I would like to thank you for these videos. I am going to take calculus this fall, and through my subscription to your channel, I have become more comfortable with integrals and derivatives.
I solved the integral by doing by integrating by parts with u = e^arccosx and dv = dx. After integrating by parts twice, the original integral returned. Overall, it's another way to do it.
@@simohayha6031 in my high school books they didn't use the sec notation. They used cos^(-1) and arccos. In my experience it is difficult to know which notation people use that is why I use sec, and arccos, never cos^(-1)
It takes quite a bit of work...you can read up on Liouville's Theorem on elementary anti-derivatives. To understand the proof of Liouville's Theorem you need to know some ring/field/Galois Theory.
Thanks. I'm just a stupid physicist and that is something where there was not much emphasis in our courses. Also, my Galois Theory course was quite a time ago...
Hahahaha well there are two papers that I found to be quite useful in explaining how to rigorously define an 'elementary' function and then prove that elementary functions need not have elementary anti-derivatives. There's actually not too much of field/ring theory involved-just knowledege of algebraic extensions and believing that polynomials over fields can be factored into a product of irreducibles. The Galois Theory only features in one part of the proof of Liouville's Theorem and there all you need to remember is that if you extend to the splitting field of some polynomial and consider the Galois Group of this extension, if something is fixed by all automorphisms in the Galois Group then it must actually be in the base field. Here are the papers: wstein.org/edu/winter06/20b/Conrad.pdf and www.dm.unipi.it/pages/gianni/public_html/Alg-Comp/rosenlicht.pdf
This video is perfect !!! I like it. Your explanation is very clear and integration by parts is an easy step. It's quite good of you to let do our work too. Thank you mate.
cos(u) = x ( in red pen) we put cos(u) = u and then the answer like this-; 1/2 * e ^ cos-1(x) * x - 1/2 * e ^ cos-1(x) * sin(cos-1(x)) I am waiting for your replay Thanks:)
Thanks for letting me know about non-elementary integrals ! ☺ Who are here seeing only on the board without listening his explanation ! Sorry , I am unable to listen ur words
Yes, I was able to do the first. It looks a lot to an exercise that I did in my school, in which you have to demonstrate that both functions sen(x) and e^x are orthogonals, you only have to pass from integral of e^arccos(x) to integral of -(e^u)sen(u) and is almost the same. You can also use Euler's formulas to solve it :). Regards.
I'm just guessing but maybe you can show that your integral has properties that no combination of elementary function can have. In any case that would use complex algebra, since this is not an analysis problem but an algebra problem.
+FortNikitaBullion It's a theorem of Joseph Fels Ritt, a differential algebraist from the 1950s. He is technically my math-PhD "grandfather": i.e. my PhD advisor's PhD advisor. Look up Liouville extensions (integrals and exponentials of integrals)
I'm wondering if it is possible to attempt an ansatz solution? If we just assume that [f(x)exp(arccos(x))] ' = exp[arccos(x) it seems that we should be able to solve for the solution by finding f(x) (which appears to be an ODE). Though I'm not able to solve it this way...... Any suggestions? Your way works (obviously), but I am always looking for other solutions...... thank you for posting!
But in sin x /x can't you just substitute sin x with its Maclaurin representation then divide by x so that you get the new integral as integral(1-x^2/3!+x^4/5!-...) so that you can get an approximation of the integral?
It's possible to rewrite it as the integral of x+i*sqrt(1-x^2). Not necessarily easier, but it would be interesting to work out the equivalence between those two answers.
This made me think: why do we choose "non-elementary" to mean using only the operations that we define it to mean? It means any irrational number that an integral evaluates to which we cannot describe with any normal functions, but then why does exponentiation with base "e" count as elementary? It can't be defined any better than the result of one of these integrals, or a the limit of a function where we can never actually write down the entire number. I don't see anything "elementary" about it since you need to understand some calculus (integrals or limits) to measure it.
Is there an alternative way of solving the integral e^(cos x) dx? Polar coordinates? Parametric? Complex? Cylindrical? Spherical? I'm trying, but it's hard. If you graph this function it kinda looks like a horizontally squished spiral.
For a definite integral over (-pi, pi), you can write the integral in terms of the modified bessel function. The indefinite integral has no answer, even in "exotic" functions.
Wouldn't replacing sin(u) by its complex exponential representation be simpler that integration by parts ? i guess it depends what you're comfortable with. Of course if we're allowed complex numbers it would be even more direct to just write acos as a complex logarithm.
The right integral does not have an answer if we test by differentiating the reult as a quotient. But supposing that is not the correct method may there is another way?
Manuel Odabashian It turns out, there is no other way to find the indefinite integral of e^(cos(x)). It's called non-elementary, meaning there is no way to represent it using elementary functions. To be clear, the function is still integrable, in that we can compute a definite integral with given bounds (at least using a computer).
Only reason I didnt understand this is because I havent yet learned shit on Integrals, and very few of Trigonometry. But anyway I found the explanation very clear
It’s non-elementary as a real answer. If you want it as complex, it will still be non-elementary but has a special function called dilogarithm integral function denoted as Li_2(x). The answer can checked in the integral calculator website.
I mean kinda sad though I feel like e^x sinx is the actual content vs just substituting u back into x, so it kinda feels like you just skipped the whole thing. Not to complain or anything....
2.16 wait what? I mean, isnt cos^(-1) x = 1/cos x? Just like cos^2 x = (cos x)^2... Is it me or this notation for whats known as arccos x is really weird?
It’s a weird notation the way that was set up in calculators and textbooks. This notation is an inverse cosine function, which is another way to say is arccosine. This is the only exception that -1 exponent looks a bit weird. All others is just the way we used in algebraic notation.
Use IBP by splitting x and xe^(x^2), it will turn out to be an imaginary error function denoted as erfi(x) since it’s non-elementary. The answer is (2xe^(x^2)-sqrt(pi)erfi(x))/4+C.
I admit i just thought "well arc functions can almost always be integrated by a u-sub. so let´s try that"^^ and it worked. obviously i have no Idea if my result is right, but at least it came rather easily after the u-sub^^ calling arccos=u meaning x=cos(u) just taking the derivative with respect to u instead of the usual x and i was left with IBP and well after that i just resubstituted. and while it´s a fuck show i got a result^^ I got e^arccos(x) (sin(arccos(x)+x)/2 and well at least my Idea was right it seems^^ well i messed up a - it seems. Didn´t get the Idea to workout the square root. but well not bad for a five minute try at work^^ (it´s a dead shift. Evening shift in callcenter is no fun. thank god they didn´t block youtube)
I love your attitude when you address the complainers. Hahaha
thanks!
swears it so funny
@@blackpenredpen waiting for 800 k subs!
instaBlaster...
DAMN IT. I stayed up all night to do e ^ cosx and it turns out it is non-elementary!!!
lol
You can solve it.......numerically.
Sorry, Goddard do you refer to Taylor series ?
@@howardman3926 oh yeah yeah
Same ;-;
That u sub for arccos was one of the most beautiful things I've seen
I would love to hear your thoughts on non-elementary integrals! Real mysteries.
love these videos. thank you for making them.
Thank you.
That trig at the end was so cool
I would like to thank you for these videos. I am going to take calculus this fall, and through my subscription to your channel, I have become more comfortable with integrals and derivatives.
You're very welcome. I am glad that these videos have been helping you. More are coming. Best of luck to you in your study
I solved the integral by doing by integrating by parts with u = e^arccosx and dv = dx. After integrating by parts twice, the original integral returned. Overall, it's another way to do it.
I did it that way too baby
That's incredibly cool.
I hate the cos^(-1)(×) notattion. Never sure if they mean arccos or 1/cos
If they mean 1/cos(x) they'll write sec(x). That's how I judge it.
@@simohayha6031 in my high school books they didn't use the sec notation. They used cos^(-1) and arccos. In my experience it is difficult to know which notation people use that is why I use sec, and arccos, never cos^(-1)
1/cos is sec not cos^-1
@@simohayha6031 in the french system, sec is never used (I live in Lebanon)
The problem is that they borrow the notation f^(-1)(x). That confuses students all the time. Does it mean inverse? Reciprocal? Both??
Holy shit is that a thermal detonator???
how can you SHOW that a function doesn't have an elementary solution?
It takes quite a bit of work...you can read up on Liouville's Theorem on elementary anti-derivatives. To understand the proof of Liouville's Theorem you need to know some ring/field/Galois Theory.
Thanks. I'm just a stupid physicist and that is something where there was not much emphasis in our courses.
Also, my Galois Theory course was quite a time ago...
Hahahaha well there are two papers that I found to be quite useful in explaining how to rigorously define an 'elementary' function and then prove that elementary functions need not have elementary anti-derivatives. There's actually not too much of field/ring theory involved-just knowledege of algebraic extensions and believing that polynomials over fields can be factored into a product of irreducibles. The Galois Theory only features in one part of the proof of Liouville's Theorem and there all you need to remember is that if you extend to the splitting field of some polynomial and consider the Galois Group of this extension, if something is fixed by all automorphisms in the Galois Group then it must actually be in the base field. Here are the papers: wstein.org/edu/winter06/20b/Conrad.pdf and www.dm.unipi.it/pages/gianni/public_html/Alg-Comp/rosenlicht.pdf
I should've said "whenever you extend to the splitting field of some irreducible polynomial" (or essentially whenever you have a Galois extension).
Thanks.
THIS PERSON IS SO LAZY HE IS NOT DOING ALL THE STEPS!
sorry just had to do that :P
I LOVE YOUR COMMENT
No he is NOT lazy. He did a great job by not just spoon-feeding lazy headed people out there.
+Taslim Torabally Either you didn't read the comment properly or you're extremely fun at parties
I think it was to make us try solving it
And it's great to me 😎
Incredible, thanks to you I'm learning how to do Trigonometric Integration... Amazing.
Maths are beautiful.
Tysm for teaching us all. These cool methods. I really like the one where you figured out sin(arccos x), I will be sure to use it in my Calc class
In "e^Arccosx" case, you can use Complex analysis
2 watched videos were enough to subscribe. Great content!
If you want to practice for integration competition.
Visit our channel to find great integrals.
This video is perfect !!! I like it. Your explanation is very clear and integration by parts is an easy step. It's quite good of you to let do our work too. Thank you mate.
*I paused it. I tried it. And I did it. That's a first for me!*
Good stuff, love the 2nd video as well where you showed the 3 stop-steps, makes life so much easier hahaa
Master of markers 😂😂😂
lol! Thank you!
cos(u) = x ( in red pen)
we put cos(u) = u and then the answer like this-;
1/2 * e ^ cos-1(x) * x - 1/2 * e ^ cos-1(x) * sin(cos-1(x))
I am waiting for your replay
Thanks:)
what?????????????????????????? can we integrate e^cos(x)???????????????????
Thanks for letting me know about non-elementary integrals ! ☺
Who are here seeing only on the board without listening his explanation !
Sorry , I am unable to listen ur words
I love the way you tought
Thank you for the video! Nicely done
great!! eloquent...understood!! greetings from Colombia
Thank you so much boss.It really helped me.
I really enjoy that man! Cheers
Yes, I was able to do the first. It looks a lot to an exercise that I did in my school, in which you have to demonstrate that both functions sen(x) and e^x are orthogonals, you only have to pass from integral of e^arccos(x) to integral of -(e^u)sen(u) and is almost the same. You can also use Euler's formulas to solve it :). Regards.
How does one prove that an integral results in a non-elementary function?
Good question. I wonder if a formula exists that serves to find non-elementary integrals.
I'm just guessing but maybe you can show that your integral has properties that no combination of elementary function can have. In any case that would use complex algebra, since this is not an analysis problem but an algebra problem.
+FortNikitaBullion It's a theorem of Joseph Fels Ritt, a differential algebraist from the 1950s. He is technically my math-PhD "grandfather": i.e. my PhD advisor's PhD advisor. Look up Liouville extensions (integrals and exponentials of integrals)
en.wikipedia.org/wiki/Risch_algorithm
FortNikitaBullion Use Liouville's theorem
I like your videos and your way to explain 🤜🏻👍🏻
Nice video man!
I'm wondering if it is possible to attempt an ansatz solution? If we just assume that [f(x)exp(arccos(x))] ' = exp[arccos(x) it seems that we should be able to solve for the solution by finding f(x) (which appears to be an ODE). Though I'm not able to solve it this way...... Any suggestions? Your way works (obviously), but I am always looking for other solutions...... thank you for posting!
But in sin x /x can't you just substitute sin x with its Maclaurin representation then divide by x so that you get the new integral as integral(1-x^2/3!+x^4/5!-...) so that you can get an approximation of the integral?
We don't have ELEMENTARY solutions... your solution is not an elementary one (morever, solution by series are approximation)
The integral of sin(x)/x is non-elementary but has a special function for this as a sine integral, which is denoted as Si(x)+C.
@3:58 where do you get the 1/2u? the integral of e^u=e^u
Because an integral of a product of 2 functions is not the same as the product of 2 antiderivates. e^u*sin (u) requires integration by parts
I liked the step with the triangle the most :)
It's possible to rewrite it as the integral of x+i*sqrt(1-x^2). Not necessarily easier, but it would be interesting to work out the equivalence between those two answers.
it was easier to use the complex definition of sin
Why not to use church definition of sin?
@@NoNameAtAll2 lmao
could you please make a video on why these integrals don't work? I really wanna understand
No ,e to the power X squared has a solution which can be done by Gaussian integrals provided limits are known to us.
he knows... the function itself has no antiderivative in elementary functions. the question posed had no bounds
cos is exponential form, therefore cos^-1 is logarithmic form.
I used the DI method and arrived at the same answer
This made me think: why do we choose "non-elementary" to mean using only the operations that we define it to mean? It means any irrational number that an integral evaluates to which we cannot describe with any normal functions, but then why does exponentiation with base "e" count as elementary? It can't be defined any better than the result of one of these integrals, or a the limit of a function where we can never actually write down the entire number. I don't see anything "elementary" about it since you need to understand some calculus (integrals or limits) to measure it.
This integral can be calculated only by parts
There are two ways of choosing the parts
Discuss the convergence of series whose nth term is (ni/(2n)i)x^n
You should have clarified the question of "is it doable?" as "can the solution be written in closed form? Does the integral even exist?"
The integral doesn’t work but there are solutions using Taylor series
thank u Sir , you explain step by step
And I have to add, the derivatives of them are making some sense. Could it be that there is some Differential equation out there that solves this?
Is there an alternative way of solving the integral e^(cos x) dx? Polar coordinates? Parametric? Complex? Cylindrical? Spherical? I'm trying, but it's hard. If you graph this function it kinda looks like a horizontally squished spiral.
For a definite integral over (-pi, pi), you can write the integral in terms of the modified bessel function. The indefinite integral has no answer, even in "exotic" functions.
So is there some way to get an approximation of a non-elementary integral?
You may use inf. series
in this case use e^z = sum from 0 to inf if z^n/n! where z is cos
Is in general integral of e to the power of any symmetric function without integral function?
Wouldn't replacing sin(u) by its complex exponential representation be simpler that integration by parts ? i guess it depends what you're comfortable with. Of course if we're allowed complex numbers it would be even more direct to just write acos as a complex logarithm.
You should do a reddit group
can you show us the answer to some of the non-elementary functions? do you have some video on i-functions or error function?
If you want to practice for integration competition.
Visit our channel to find great integrals.
thanks!!
You have only approximation solution of non-elementary integral.
Can be integral: x.e^cos(x) ,please.
The right integral does not have an answer if we test by differentiating the reult as a quotient. But supposing that is not the correct method may there is another way?
Manuel Odabashian It turns out, there is no other way to find the indefinite integral of e^(cos(x)). It's called non-elementary, meaning there is no way to represent it using elementary functions.
To be clear, the function is still integrable, in that we can compute a definite integral with given bounds (at least using a computer).
i love this Poké Ball
lol!
you are integration talent
You're genius bro
holy s** this is a MIT integral problem??? I just got a very similar exercise in my first exam of Integral Calculus (no joke) 0.o
If one can not solve a problem, then it's non-elementary. #Confirmed
any reason as to why isn't it integrable sir
Could you not just have substituted x directly at 4:35, since we knew that cos(u)=x?
Pierre-Bob Kjellén yes
Thanks!!! You save my life!!
If it worked on arc cosine x, I'm guessing it will also work on other five inverse functions.
Unfortunately this only works for arcsine and arccosine
Guessing is not enough in math. You must prove it.
why am i watching this im only 15 and i can hardly follow whats going on... still really interesting though
ItsRapid Gaming Stay interested. If a problem is too hard, that means there must be an easier one, right?
Schiff Granger :)
if the second one cannot be integrated that means ,it doesn't have a function, or this kind of rate of change in nature dosent exist
It means it doesn't have a function that can be written in our usual notation. Such a rate of change may exist in nature.
Only reason I didnt understand this is because I havent yet learned shit on Integrals, and very few of Trigonometry. But anyway I found the explanation very clear
Is there any method to predict which integral is not elementary?
Kaivaly Daga no elementary method!
I understand the steps you took if x is restricted to -1 to 1, but you made no mention of this. How do we know your steps are valid for all x?
trucid2 It's not valid for all x... only for -1 to 1, since it is the domain of this function.
why can't we do it???
what is the integral of tan inverse (x) over x please solve it
It’s non-elementary as a real answer. If you want it as complex, it will still be non-elementary but has a special function called dilogarithm integral function denoted as Li_2(x). The answer can checked in the integral calculator website.
inst the derivative of cos(u) = -sin(u)*u' since u is a function of x?
Alonsinho Gomez He did it! u' = du ;-)
Yes, it is. When you differentiate the left side, it will be -sin(u)*du/dx. Multiplying by dx gives the answer to both sides, which is -sin(u)du=dx.
But you can solve non elementary integrals with Taylor Series...
I mean kinda sad though I feel like e^x sinx is the actual content vs just substituting u back into x, so it kinda feels like you just skipped the whole thing. Not to complain or anything....
Bruhh you explain an easy thing soo much 😅
integral of (x e^sinx)
Are you from the planet of Ood?
2.16 wait what? I mean, isnt cos^(-1) x = 1/cos x? Just like cos^2 x = (cos x)^2... Is it me or this notation for whats known as arccos x is really weird?
cos^-1 = arccos. 1/cos = sec
It’s a weird notation the way that was set up in calculators and textbooks. This notation is an inverse cosine function, which is another way to say is arccosine. This is the only exception that -1 exponent looks a bit weird. All others is just the way we used in algebraic notation.
when you say integral e^x^2 dx is non-elementary :
do you mean (e^x)^2 or e^(x^2)?i
Yuval Paz he means e^(x^2)
ok, make sense
Yuval Paz the integral of e^(-x^2) is solvable tho. look it up on mit courseware
@@Someone-cr8cj is solvable if integrated up to infty. The general integral is a special function, the error function
I did it with some little help from my book :) although I wanna know why then other integral is non elementary
If I have integral e^-x cos x dx, i need help please :(
Itzel Sustaita By parts...
Sir how to solve (x^2)e^(x^2)
Use IBP by splitting x and xe^(x^2), it will turn out to be an imaginary error function denoted as erfi(x) since it’s non-elementary. The answer is (2xe^(x^2)-sqrt(pi)erfi(x))/4+C.
Lambert W function?
Integration by parts isn't necessary, complex number can be used.
Y U NO factor out 1/2 e^u :D
Sir I got an answer for the second question.
Ans: X+c
Do you get e^cosx when you take the derivative of x+c ?
I don’t know why but I cracked up after reading both those comments
@@sciuresci1403 just 1
Can we integrate e^sin(x)?
It can't. For the same reason of e^cosx
@@angelfernandovasquezporras1144 and what is the reason that makes e^cosx non integrable?
im dumb asf. buy i enjoy watching this.
(Cos(x))^(-1) is not arccos
It's 1/cos(x)
But cos^(-1)(x) is indeed arccos(x)
yeah that's why I don't really understand why there is sqrt(1-x^2) instead of tan(x)
Cos^(-1) (x) is Arccos(x).
😅
bee or be or b or honey bee?
how to integrate cos(e^x)
Do a u-sub:
u=e^x
du=udx
dx=du/u
You will get the integral of cos(u)/u, which has a vertical asymptote at u=0, and so is non-elementary.
how u written there 1÷2 there
?
interesting.good example.
Integration makes easy
Bhai ye formulae h aisey question Ni ate bhai 2 no ke liye bhi
really nice beard on that video
I admit i just thought "well arc functions can almost always be integrated by a u-sub. so let´s try that"^^ and it worked. obviously i have no Idea if my result is right, but at least it came rather easily after the u-sub^^ calling arccos=u meaning x=cos(u) just taking the derivative with respect to u instead of the usual x and i was left with IBP and well after that i just resubstituted. and while it´s a fuck show i got a result^^ I got e^arccos(x) (sin(arccos(x)+x)/2 and well at least my Idea was right it seems^^ well i messed up a - it seems. Didn´t get the Idea to workout the square root. but well not bad for a five minute try at work^^ (it´s a dead shift. Evening shift in callcenter is no fun. thank god they didn´t block youtube)
what the fuck am I doing watching this I havent taken calc since high school
Does not have an answer… *yet*