integral of sqrt(x^2+1), with Euler Substitution, math for fun
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- Опубликовано: 4 окт 2024
- integral of sqrt(x^2+1) with trig sub: • integral of sqrt(1+x^2...
wikipedia Euler Sub: en.wikipedia.o...
integral of sqrt(1+x^2), integral of sqrt(1+x^2),
integral of sqrt(x^2+1), with Euler's Substitution for integrals,
blackpenredpen,
math for fun
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trigsubs are way easier typically but I may do it this way in my D.E. class just to make my teacher think im smarter lol.
that works!
Here after Euler's substitution and some linear properties you will get integal of power function
To see that not always trig subsittutions are faster calculate integral
Int(\frac{dx}{x^2(4x^2-3)^2\sqrt{x^2-1}}) or
Int(\frac{dx}{x\sqrt{2x^2-2x+1}})
If we use first Euler substitution for integral Int(\frac{dx}{x^2(4x^2-3)^2\sqrt{x^2-1}}) we will get integral which can be
easily calculated mentally but if we use third substitution (with the roots) we should use Ostrogradsky method for isolating rational part of integral and calculate twelve coefficients
If you still want trig substitutions you need three substitutions (actually two of them are inverse trig substitution)
Yeah is there someting easier then power rule and linearity of an integral , because it we will get after Euler substitution in this integral
Jacek Soplica you do LaTeX right?
I use multiple u-subs on the exam and got the exact same answer... but that's too much subs and after i submitted my answer i just realised that trig-sub 'exists' 😂
And all of you hating and saying why don't you do trig sub. This is the beauty of math. 1. Two different methods arrive at the same solution and 2. There exists a much more elegant approach. Now imagine this was the way you were taught and someone showed you trig sub. What would you say?
"That's pretty awesome!" (but then again, I do appreciate the fact that there are multiple ways to skin mathematical cats)
Swedish Math Tutor I would say, why didn’t you tell me this first?
id say: you cheater!
Also, this sub works always, not the case with trig sub, where you have troubles with domain if you put x=tan(theta) for example.
Rewriting (1/t^2 - t^2) as a difference of two squares, then multiplying top and bottom of the fraction by the conjugate, only to create a difference of two square conjugates is beautiful and amazingly clever.
Greg Brown thank you!!!!
Sí, se soltó la greña lmL
@@allaincumming6313 "Yes, the lmL hair was released" thanks google translate
@@ekxo1126 I wonder where can I buy the rumored lmL hair...
@@ekxo1126 I'm a native spanish speaker, that's indeed what it means. The commentary itself is what doesn't make sense lol
This Euler's sub method of solving this rather simple integral was really fascinating and I just loved the algebra involved here. The ugly radicals just vanish like pure magic. Who said that Math is not magic? You just proved that math can be magical and beautiful. Thanks for this awesome substitution. It can make anyone's day (anyone who likes math) beautiful. In a parallel universe, I hope I come across a math teacher like you in real life instead of the virtual interaction here. I would surely feel blessed then.
What's is the "ugly radical"?
@@jif7707 ur mum
*Try to subtitute x by the hyperblic sinus : sinh(t)*
as a remind, sinh(t) = (exp(t)-exp(-t))/2 ; cosh(t) = (exp(t)+exp(-t))/2 ; sinh'(t)= cosh(t)
then if x = sinh(t) , we have t=ln(x+sqrt(1+x²)) _(easy to obtain in two lines!)_
So, *1+x² = 1+sinh(t)² = cosh(t)²* and *dx= cosh(t) dt*
let's note int(f(x)dx ,0 ,a ) the integral of a function f(x) between 0 and a
then int(sqrt(1+x²) dx ,0,a) = int( cosh(t)² dt, 0, ln(a+sqrt(1+a²))
*## i will note b := ln(a+sqrt(1+a²) which means that sinh(b) = a##*
But we have cosh(t)² = (exp(2t)+exp(-2t)+2) / 4
so int( cosh(t)² dt, 0, b) = [exp(2t)/8 - exp(-2t) /8 + t/2] # between 0 and b
= exp(2b)/8 - exp(-2b)/8 +b/2
Moreover exp(2b)-exp(-2b) = (exp(b)+exp(-b)) (exp(b)-exp(-b)) = 4 sinh(b) cosh(b) and we know that cosh(b) = sqrt(1+sinh(b)²)
then int( cosh(t)² dt, 0, b) = ( sinh(b) . sqrt(1+sinh(b)²) ) /2 + b/2 .
As we have a = sinh(b) then : int( cosh(t)² dt, 0, b) = (a . sqrt(1+a²))/2 + ln(a+sqrt(1+a²))/2
Which means that the integral function of sqrt(1+x²) is *(x sqrt(1+x²))/2 + ln(x+sqrt(1+x²))/2* AS FOUND IN THE VIDEO
Notice that if you deal well with hyberbolic trigonometry this way of calculating the integral is faster and easier !!
In this case: Hyperbolic sub>Trig sub>Euler sub
You are amazing and I was wondering why he didn't use the change X = SINH(t)
Wow U are genius
Another substitution that works is x = tan t
Nah tany = x is way faster
Excellent video! I prefer x=sinh(t). Hyperbolic trig sub is an elegant approach, and is a middle ground between Eulers trick, and classic trig sub.
Not really
When you trying to find the easiest way to integrate a problem and you found this guy who makes it harder. I know how to solve this problem and after I watched this video I dont know how to start anymore.
😂😂😂😂😂😂😂 LMAO
sounds like you have trouble organising your knowledge on integral calculus.
This is something I have given a lot of thought to. Want some help?
@@maalikserebryakov Hi, thanks for your offer. I’m just joking. I’ve passed the class and moved on for a while. Something I will never look back :)
Genuinely nice explanation! This is amazing.
Math is beautiful and magic, you are an amazing magician. Kudos!!
This made me appreciate trig sub
Trig subs can be avoided in many cases. Particularly with radical expressions with x2 - 1, x^2 +1 and 1-x^2 in the radicand. Do not use trig for them
We had a sir who taught it like this
x/2(question)+constant/2(integral of reciprocal of question)
looks like we are now black pen, red pen and blue pen!
Diamond Dave yup!
Diamond Dave there was already a video with a green pen included. I guess crazy things are going on these days..
purple pen too
@@AlgyCuber blackpenredpenbluepengreenpenpurplepen
@@slippygames3519 RGB pen
Little note : 1-t^2 / 2t , is actually the formula (when you let t=tan(theta) ) for 1/tan(2*theta) . This method thus shares a relation with trig sub anyway :)
Very nice video!
Muy bien, ese cambio de Euler. Otra forma más de para completar la derivada ¡ Le felicito !
t^-2-t^2 = (t^-1-t)(t^-1+t) = 2x(t^-1+t) = 2x(t^-1-t+2t) = 2x(2x+2t) = 4x(x+t) = 4x(x^21)^(1/2)
Easier to do difference of squares while still in t.
I did the integral by using )complex substitution where I put x =isin(theta) and it worked!!
I can understand this. No wonder I'm alone.
lol
It’s not hard to understand, it’s just a substitution with lots of algebra
we are with you
Chapeau ! 😊
On peut aussi faire X = Tan (téta) ... par la substitution trigonométrique.
You made this super easy sums look complicated and tough
Every time I see Euler in the title I'm like "Daaaamn he EVEN has a substitution????"
Ohh! Thank you so much for this video that answered my question. Good job bro, keep it up!
very interesting this substitution for the calculation of this integral. had solved this integral by trigonometric substitution x = tgO.
Very cool thank you BPRP!
when do you specifically need to use euler's sub? That'd be interesting :)
It was a question sent by one of my subscribers
You can use it whenever there is a specific composition of a quadratic inside a radical
But when the quadratic contains the x term, trigsub will no longer work. So you will use euler sub then.
nice !you can also : sub x with tan(u) then you integrate by parts : 1/cos^3(x) !
Yup! That's trig sub!
khiari youssef no parts needed...
@@fgdhlololo1887 parts is usually used to compute the integral of 1/cos^3(x)
Then seems to be way easier.
i never knew this method for integration anyway i gain another weapon in my magical pocket
Thanks bprp !
Yastangent
The best way to solve this integral is by setting x = sinh(t). Then we have Integral (cosh(t)^2 dt and here cosh(t)^2 = 1/2(1 + cosh(2t)) etc.
Wow, you were right. That WAS really cool!
Alex Behlen I am glad that you like it!!!
In polish and russian schools
Euler's substitutions was standard ones
f.e can be found in russian textbook
Курс дифференциального и интегрального исчисления Григо́рий Миха́йлович Фихтенго́льц
with short geometric interpretation
Two minor issues... 1) You never justify swapping the order of the absolute-value and -1 power at 17:12. 2) At 19:38, the reason why the inside is positive and you can drop the absolute-value is not complete. If you have (A+B) and A > B, you cannot conclude (A+B) > 0. Counterexample: A=1, B=-2. (The missing reason is that if x is negative, the A portion is greater than -x to get the sum to be positive.)
1st the reason he can brirng the -1 to the t as a power is because of a ln property. 2nd it's different because you square x and add 1 inside a square root and subtract x. if you just have sqrt(x^2)-x then it's zero. But instead we have the +1 so sqrt(x^2+1)-x is always positive. You're example doesn't consider the square and sqrt.
excellent job! this was a beautiful integration.
The best method is not either trig sub or euler sub. The best method is integration by parts. Let u =(x^2+1)^1/2 and dv = dx. then a little add and subtract will give the original integral plus a familiar integral.
Give it a try.
Never seen that method! Very interesting. Would like to see the idea behind it
The idea behind it is the assumption that there exists a linear equivalent for any nonlinear one with the incorporation of another variable in the addition format (if you multiply t by x, you only make it more complicated) The question is: under what circumstances this Euler method is more efficient that simple trig substitution? Euler is well-known for creating such equivalence. Consider the famous Euler formula in the complex plane which maps a complex exponential to a complex trigonometric form. That formula was built on the same pattern of logic. Because integration is such an art, transformations are central to finding anti-derivative of complicated integrands.
Muito obrigado, acho q agr vou conseguir resolver uma integral q eu tô penando a quase 2,5 semanas com isso que acabo de conhecer (substituição de Euler)
Qual?
Geez...you have the patience of a saint!
How good you are changing the markers in the hand.
Never thought I'd say that I'd prefer to do trigonometric substitution, but here we are.
Trig sub is not useful when there are many other terms apart from the radical
This method works always to simplify the integrand
You are awesome ,dude.Thx for this video!!
easier way: just replace x with sinh (t)
Or tan(t)
@@nejlaakyuz4025 then you have to integrate sec (x)
@@adi-sngh integrate of secant is "known", this channel has a video on it
@@me_hanics it's ln(tan(X)+sec(X)) it's a pretty standard integral should be remembered probably
If x^2 + 2 then your method might not work if you sub sinh(x), Euler method seems to be working in x^2 + a, maybe x^3 + a, not sure yet
Hey buddy your are a best mathematics teacher
I like the video. Good job. Sometimes you could be too fast but it's fine because I can stop video for one or two seconds. Reminds me my old good times when I was a student. :)
Mirek Soja thank you. Yea pause the video whenever you need to. Hopefully overall is good.
I think he chose wrong example
If he had chosen Int(sqrt(x^2-1),x) as an example he would have shown two Euler substitutions
which cover all integrals in the form \int R(x,\sqrt{ax^2+bx+c})dx
Substitution with leading coefficient he showed but substitution with the roots is missing
If the purpose of this video is calculating this integral then video is ok
If the purpose of this video is to show another substitution which is less known in US
then video is not finished
This is much more interesting because if you have the equation of the circel Y^2 + X^2 = r^2 then as you write it as a function
f(x) = (r^2 -x^2)^(1/2) you can intregrate the Circel if you calculate the intergral.
Thank you so much!! I love this vedio
well done, Steve. You pronounced Euler in the correct way. :)
hmm if you set x=cot t, then ta
n (t/2) = w, you get that x=(1-w^2)/2w
So it's equivalent to doing a trig-sub followed by a Weirstrass substitution. Kinda sorta
The frequency of advertising interruptions for an educational feature is disgraceful. I would remember who the advertisers are - not to buy their products but to avoid them!
This was great. Thanks.
Very interesting, I've never seen this method before! However, I think I'll stick to trig-sub. This looks hard!
Thank you so much!
what an amazing channel i just found !
Glad you like it!
We learn it since it is universal. Trig sub doesn't always work if you have other things apart the sqrt in the integral.
I see. Thank you. This will help me immensely
Damn, that's crazy. No wonder I had so much trouble solving this on my own.
And that's why the hyperpolic functions exist. You'll get the solution a lot easier with 'em.
Gigi 96 yes
At 17:00 I would have just put in t=sqrt(x^2+1) - x and be done with it!! Other than that great video and thanks for showing us that Euler rules!! :-)
This Euler fella his name‘s all over mathematics
天才やん
That was a real tour de force right there! Great video!
1SaneManiac thank you!
easier method: Write sqrt(x^2+1)= (x^2+1)/sqrt(x^2+1). You get a integral that has a simple formula(that ln|x+sqrt(x^2+1)|) and also a integral that you solve using partitions. In final you will get that the double of that integral equals sth you know and it will be that thing/2
Love these vids but this is rather hard and convoluted way for a simple integral which reduces to solving /(sec x)^3 dx using
integration by parts.
I forgot all about the existence of Euler substitution. Maybe we spent a day on it in class.
Thank you so much
amazing video tanks!
Great Mr 👍
In my opinion, it is more important for students to learn
mathematical modelisation of real life problems, including variables identification, and finding out the general equation of a a specific problem, then let the computer do what it was built for: Tedious calculations....(I was able to calculate this integral on my smartphone within 10 seconds using a TI-89 emulator)....I still wonder if there is a physical phenomena that obeys this law ?
I think he should show how to get this substitution
There are at least two ways to get it
1. Secant line of curve y^2=ax^2+bx+c
2. Right triangle with sides labeled as in inverse trigonometric substitution
PLEASE FACTOR that 1/2
thank you :)
1/3*3/2
Maybe Integration by parts is another way that can solve in a more faster way,because in the process ,add 1 and minus 1 could make a difference.
Euler freaking Demi God!
Just one word...wow!
This stuff is all great fun and good brain exercise but without value in the world of work. Students should understand that no one will offer them money to solve complicated indefinite integrals, just as no one will offer them money to solve crossword puzzles. That said, I applaud Prof. Redblack's teaching style.
Why do you think Engineers learn so much mathematics if they never have to solve any complicated indefinite integrals? Your logic does not make sense.
Trigonometric substitutions are easier, but sometimes there is not an easy defined method to solve a problem, so they need to be solved creatively in more round-about ways.
When engineers build a bridge, do they eyeball it, or do the mathematics to make sure it can hold the weight of the vehicles? The engineers are liable if anything goes wrong. so they need to crunch numbers first so they don't spend all their time re-stitching their mistakes together.
In my 30+ years as a practicing civil and mechanical engineer, I never once did an indefinite integral. Bridges and other structures are designed using advanced software based on numerical methods that provide approximate solutions to the applicable differential equations. Indefinite integrals, LaPlace transforms, and lots more of the fun stuff we see here are not involved.
@warren Gibson its not always about making money, however... if you know how to do these integrals, maybe you can put them on the internet (as such) and make money off it. theres a way to make money doing anything. however my initial point is that the end goal of life is not to make money... but to enjoy life. Newton didn't need money, he simply enjoyed being alone all day and doing this.
nice, solving integrals is allways fun!
Smart approach but perhaps a little quicker by an 'x = sht' substitution
While making the initial substitution for X+t, couldn't we just integrate Xdx and tdx in parts and then substitute dx with the dt form? That would simplify things I guess. We won't need to substitute t with X form for atleast one part. I'm probably not thinking this through and might be overlooking something.
I was going to say to the publisher, "hey bro, you just missed the beauty of this method: first part is just (x^2)/2, and you do just t*dx part". I said let me first scan through the comments not to repeat the same thing. then I saw your comment. You are damn right, that is the way to do this integration.
No you should express x and sqrt as a function of t
and then differentiate x with respect of t
Keep going, you’re awesome 🌹
You can also solve this really easily by integration by parts, and it also has a shorter answer... please tell me if you want to know the solution.(it involves recursive integrals)
Yes
That's really nice.
Would you give some questions where I can try to use it
You are amazing!!!!!
Why are you able to assume that X only has real values like that?
Because he's chosen the reals as the domain for this function. He could have extended it to all complex numbers if he'd wanted
Very good!
HI was it possible to simply use x=sh(t) ?
At 4:05 you could have made things simpler by separating the integral into intgl( x dx) + intgl( t dx)
Just assume x=sinh(t) (hyperbolic sine: sinh(x)=(e^x-e^-x)/2, hyperbolic cosine: cosh(x) = (e^x+e^-x )/2) . Then dx=cosh(t)dx and sqrt(1+x^2) = cosh(t).
Fiinita la comedia!
Desde luego que con la sustitución hiperbólica se consigue resolver más rápido esta integral. El solamente está mostrando otro camino para resolverlo. Obviamente también se puede usar la sustitución Trigonométrica. Es interesante, porque a los estudiantes se le debe mostrar todas las posibilidades existentes para posteriormente estar en condiciones de resolver otras integrales.
NO ES UN CONCURSO DE VELOCIDAD, SINO DE CONOCIMIENTO Y EL SABER NO OCUPA LUGAR
@@joserubenalcarazmorinigo9540 Solamente quería mostrar un otro método para obtener la integral, que (para ser honesto) me gusta más. A propósito, las funciones hiperbólicas no son realmente trigonométricas.
Beautiful
Should I show a real elegant way of taking this integral? I think the way above is really insane. Just for the audience: it should be done in three simple steps just with one substitution. The way Boris Demidovich was doing it in 1970's.
Demidovich must have come from Eastern European schools as a student in his younger days. Western European instructors seem to rely to heavily on trig sub.
I do have one question
What made you think in the first place that you have to choose x+t instead of only t ??
Otherwise it was amazing algebra and magic way of thinking ☺
Cause it works. He didn't think of it,it's a well-known sub.
Oiler's Substitution 🔥
¿cuál de las 3 sustituciones Euler es mejor? ¿algún consejo?
just use x = sh(u) ==> dx = ch(u)du ==> integrel(ch²(u) , du )
Integration of root(x+root x power2+1) solve it sir
Why does the initial equation equal X+t?
Jacob Anderson he just declared it is. You're allowed to do that when you are defining a new variable t.
Jacob Anderson because it must be something bigger than x
Even negatively bigger 😂
The wording would be "select t such that sqrt(x2+1) = x+t".
I got this in 3 minutes by rationalizing
Good job.
saikou xs good job you want a cookie?
That is Magic !
How many of these types of problems would be on a test?
It's over complicated we can solve it by parts, infact integral of sqrt(x2+A2) has a standard formula
Did you miss the point of the video lol
can you make a video demonstrating the Euler substitution???
I don't want to memorize without understanding why is that possible, but I can't find the demonstration on Wikipedia ... :(
Говоришь по русский ?
Hablas espanol ?
They have recorded video about Euler's substitutions
si hablo español, como lo notaste? :O
Sprichst du deutsch ?
I found Euler's algebra book in which is rationalization of square root of quadratic
The guy whose video i had watched changed state of his videos to private
but russian is even better because you have video on youtube and also quite good book
Курс дифференциального и интегрального исчисления Фихтенгольц Г М
Euler's book
Leonheardi Euleri opera omnia
www.math.uni-bielefeld.de/~sieben/Euler_Algebra.ocr.pdf
page 349
thanks :D
Watch also separation of rational part of the integral of rational function
(Ostrogradsky method of undetermined coeffincients ) because it can be useful after Euler substitution
Your analysis of Euler substitution start with cutting curve y^2=ax^2+bx+c with secant line
Tnx sir ❤️
My calculator said ½x√(x²+1)+½asinh(x)
So asinh(x)=ln(√(x²+1)+x?
(asinh= area sinh=inverse sinh)
I can't hear the words at 0:41 : "and I'm just going to say that ??? is one".
I tried to derive that and I could not get to the integrand. Also, I tried to use the x=tan(t) and x=sinh(t) substitutions to solve the integral and I got stuck. Please help.
How were you stuck? Maybe we could email
I got (sqrr(x^2+1))/(1+2x)+c.I wasn't even nearly close.
Could you show and prove ds^2=1/y^2(dx^2+dy^2)....
All in all .. thanx