deal with cotangent? it equals to so, this is a funny integral, what is cotangent squared equals to? |
| 18:31 | negative cosecant?
you are not forgetting the one minus something? yes, he is there are two ways of getting to this formula this is equals to one minus cosecant squared of theta d-theta right? there are two ways of getting there one is remembering this one is just looking at this the other one is just taking these guys |
| 19:01 | so this is the same
by, you know, one minus sine squared of theta
and then divide it by sine squared of theta
and then, as you see, you get one over sine squared is cosecant, right?
minus one so, this is cosecant squared minus one the other way around according to my formula this is cosecant squared minus one |
| 19:30 | correct?
this formula says one over sine squared that's cosecant squared minus one so that's the correct formula ok, and now we are lucky because we have an antiderivative we have a formula for this cosecant squared and this is easy to do because this is just d-theta so this is just theta and, what is integral of cosecant squared? |
| 20:05 | cotangent?
negative cotangent is equal to cosecant squared? so, that's it, that's the answer ah, ok, ok, Im not done because... I still have to solve for x, right? my problem started with x, correct? my problem started with x and all that I have here is theta |
| 20:32 | so I have to go back to x
how do I go back to x?
I have to solve for it here so this is gonna tell me that sine of theta is equal to x over three and now, how do I get rid of sine? yes, arcsine so, theta is gonna be equals to arcsine of x over three so what I can do here is write |
| 21:02 | this is minus cotangent of arcsine of x over three minus arcsine of x over three plus a constant that's my answer so this integral that looks horrible is much easier If I change it to that I mean, "easier" |
| 21:30 | Im not saying it is easy easy
but it's easier
ok, questions?
ok, he wants me to do partial fractions |
| 22:07 | you want another one like this?
another one like this and then partial fractions then at least one partial fraction he said, he said at least one ok let's see |
| 23:05 | ok, so let's do another with a different sign so that the change of coordinates is different ok, so let's do problem number 16) because 16 is the definite integral from zero to two square root three of x cube over square, uhm, no, I don't like that one |
| 23:37 | I dont like it because it has minus and I don't want to do minus I want to do the other I want to do 18) so, 18) is the integral of x cube over square root of x squared plus one dx ok this is easier, I hope |
| 24:00 | so, what is that change of coordinates?
what is it? how do you want to integrate that? maybe you don't wanna do what I want how do you want to do it? see, I have mixed feelings right now, I have mixed feelings |
| 24:31 | do I tell you my mixed feelings?
I kind of want to do it like by parts, and then and then Im going to need a substitution trigonometric substitution but, I don't know if I can go straight to trigonometric substitution so I have these two problems I kind of want to do it by parts but I also want to do it by using a trigonometric substitution |
| 25:00 | I don't know
all right, what is my change?
x equals tangent? beautiful, people, I see a plus I said in all this I don't have a problem because this seems one this is one, so this seems it equals to tangent and let's hope for the best all right, so what is the derivative of tangent? |
| 25:33 | secant squared
all right so now I go here
and then I want to change
my x is gonna become tangent
cube
and Im not going to worry about that now
now, over here I just going to put integral or square root of secant square
of theta plus one
and my dx
uhm, what am I doing?
|
| 26:02 | tangent square
of theta plus one
and over here I have
secant square d-theta
correct?
Check on me because then I get a mistake and I forget what I am doing. is it right? Is it right? all right, so |
| 26:30 | what is tangent squared plus one?
so how does this reduce? it looks like tangent cube theta and what is in the denominator? ok, secant square of the theta and then? |
| 27:03 | ok, then I have the following
I leave my tangent there, I don't care
and then Im gonna take the square root of this secant
which is gonna become just secant
yes?
ok, so far this integral looks like tangent cube of theta secant of theta d-theta |
| 27:33 | yes?
any help? what is next? ok, first of all, can I use, what is the derivative of tangent? secant square, so this is not gonna help me |
| 28:00 | correct?
it's not gonna help because to be able to do substitution for here I will need to have a square here, and I don't have it, right? so I cannot do that so I cannot just integrate this tangent what can I do there? |
| 28:35 | tangent square of theta
and then times tangent
secant of theta
and then?
ok, secant square so, you are telling me I write secant square, uhmm tangent square equals secant square minus one |
| 29:06 | now I understand so what he says is that I need to use my formula tangent square equals secant square minus one so, let me put it there so he says use your formula secant square theta tangent square equals to secant square of theta minus one and then I guess he wants to put my secant here and my tangent here |
| 29:43 | secant cube theta
yes?
this? call this your "u"? |
| 30:00 | ok, I don't understand call u equals to secant of theta ok, then du is equals to secant theta tangent theta d-theta ok, and then you think it is easy because now what you are gonna do over here its gonna be u square minus one times du |
| 30:31 | and that's a trivial integral
but, don't go so fast because you have to go backwards
right?
do you understand the problem? now we have to integrate that and then you have to go backwards because there are several changes here so, one of the changes is that Im going to integrate and this is gonna give me u cube over three minus u |
| 31:01 | plus a constant
but, I'm not done, I'm not done
because u is secant of theta
right?
agree? so this guy has to be replaced by secant of theta so, go back here and say this is equal to secant cube of theta over three minus secant of theta plus C am I done now? no! because what is theta? |
| 31:33 | from this one, what is theta?
exactly, theta is equal to arctan of x so now, I have to replace my theta over here so I have to say this is one third secant cube of arctan of x minus secant |
| 32:01 | of arctan of x
plus
great
yes?
here? to the right? which? this? |
| 32:30 | because this formula we know
we know that tangent
one plus tangent square is secant square
that we know
we have it over here
one plus tangent square is secant square
ok
so, this is clear?
so we are done with this well, I am done, you have to do more problems but, this is trigonometric substitution and you got an idea |
| 33:00 | of what is that you have to do
when you have this kind of problem
yes?
this that I wrote over here is very helpful because it tells you what to do and then just hope for the best .... all right, so now I want to do partial fractions |
| 33:35 | so this is another business I want to write here |
| 34:05 | all right, so partial fractions well, this class, this is divided in two kinds but Im gonna talk about the first kind today so, we are gonna try to solve integrals of this type |
| 34:31 | I want to solve an integral
that says for example
I don't know, let's say 2x plus five
over x minus 3
times x plus seven
dx
ok?
so you know that is partial fractions because it looks like it it looks like a product this is easiest |
| 35:00 | it looks like a product of linear factors
in the denominator and it has something on top
a polynomial whose degree has to be smaller than the degree of this
so what you have in the denominator is something of degree two
and here you have something of degree one
if you have the same degree, then you'd have to divide
and then you'd have to long division
and take it to this form, right?
(...) So what do we do? |
| 35:30 | we are going to undo this, kind of undo it so what we do is we write, let's forget about this integral and let's go to the inside so this is 2x plus five over x minus three x plus seven and what I want to do is I want to divide it in two integrals so I want to say this is equals to A over A, I dont know that A is but Im gonna figure it out |
| 36:01 | A over x minus three plus B over the other factor that is x plus seven these have to irreducible meaning you cannot reduce these polynomials any more so these are irreducible polynomials it cannot be factor out any more that is the most it can be factored out and now what I want do here |
| 36:32 | is I want to take common denominator and If I want to take common denominator as you know, that is what I'm saying, Im undoing it and doing it again so, common denominator is the product of these two so this is gonna be again x minus three times x plus seven and then I check this common denominator with the first part and I say what is missing oh, this guy is missing |
| 37:01 | so this should be A times x plus seven
plus, If I compare this with this
what is missing?
x minus three so this should be plus B times x minus three and then Im going to foil the top and see what I get any questions? clear? |
| 37:33 | let me know if Im confusing you all right, so, let's take Id like to write over here what I had so I have two x plus five over x minus three times x plus seven and on this side I have the common denominator so Im going to foil it and its going to look like this going to look like A x plus seven A |
| 38:01 | plus B x minus three B
over the denominator
yes?
now what I am going to do it is Im going to put together everything that has x so this looks like x times A plus B and then everything that doesn't have x goes together plus seven A minus three B |
| 38:32 | and then I am dividing over this product
yes?
yes? I don't know why but student finds difficult, but I don't understand why you just have to understand that this side is equal to this side correct? If two things are equal, then whatever has x, the coefficients |
| 39:00 | so I can only compare as they told you in elementary school or kindergarten
apples with apples, pearls with pearls, oranges with oranges
so, if this two are equal
that means that two have to be equal to A plus B
so this is like that
A plus B has to be equal to two
yes?
and then for this guy over here seven A minus three B |
| 39:33 | has to be equal to what?
to five and then this is like the kind of linear systems that you are used to from the times when you were even younger, right? and then, how do you wanna solve it? That depends on your taste. So, what do you want me to do? |
| 40:01 | I mean, do you want me to multiply by three
cancel these factors and solve for A
or you want me to do substitution
here solve for B and plug in here
what do you want?
solve for B? all right. ok, so from this one Im going to say that this equation over here, B is equal to two minus A. yes? and then I am going to replace this equation |
| 40:32 | in the one below
and this is gonna say seven A
minus three times
instead of B I write this, two minus A
so this is equal to five
so, ok, this is just algebra
so what do I get?
Im going to get I get seven A minus six plus three A |
| 41:04 | is that right?
equal to five yes? all right, so add whatever have that and I get ten A equal to eleven meaning A is equal to eleven over ten and then I solve for B so B is going to be equal to two minus |
| 41:33 | eleven over ten
so, what is this?
twenty minus eleven nine over ten yes? ok, so now everything is easy my integral became this so my integral that look like two x plus five |
| 42:00 | over x
x minus three x plus seven
becomes two integrals
become the integral of
I just have to remember what was A
so A is over x minus three
so eleven tenths over x minus three dx
plus B, which is what?
nine over ten |
| 42:31 | over x plus seven dx
and this are very nice integrals
because this is equal to what?
the first integral is equal to what? ln of x minus three plus nine tenths of ln of x plus seven plus C |
| 43:00 | nice and easy
right?
questions? questions? ok I am supposed to leave at 10:30 so, I have 10 minutes |