how Weierstrass would integrate csc(x)

[@youtubeviolatedme7123]

The only sneaky substitution I know is replacing a mic with a pokéball

[@rzno3414]

You win the comment section

[@aravinds3846]

Lol

[@BlokenArrow]

Or replacing a politician with a paramecium. Same level of intelligence.

[@heydilanschuastedopinho6682]

Minor mistake: t=tan(x/2), not arctan, so the -1 above the tan on the final answer is a mistake. Apart from it, perfect video as aways.

[@blackpenredpen]

Integral of 1/(2+cos(x)), kzbin.info/www/bejne/rJaYY2qojs6Mmrc My "FAST" channel: kzbin.info/www/bejne/iKPLk4pvoLibe7c

[@drenzine]

"is t=tan(x/2) obvious to you?" No what da hec is that

[@drenzine]

Wow, i got hearted, thanks so much! Math is amazing just pointing that out

[@noahtaul]

Totes unrelated, but in the class I TAd for, the students misheard the teacher call this “tricky substitution” so they called it “kinky substitution” and I didn’t want to correct them

[@swift3564]

Weierstrass sub is my favorite!

[@SidneySilvaCarnavaleney]

Prezados nobres amigos(as), professores,(as),alunos(as), com meu respeito a todos aqui presente, estou enviando minha "Tese" para a apreciação de todos; que pi é Racional e Irreversível, (3,15), nesta minha "Tese" tem um fator muito importante a ser respeitado, Não pode ser simplificado, Não pode ser arredondado, não pode ser aproximado, não pode ser fatorado, tem que ser exato para os cálculos do Universo da Matemática, o autor Sr Sidney Silva. www.portaldoslivreiros.com.br/livro.asp?codigo=4410831&titulo=A+Ousadia+do+Pi++Ser+Racional : www.estantevirtual.com.br/mod_perl/info.cgi?livro=2655536635 aeditora.com.br/produto/a-ousadia-do-%cf%80-ser-racional/ www.amazon.com.br/dp/655861281X?ref=myi_title_dp aeditora.com.br/produto/a-ousadia-do-%cf%80-ser-racional/ Onde encontrar minha obra, segue os links acima em epígrafe, e adquirir uma obra onde sua leitura é bem simples e objetiva provando a Racionalidade de Pi, para saber se é Racional ou não, compre minha obra, gratidão sempre o autor Sr Sidney Silva.

[@hassanniaz7583]

Same here. I learned it recently and love this technique

[@GaryTugan]

At the end... t = tan (x/2), right? Not inv tan 😃👍

[@blackpenredpen]

Yes, it was my mistake and my edit didn’t show up.

[@GaryTugan]

@@blackpenredpen still an awesome vid :) Cool beard too! (Haven’t seen your vids in awhile even tho I have full notifications on)

[@j10001]

Can you pin Gary’s comment for future viewers to see first?

[@andreumora2878]

The result shouldn’t be: ln/tg(x/2)/ instead of inverse tg?

[@blackpenredpen]

Oh man, I edited that part but my little annotation didn’t come up to block the -1. Thanks for pointing out.

[@andreumora2878]

@@blackpenredpen i thought that,anyway, really good content as always

[@orenfivel6247]

@@blackpenredpen The sneakiest -1 ever LOL GOOD VIDEO

[@dealva2035]

@@orenfivel6247 haha.. when i watch it, i got stunned for sec bcz -1 . like, where it come from

[@natasha_georgieva]

Weierstrass Sub was in fact invented by Euler, almost 100 years earlier. As usual, "no scientific discovery is named after its original discoverer". Stigler's law of eponymy (applicable to itself). Spivak is all the more superfluous here ))

[@jrtrct9097]

Having not done any research at all, I suspect that this is yet another case of "Euler discovered so many things that we have to name it after the next person to discover it."

[@DavesMathVideos]

Euler already had a substitution technique named after him. Namely, one which simplifies integrals of certain quadratics.

[@angelmendez-rivera351]

@@jrtrct9097 Haha, yes

[@quercus_opuntia]

I'm only in Calc 1 and even though most parts of your videos are confusing as of now I still love watching

[@aniruddhaghosh4793]

Studied this in the calculus book by I A Maron during 12th grade, it's referred to as the "universal substitution" in there.

[@jkid1134]

I just saw this same (x^2+y^2)/xy idea in some completely unrelated algebra video, neat to see again.

[@robsbackyardastrophotograp8885]

This must have been the video you mentioned live last night! So gonna use this method from now on- thanks!

[@monoastro]

inb4 thousands of comments ask about the inverse tangent in the last step

[@hesp9942]

This is the first time I've seen something like this; I'll probably end up using that substitution way too often just because it's possible

[@hassanniaz7583]

In Pakistan, along with sin x, cos x, dx we also use tan x= 2t/(1-(t^2)) quick proof: tan x = tan (2*x/2) = 2 tan (x/2)/(1-(tan x/2)^2) by double angle property ☝ = 2t/(1-(t^2)) since t=tan x/2 Edit: it was a slight error of signs

[@damianbla4469]

You could also do this that way: tan(x) = sin(x) / cos(x) = = [2t / (1 + t^2)] / [(1 - t^2) / (1 + t^2)] = 2t / (1 - t^2)

[@hassanniaz7583]

@@damianbla4469 Yeah it's a good approach

[@jackychanmaths]

You can use the substitution u=cos 2x to integrate csc 2x without anything related to Weierstrass substitution Also, I prefer using identities instead of drawing right-angled triangles to find other trigonometric function values to ensure that the signs of the values are correct and to make the solution very much more formal

[@stevemonkey6666]

I like the passion you display here 👍

[@tahabukhari3189]

I think that we use tan(x/2) because: Firstly, it is the only trigonometric function, other than cotx, with an infinite range (from minus infinity to positive infinity). Secondly, tangent is periodic with a period of pi. So the derivation of the standard Weirstrass formulae, that you derived in an older video, using a right angle triangle, "x/2" can range from 0 to pi and hence tan(x/2) with cover all the possible values for the substitution 't' and hence doesn't restrict the domain of the integral. This is what I think but if I have made a mistake then please do correct me. Thank you.

[@Awkwkwks]

That sneaky inverse tangent at the end

[@AstroB7]

I love the chen lu ! It’s my life :)

[@mathswithmartin6708]

Literally just taught this topic to my class; I’ll be sending them the link 👍

[@shiina_mahiru_9067]

I think of it as sort of an inverse of the trig sub. In trig sub, you sub a variable with a trig function. For Weierstrass sub, you sub a trig function with a "polynomial" (or rational function, to be precise). If you are a good observer, you might see that under the Weierstrass sub, the formula of sin and cos are precisely the "standard" rational parametrization of the unit circle, and I guess Weierstrass saw the connection between it and the double angle formula., somehow

[@ralfbodemann1542]

This is one of the best presumptions how Weierstrass may have come to the idea of his really weird substitution I've ever heard or read!

[@rezamiau]

I think this channel is way better than the other, by the way this method was great. And if you notice, you should probably see the inverse sign, in your final answer, which is wrong.

[@drv255]

Just use cos(2f) = (1 -tan²(f))/ (1+ tan²(f)) then simplify and use 1 + tan²(f) = sec²(f) in numerator.

[@andreacosta2238]

So you did make it grow in the end, good job.

[@ripudamansingh2]

Bruh, I thought this was a standard substitution taught everywhere. I didn't know it had a special name.

[@GaussianEntity]

It has the same energy as sin(x) = x for small x

[@jamescollier3]

Philly steak and cheese is my favourite sub

[@aradhya_purohit]

Priceless

[@Maou3]

I think the intuition for this substitution comes from a combination of tricky trig identities. 1/(2+cosx) = sec^2(x/2)/(tan^2(x/2)+3) Then, the substitution of tan(x/2) is obvious.

[@MichaelRothwell1]

Yes, if you substitute cosx=1-2sin^2(x/2) to get denominator 3-2sin^2(x/2) then divide top & bottom by cos^2(x/2) to get denominator 3sec^2(x/2)-2tan^2(x/2) and finally use sec^2(x/2)=1+tan^2(x/2).

[@MichaelRothwell1]

Here is an alternative method, which easily generalises to any 1/(a+bcosx): in the denominator, substitute 2=2(cos²(x/2)+sin²(x/2)), cos(x)=cos²(x/2)-sin²(x/2), so 2+cosx=3cos²(x/2)+sin²(x/2), then divide top and bottom by cos²(x/2). I prefer this method because we immediately replace each term in the denominator by a homogenous quadratic expression in cos(x/2) and sin(x/2), so when we divide top and bottom we are guaranteed to get and expression in tan(x/2) in the denominator, and of course sec²(x/2) in the numerator, which is essentially the derivative of tan(x/2).

[@usernameisamyth]

Mom : Coffee? Me : 9:56 ?

[@earnstein7607]

Thanks. Brilliant has been so useful ♥️

[@ApplePotato]

This is just reverse of trig substitution. But I think the trickiest part is to realize you need to use t = tan(x/2) instead of t = tan(x). This is because t = tan(x/2) eliminates the radicals from sin(x) cos(x) and makes t continuous on interval (-Pi, Pi).

[@NadiehFan]

Sure. But when your integrand is a rational function of *even* powers of sine and cosine it is often easier to use t = tan(x).

[@tacheboy]

12:55 should be tan(x/2) I think?? pls tell

[@pixoncillo1]

Ah, Calculus by Spivak, what a beautiful read.

[@alxjones]

If you haven't, read his sequel, Calculus on Manifolds! One of my favorite texts of all time.

[@mahmoudaboualfa5136]

We took it last year. But we called it Bioche General (BG). If the first three rules (B1, B2, B3) didn't work, then we take t=tan(x/2) and do the calculations.

[@MichaelRothwell1]

What are the first three?

[@mahmoudaboualfa5136]

@@MichaelRothwell1 For integral f(x) dx (for trigonometric integrals) B1: substitution of x by pi-x (II quadrant) d(pi-x)=-dx If f(pi-x) d(pi-x)=f(x) dx (Doesn't change sign) Then t=cos(x) B2: substitution of x by pi+x (III quadrant) d(pi+x)=dx If f(pi+x) d(pi+x)=f(x) dx (Doesn't change sign) Then t=sinx B3: substitution of x by -x (IV quadrant) d(-x)=-dx If f(-x) d(-x)=f(x) dx (Doesn't change sign) Then t=tanx If else fails, then BG: t=tan(x/2) You have to know all of the trigonometric relations in all quadrants and their signs (positive or negative). If one of them works, then you can substitute t by its corresponding relation, then solve the integral and reintroduce the relation that has x to get the final result. Hope this helped.

[@holyshit922]

This substitution is closely related with Euler's substitution Assume that x is in first quadrant From Pythagorean identity we have sin^2x=1-cos^2x On the right hand side we have difference of two squares which will be useful From the Euler substitution (with the roots) we have sqrt(1-cos^2(x))=(1+cos(x))t With assumption that x is in first quadrant (for sign) and with Pythagorean identity we have sin(x)=(1+cos(x))t

[@holyshit922]

We know that tan(a) is the slope of the line and if we find slope of angle bisector we will express tan(a/2) with sine and cosine How we find angle bisector ? There are few ways f.e. we can follow the construction Lets remind steps of construction Assume that vertex of the angle is A 1. On the one of the rays we choose point D 2. We draw the arc from A with radius AD until it cross the other ray of he angle and name that point E Now we have isosceles triangle ADE If we draw perpendicular bisector of DE it will also pass through the vertex A because ADE is isosceles This perpendicular bisector will divide isosceles triangle ADE to two congruent right triangles ADM , AEM Assume that we have given equations of the lines which contain rays of an angle sin(a)x - cos(a)y=0 y = 0 Assume that vertex of the angle is A = (0,0) Lets choose point D = (1,sin(a)/cos(a)) Radius |AD| = sqrt((1-0)^2+(sin(a)/cos(a))^2) |AD|=sqrt(1+sin^2(a)/cos^2(a)) |AD|=sqrt((cos^2(a)+sin^2(a))/cos^2(a)) |AD|=sqrt(1/cos^2(a)) |AD|=1/cos(a) x^2+y^2=1/cos^2(a) y=0 x^2=1/cos^2(a) Now we choose point E = (1/cos(a),0) Equation for line perpendicular to the line AB and passing through point C is y - yC = - (xB - xA)/(yB - yA)(x - xC) y = - (1/cos(a) - 1)/(0 - sin(a)/cos(a)) x y = - (1/cos(a) - 1)/(- sin(a)/cos(a)) x y = (1/cos(a) - 1)/( sin(a)/cos(a)) x y = (1/cos(a) - cos(a)/cos(a))/( sin(a)/cos(a)) x y = ((1 - cos(a))/cos(a))/( sin(a)/cos(a)) x y = ((1 - cos(a))/sin(a))x tan(a/2) = (1 - cos(a))/sin(a) so if we want to write this substitution with original cosines and sines we will have t = (1 - cos(x))/sin(x)

[@kaurapriyanshu8370]

Sir love from India 🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳❤️💕❤️💕💗💐💐💐

[@DavesMathVideos]

Ha, I just made a video about this a few days ago! The way I see it is that if you have a function of f(sin(x), cos(x)) we can thing of this as some function that is parameterized along the unit circle. By switching variables, you are essentially transforming the function into something more managable! kzbin.info/www/bejne/nZWxlZuveNeZhM0

[@JerrysPlace]

I saw your video. Explanation isn't bad but your sound quality isn't so good.

[@DavesMathVideos]

@@JerrysPlace Yeah, I should really get a webcam and a mic. I was in the midst of setting up a place in my apartment where I could record live, but certain things.... happened. Besides that, my day job is actually teaching math and doing IT at a high school, so that keeps me pretty busy as well.

[@carterwoodson8818]

It is such a sneaky substitution very true!!

[@theimmux3034]

The return of the King

[@kormosmate2]

There is also a hyperbolic version of this.

[@DarkOceanShark]

Would you mind sharing that version with me, please?

[@kormosmate2]

@@DarkOceanShark There's a seperate section for it on the wikipedia page of the topic. Although a little short, but basically it's almost the same, only the signs are different as with most hyperbolic identities. The derivation process is also almost the same. Just use the hyp.version of the identities shown in the video.

[@DarkOceanShark]

@@kormosmate2 Thank you Mr. Kormos, that was helpful.

[@leepeel6529]

Man, the formulae on your tshirt are upside down. Awesome. Where can I buy such a thing?

[@ahmadelsonbaty2922]

In fact this strange substitution comes very naturally in studying the rational points on the unit circle.

[@QwertyUiop-gv1dv]

This was a 12th grade example of our GHSEB textbook in India with same substitution 😀😀

[@victorserras]

This result feels like you hacked mathematics

[@peterchan6082]

Got you again at 12:53 . . . while saying log[tan(x/2)] you wrote log[arctan(x/2)]

[@blackpenredpen]

Yea. I actually edited it that’s why you saw a small jump cut there but my edit didn’t come up to block the -1 😆

[@thomasblackwell9507]

When do you know to use this substitution? For example it is rather obvious when to use a Trig. Sub. However, what are the indicators to use the Weierstrass substitution?

[@ralfbodemann1542]

You can use the t-substitution whenever there is a linear combination of sine, cosine and a constant in the denominator, but no higher degrees of sine or cosine. When you want to integrate functions like 1/(5 + 3*cosx) or 1/(3-2*sinx), Weierstrass substitution will be a good choice. (Although there are other solutions paths. But once you understood and memorized the Weierstrass substitution formulae, this method is faster and more elegant.)

[@evceteri]

I didnt notice the pokeball until the end of the video

[@nimmira]

hmm why arctan at the end?

[@blackpenredpen]

Sorry, it was a mistake.

[@nimmira]

@@blackpenredpen :D phew! for a moment I thought I need to re-study all my calculus!

[@fordtimelord8673]

I think it would be very intuitive and instructive if you geometrically derived sin x and cos x in terms of T=tan x/2 on the unit circle.

[@NadiehFan]

This is done in the Wikipedia article on the Weierstrass substitution. The straight line through (−1, 0) with slope t = tan ½φ, −½π < ½φ < ½π also intersects the unit circle in the point ((1−t²)/(1+t²), 2t/(1+t²)) as we can check by solving y = t(x + 1), x² + y² = 1 for x and y expressed in t. But it also clear from the inscribed angle theorem for a circle that the coordinates of the second point of intersection are (cos φ , sin φ) so we have cos φ = (1−t²)/(1+t²), sin φ = 2t/(1+t²). Furthermore we can see in the diagram that tan ½φ = (sin φ)/(1 + cos φ) = (1 − cos φ)/(sin φ) from which we can also express cos φ and sin φ in tan ½φ. In fact this is even easier since we only have to solve a linear system. To see this, put tan ½φ = t, sin φ = s, cos φ = c, then we have ts + c = 1, s − tc = t which is a linear system in s and c from which we get s = 2t/(1+t²), c = (1−t²)/(1+t²).

[@rashmithmr1117]

His chart is very helpful 😂😂

[@shreyasparameshwaran5421]

Do u know where we can get it???

[@shreyasparameshwaran5421]

Or is it available only in the vid

[@bu4771]

Thank you

[@manosxa]

would you be able to solve exactly the d equation y ' ' = A*sqrt(1+y'(0)^2) for x Ε (0, L)

[@angelmendez-rivera351]

The notation you are using to denote this equation is a little too confusing. Did you mean to write y''(x) = A·sqrt[1 + y'(x)^2], or did you actually mean to write y''(x) = A·sqrt[1 + y'(0)^2]? Because the answer to the question varies drastically between the two. Also, what does "x E (0, L)" mean? Does this mean "x is in the interval (0, L)"? If so, what is L, and what relevance does it have to the problem? These questions need to be answered before one can even attempt to solve the equation.

[@ecavero1]

The answer to the last integral is tangent; not inverse tangent!

[@dovidglass5445]

Shouldn't the thumbnail read t=tan(x/2)? At the moment it says t=sin(x/2).

[@blackpenredpen]

U r right. Thanks!

[@dovidglass5445]

@@blackpenredpen You're welcome! :)

[@jamesdiamond9907]

Can you please make a video on rationalizing the denominator of (1/pi)? Cause if we multiply denominator and numinator by "pi", denominator is still irrational. Love from india❤

[@blackpenredpen]

(pi)^-1

[@angelmendez-rivera351]

1/π cannot be rationalized, because π is a transcendental constant.

[@Dionisi0]

Who is Chen Lu?

[@mohitsingh4744]

Hi, I have a question about Euler's method (for approximation). In our calculus class, we went over how we can divide it into steps, but these were all a finite number of steps. Is it possible to take an infinite number of steps by setting the number of steps = b and then setting lim b->infinity? Sorry if my vocab is bad I'm still a beginner in calculus but if you want elaboration please let me know!

[@MichaelRothwell1]

Yes, you can make the steps smaller and smaller and see where your answer is heading (if that is what you meant) . Calculus is all about handling the infinitely small by taking limits.

[@mohitsingh4744]

@@MichaelRothwell1 Thank you for your response! Do you know how I could actually apply this to a problem because I can't seem to find any examples online? I know that it is possible to just take the integral of the differential equation but I thought this method might be an alternative. Thank you again!

[@MichaelRothwell1]

@@mohitsingh4744 The method you suggest could work in theory, but I have never seen this method proposed or used. However, I believe you could use it to solve dy/dx=y.

[@ZyroZoro]

This is so cool :O

[@frozenmoon998]

I bet that guy was so sneaky, he substituted some single variable 5th grade equations to be equal to hard abstract algebra problems. In any event, bprp can make complicated people/problems look easy and enjoyable ^^

[@shivansh668]

I heard about this first time 😐

[@kabsantoor3251]

Why is absolute value taken for log

[@MichaelRothwell1]

In case the argument is negative.

[@ψωςεψωςε]

What is the difference between weierstrass sub and bioche sub

[@defrom7147]

Weiersstraas sub is one of the Bioche rules

[@Iamnotyou29]

Why that was tan-1???

[@akotosiLord]

This is not connected in the video but, can you pls show the solution in find a,b, and c. a+b+c=12 abc=48

[@novidsonmychanneljustcomme5753]

If you want to get explicit numbers as a solution (with no variables left), you need a third independent equation expressing a relation between a, b and c. Otherwise it's not solvable.

[@akotosiLord]

@@novidsonmychanneljustcomme5753 Hello, thank you for noticing. The third equation is ab+bc+ac=44 Can I solve it without transforming into a polynomial?

[@novidsonmychanneljustcomme5753]

@@akotosiLord Well OK, here we go: (x+a)*(x+b)*(x+c)=x^3+(a+b+c)*x^2+(ab+bc+ac)*x+abc -> The coefficients of this general 3rd degree polynomial do exactly have the form of your equations. So in other words, you can get a, b and c by finding the zeroes of the polynomial. In this case the polynomial would be x^2+12*x^2+44*x+48. By guessing you find that x1=2 and so you can do the polynomial division with (x+2). After that you get a quadratic polynomial which gives you the other zeroes x2=4 and x3=6. And these three zeroes are also the solutions for a, b and c. (The order doesn't matter because you only got addition and multiplication in your equations where you can change the order of the numbers within addition or multiplication as you like.) So in order to give an explicit answer to your question: The solutions to your system of equations are a=2, b=4 and c=6. (But it also could be like a=4, b=6 and c=2 and so on...)

[@akotosiLord]

@@novidsonmychanneljustcomme5753 Hello, thank you for the solution but that is my original given. x³+12x²+44x+48 I am trying to solve for roots without using the usual way. I am finding a clue so that I can make a cubic formula for solving roots. P.S. I already found one but it is so long. I am trying to make a short and easy one

[@novidsonmychanneljustcomme5753]

@@akotosiLord Well OK, good luck then. ;-) The "long" formula you mentioned must be Cardano's method and besides this the polynomial division is the only other general way I know to solve problems like that. Not meant to offend you, but tbh I doubt that you will be able to find a shorter formula than Cardano's one. (When I used to read about all this a little more in detail I accepted for myself that this should be the simplest possible way.) There is still an approximation way called Newton's method which is applicable not only for this kind of functions, but the results will always stay approximations, no matter how often you iterate. And btw it is still possible to solve 4th degree polynomials in general (even though the formula gets even much more complicated than the one for 3rd degree), but from 5th degree on there is no general solution possible anymore.

[@markpage2779]

Brilliant :)

[@laurentiusmichaelgeorge1118]

Just misread the title as 'weirdass sub'. Probably have had a little bit too much of that GTA V meme.

[@birdboat5647]

I love SO(2) :)

[@AhmadAhmad-qx6fp]

Weierstrass was also a lawyer... so in terms of being sneaky? Yeah! It follows naturally

[@BalaMurugan-hk2gd]

Hey dude what's your name.your ideas are great i love you from India

[@nic741]

How long do you think the cool beard’s going to grow?

[@motekkz9997]

Human brain is limitless wow

[@oa213]

Legend

[@aak-agami9734]

Undo the chandu

[@tomatrix7525]

Bprp, you pronounce the W as a V since it is german...So it’s like ‘Veierstraß’

[@blackpenredpen]

Oh?! I didn’t know!

[@YKj3119]

Could talk about argument of 0

[@angelmendez-rivera351]

arg(0) is undefined.

[@sirllamaiii9708]

ayyy im early to my math lesson for once

[@fatihsinanesen]

Hold on! A beard is talking.

[@the_hasnat]

Assalmu Alaikum!

[@rfzafar]

proof of fermats last theorem .......

[@jomelherras8910]

the beard thoooooo.

[@nathannguyen7449]

Nice

[@Dionisi0]

U made a mistake at the end bUddy

[@theboss9600]

Just 10 comments lol

[@SuperOkitos]

Sos muy cra chino.

[@6612770]

Oopsy!

[@johnwick5901]

You definitely should shave

[@sueyibaslanli3519]

10th

[@andrewli8901]

6th

[@duynguyencong6990]

ho

[@dilemmacubing]

first

[@anushrao882]

51th Yay

[@akagamishanks-q3o]

Please cut your beard. It doesnt look good at all.

[@chriswong7928]

get your beards off

[@samiunalimsaadofficial]

Nice