Elementary vs. Non-Elementary integral battles! (beyond regular calculus)
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- čas přidán 8. 09. 2024
- Integration techniques required: integration by parts, u sub, trig sub, partial fraction, algebra, experience, and patience.
Check out these 8 special functions: • how WolframAlpha defin...
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Battle 1, integral of cos(x^2) vs integral of cos(ln(x)), 1:00
Battle 2, integral of ln(1-x^2) vs integral of ln(1-e^x), 7:55
Battle 3, integral of x^(x/ln(x)) vs integral of x^x, 16:23
Battle 4, integral of x*sqrt(x^3+4) vs integral of x*sqrt(x^4+4), 19:29
Battle 5, integral of x/ln(x) vs integral of ln(x)/x, 32:25
Battle 6, integral of ln(ln(x)) vs integral of sqrt(x*sqrt(x)), 34:00
Battle 7, integral of sqrt(sin(x)) vs integral of sin(sqrt(x)), 36:13
Battle 8, integral of sqrt(tan(x)) vs integral of tan(sqrt(x)), 40:52
Battle 9, integral of tan^-1(x) vs integral of sin^-1(x)/cos^-1(x), 59:13
Battle 10, integral of 1/(1-x^2)^(2/3) vs integral of 1/(1-x^2)^(3/2), 1:04:23
subscribe to @blackpenredpen for more integration videos.
Battle 1, integral of cos(x^2) vs integral of cos(ln(x)), @1:00
Battle 2, integral of ln(1-x^2) vs integral of ln(1-e^x), @7:55
Battle 3, integral of x^(x/ln(x)) vs integral of x^x, @16:23
Battle 4, integral of x*sqrt(x^3+4) vs integral of x*sqrt(x^4+4), @19:29
Battle 5, integral of x/ln(x) vs integral of ln(x)/x, @32:25
Battle 6, integral of ln(ln(x)) vs integral of sqrt(x*sqrt(x)), @34:00
Battle 7, integral of sqrt(sin(x)) vs integral of sin(sqrt(x)), @36:13
Battle 8, integral of sqrt(tan(x)) vs integral of tan(sqrt(x)), @40:52
Battle 9, integral of tan^-1(x) vs integral of sin^-1(x)/cos^-1(x), @59:13
Battle 10, integral of 1/(1-x^2)^(2/3) vs integral of 1/(1-x^2)^(3/2), @1:04:23
file: docs.wixstatic.com/ugd/287ba5_3f60c34605f1494498f02a83c2e62b29.pdf
New challange for me😊
wow nice timestamp! Should be pinned yrself!
Where are those special functions?
Yale NG Which ones are you talking about? They never appeared in the video.
SIR THE RESOURCES AND LINKS TO LEARN MATHEMATICS THAT YOU SAID IN YOUR VIDEO WITH fematika ARE STILL NOT UPLOADED IN THE DESCRIPTION OF THE VIDEO , please do upload those links
Another approach to the integral of ln(1-x^2) dx would be to factor the inside and then use the product rule of logarithms to get the integral of ln(1-x) + ln(1+x) dx. It's a bit easier to solve this way.
Just a real minor point of #4: you could also do a hyperbolic trig substitution instead, and you'd get a simple inverse hyperbolic sine term in the final answer instead of the natural logarithm. That natural logarithm is also convertible to the inverse hyperbolic sine.
Solution to integral of sqrt(tan(x)):
There's a blackpenredpen video on that + c
The way I like to think about the Integral of cos(x^2): with some clever substitutions and Euler's formula it can be shown that it can be written in terms of the integral of e^(x^2) and since that cannot be defined in terms of elementary functions, thus the integral of cos(x^2) cannot be
heloo
To integrate arcsin(x)/arccos(x) from x = -1 to x = t < 1, let x = cos(θ). Then dx = -sin(θ) dθ. The integrand is now -arcsin(cos(θ))·sin(θ)/θ. The bounds are from θ = π to θ = arccos(t). On the interval (0, π), which is the codomain and range of arccos(t), arcsin(cos(θ)) = π/2 - θ. Therefore, the integrand is -(π/2 - θ)·sin(θ)/θ. Factoring -1 will change the bounds to run from θ = arccos(t) to θ = π, with integrand (π/2 - θ)·sin(θ)/θ. By linearity, this gives the integrals of (π/2)·sin(θ)/θ and -sin(θ). The first integral is equal to (π/2)·(Si(π) - Si(arccos(t))), and the second is equal to cos(π) - cos(arccos(t)) = -(1 + t). Then the total integral is simply equal to [(π/2)·Si(π) - 1] - (t + Si[arccos(t)]). Call (π/2)·Si(π) - 1 = C, so the integral is simply C - t - Si(arccos(t)). Done! For the record, Si(x) is defined as the integral from s = 0 to s = x of sin(s)/s.
We can extend the answer to other intervals, but this requires some caution, since arcsin(cos(θ)) = π/2 - θ is no longer true in other intervals.
On the first one, it was obvious, because cos(ln x)=(x^i+x^-i)/2. Power rule, separate real and imaginary coefficients, and put it back to trig functions. Even if you're not going to use complex numbers, you can guess the right integral because cos is like an exponential and goes well with ln and poorly with x^2.
22:21 Euler's substitution sqrt(u^2+4)=t-u would be better idea here
Last one third Euler substution (with roots) or integrating by parts also are good option
One-hour long video but u definitely spent a lot more time than that! Your effort should be appreciated! And also the patreon list grows longer everytime 😁👍
PS it's 1am here in HK and yr thumbnail looks cool with some chill 😆
Mak Vinci lollll thank you!! I prob will make another thumbnail tho. I don’t think that is that appealing lol
@@blackpenredpen Hey keep this kind of thumbnail man(but not too many), it makes others curious to press the thumbnail 😁
I want to know, how to prove that the integral of a function is not elementary, please tell
Chirayu Jain
It’s quite hard to prove it mathematically. I think we need to know Galois theory from advanced abstract algebra in order to do so. I actually don’t have experience in it unfortunately.
@@blackpenredpen, what a coincidence I started learning abstract algebra just 2 weeks before., 😁
Galios Theory, it's probably easier to just know which ones are non-elementary, rather than to prove each one individually.
Chirayu Jain You can prove the non-elementariness of an integral using the Risch algorithm.
@@japotillor That by itself doesn't disprove that there might be some weird unknown way to do an integral.
Integration of e^-xx from +inf
To -inf with pler co-ordinates
The second one was a bit over the top, ln(1-x^2)=ln((1-x)(1+x))=ln(1-x)+ln(1+x)
Either way you need to do integration by parts. Personally, I broke up the ln but if makes sense to use IBP with a bit of work extra then go for it. As long as you get an answer and understand the process
GhostyOcean no you don’t need to do integration by parts with the method he stated. After you split the ln you can split the integral and solve them both by u sub
-James- Integrating ln(u) requires integration by parts, so you are wrong.
@@-james-8343 in order to integrate ln(x) you need to do IBP unless you have the answer memorized (xln(x)-x)
Gábor Tóth Tbh, it’s just as hard if you split it. I split it, and if anything, that made it harder because you have to do IBP twice.
who else got a smile on the face at 16:15 because you have watched an old bprp video?
Sun and clouds me
Sun and clouds Nah, I still messed it up, ffs. 😂😂😂
This is the best video You have made - of those I've seen.
I was especially happy to know that ln(ln(x)) is a non-fundamental function. That question has been bothering me for years.
19:57 you can do both u-sub and trig-sub at the same time by letting x^2=2tan(theta) ;) then, xdx is nicely equal to sec^2 and the rest is just the usual
15:05 In the last two terms of that answer (before the +C) it was not necessary to use absolute value around the ln input. Respond to this comment if you can figure out why!
Number 9 is a pretty straightforward battle, once you know the formula for antiderivatives of inverse functions. As long as a function has an elementary antiderivative, its inverse has an antiderivative of the form, xf^-1(x) - F(f^-1(x)). Once you know tan(x) has antiderivative ln|sec(x)| + C, you just plug tan^-1(x) into the formula and do some trig identities on sec(tan^-1(x)) to get the same result.
You probably made that future video already, but it is interesting to point out that the most obvious attempt to antidifferentiate arcsin(x)/arccos(x) with respect to x results in the sine integral:
A basic trigonometric identity has arcsin(x)=π/2−arccos(x), from which the integrand becomes ½π/arccos(x)−1; then the substitution x=cos(y) with dx=−sin(y)dy results in the sine integral.
That is, ∫arcsin(x)/arccos(x) dx = -x−½π∫sin(y)/y dy = −x−½πSi(arccos(x))+C.
So nice to see a notification from bprp just after the first day of school :D
Thanks!!!!
I love your enthusiasm!
15:15 „Integrale für Euch“ 😂
Grüße an alle Deutsche 🇩🇪🙌🏽
SGE 1899 Hahahah yea!!! Lars helped me to translate it. : )
hahahah Ehrenmann
20:00 you can directly let x = √(2tan(theta))
You are a great teacher
I came here after the rap battle in 8 Miles😂... I am ready for the battle!!!
Would it actually be faster to integrate cos(ln(x)) by using the complex definition of the cosine? You would then need to integrate (x^i+x^-i)/2, which is just a matter of integrating polinomials.
Account Fantoccio Relatively, yes.
Battle 2: Don't use partial fraction! Use ln(ab) = lna + lnb rule first, much more simple!
That was my thought. ln(1+x) + ln(1-x)
That's what I did - got two standard ones.
Sir please make a video on ramanujan formula on finding value of pi
Oon Han has made a video on it
integrating arcsinx/arccosx is actually doable;much easier to do than the other ones mentioned as undoable previously. its just a bit of subs and ibp and using the Si function.
We presently scratch the integral, if it is a non-elementary integral.
In India we have National Teachers' Day on 5th Sept. So, Happy Teachers' Day to BPRP and all other teachers in advance.
Take my love for this channel from Bangladesh.
Lets see those special functions!
Please make a collaboration video with 3blue1brown together
YES
YES
@@azujy2959 gosh that would be so cool
On 14 September it is teacher's day in India . Please make a excellent special video on the day.
one take, with some cuts. i dig it 😁
I love these videos! Encore, encore :-)
Brilliant sir
7:55 wouldn't that be easier to just factor 1-x^2 as (1-x)(1+x) and then use the log propertry to split the ln of the product?
Here's another way to write the answer to question 2, xln(1-x^2)-2x+2tanh^-1(x)+C
√tanx i love this integral same as 1/(x^6+1)
Now solve the special function ones!
integrating ln(cos x) would be an interesting one
It will be a great pleasure to me, if you explain how to separate elementary from nonelementary ones. Does such formular exist?
Промо Risch algorithm.
What about x^dx? Can u do ir pls?
kemosabe What is that?
Correct me if I am wrong, but at 8:50, you can factor 1-x^2 and use rule of log to expand it into 2 terms?
Oh yes. Then integration by parts after that. Both work
@@blackpenredpen Right, unless one memorize that integral of ln(x) is xln(x)-x hehe
n choose k yea
the ad I had for this just said "Find your Steve" 😱😱😱
!!!
Second round:
integral 1 / (1-x^2) = arctanh x + C
Hi, cos(X square) is a function . Geogebra gives a result, if you integrate ( calculate the area) between 2 points
Why we can say that this integral does not have a result.thank you For your reply
Cent Uğurdağ Because the antiderivative of cos(x^2) is *not* the area. The antiderivative of cos(x^2) is simply another function, but the area under the curve is a number. Not remotely the same thing. Any software can calculate any area, but if you ask Geogebra to give you the antiderivative, it *cannot* and *will not* give you an answer, because there is no answer.
İ agree but want to know why there is no antiderivative of this function
Hi BPRP, and thank you for the videos :D I guess this comment will go unnoticed, but if I never ask, I'll never know :)
Why are half of these functions impossible to integrate? You just mention as a fact that it's impossible but never why. I'm not great at integration, so I don't understand _why_
Best videos sir for maths
Question 3, the absolute troll
Hey BPRP , can you make a video about Group Theory ?
Big salutation from Algeria thank you Allah blesses you
What font did you use in your document? Do you use LaTeX package or?
For #9, the ln part turned out to be ln|cos(arctan(x))|, anyone else have this??
Freddie Correct
Yall I was just vibing to the Doraemon theme song in the beginning.
which integrals are intermediate and high school?
12:30 you could just directly integrate it to 2tanh^-1(x).
instead of partial fractions.
حودا روك No, because the domain would be incorrect.
bprp: *showing 8 integral battle*
me: ...here we go again
Konstantin Cherkai 10*
Number 8 was crazy
Thumbnails are getting stronger
I never forget the chendu😆
LETS GO!
May the chenlu be with your integrals.
11:22-11:25 the integral of the thing you are saying needs partial fractions doesn't, actually, because the answer is clearly inverse hyperbolic tangent (Argthx/Argtanhx)
Serouj Ghazarian Well, that's not correct either, since the domain or arctangent is different from the domain of the function we started with. Strictly speaking, partial fractions are the only correct way to get the most general antiderivative, and this can be proven.
@@angelmendez-rivera351 ArGtanH, not arctan
@@angelmendez-rivera351 the function we started with is ln(1-x^2), which has EXACTLY the same domain as Argtanh.
58:05 is just insane lol
n choose k yea! And I didn’t do partial fractions just to save time. Lol
Number 2 way easier to write ln(1-x^2)=ln(1+x)+ln(1-x)
Wait... 1 hour 😯💚
Can you solve it
Int. (x-2)/[(x-2)^2(x+3)^7]^1/3
can show hyperbolic functions more love or not?
Good video, can you please help me with this integral
.. X*Sec(X)
Integration by parts
X take D and I sec x
Integration of secx is log|secx + tan x| and then its easy
It's non-elementary because if you try to do IBP, you get xln(abs(sec(x)+tan(x)))-integral of ln(abs(sec(x)+tan(x)))dx. Here integral of ln(abs(sec(x)+tan(x))) is non-elementary.
I like your microphone
26:25 100 Integrals #61.
My god, integral of x times the square root of (x^4 + x) is a really complicated integral. It would be even more complicated if one had to integrate sec^3(x) from scratch...
34:26
The integral of the square root of (x times the square root of x)?? The integral of the square root of (x + the square root of x)... 🙂
The integral of √(x + √x)
Or the integral of 1/√(x + √x)
Or the integral of 1/√(1 + √x)
James Oldfield Obviously, it is sqrt(x·sqrt(x)). Also, the integral of x·sqrt(x^4 + x) is non-elementary, and is also not the integral dealt with in the video, and the one in the video was actually very simple.
Also, integrating sec(x)^3 from scratch is fairly easy too.
@@angelmendez-rivera351
You must be a really smart person to find this kind of thing easy. I'm still at the level of basic integration and differentiation, power rule stuff. Like 1/cube root (9x^4) + 3x^3 + x^2. Really, really basic stuff like that...
@@angelmendez-rivera351
I was being cheeky. I know he said √(x√x). I was thinking it was easy (relatively), and that √(x + √x) would be a harder integral to do...
James Oldfield I wouldn't say I'm smart, just math savvy. Anyway, I only said it's easy because that was one of the easier integrals showed in the video. Most of the other ones were more complicated. And it doesn't have anything on the integral of sqrt[tan(x)], or even worse, the cbrt[tan(x)] integral.
The integral of sqrt(x + sqrt(x)) is indeed more complicated than the integral of sqrt(x·sqrt(x)). In fact, the integral is very clever. For example, if y = x + sqrt(x), then dy = [1 + 1/{2·sqrt(x)}]dx. Thus, sqrt(x + sqrt(x)) = sqrt(x + sqrt(x))[1 + 1/{2·sqrt(x)}] - sqrt(x + sqrt(x)/{2·sqrt(x)} = sqrt(x + sqrt(x))[1 + 1/{2·sqrt(x)}] - sqrt(sqrt(x) + 1)/2. Now one can split the integral in two parts using linearity. The integral of sqrt(x + sqrt(x))[1 + 1/{2·sqrt(x)}] can be found using the very simple substitution I already mentioned, and this integral will be equal to (2/3)·sqrt(x + sqrt(x))^3 + C. All the remains is evaluating the integral of sqrt(sqrt(x) + 1). Let z = sqrt(x) + 1, so x = (z - 1)^2, and dx = 2(z - 1)dz. This leaves the integral of 2z^(3/2) - 2z^(1/2) with respect to z. This is just a very basic power rule integral, and it gives the antiderivative (4/5)z^(5/2) - (4/3)z^(3/2) + C. Substitute back to get (4/5)·sqrt(sqrt(x) + 1)^5 - (4/3)·sqrt(sqrt(x) + 1)^3 + C. Altogether, the integral of sqrt(x + sqrt(x)) is nicely equal to (2/3)·sqrt(x + sqrt(x))^3 + (2/3)·sqrt(1 + sqrt(x))^3 - (2/5)·sqrt(1 + sqrt(x))^5 + C.
Isn't it instead of using partial fractions, Can we not have
xln (1-x²) -2x + tanh-¹ (x) +c ?
As the answer?
Battle 8 is the best integral....
Can you teach us group theory?
Hey bprp, what font do you use in your files and thumbnails?
2-nd ln(1+x)(1-x) = ln(1+x)+ ln(1-x).
BPRP is an asmr youtuber now? 58:30
Yeah mate, he’s done it before.
For no. 8, can't we split 1/(t^2-2) into partial fractions and use ln? It is much friendlier than coth. Also, why coth instead of tanh?
Yes. But it would be just longer...
@@blackpenredpen But why coth instead of say tanh? According to you they are identical...
Mohammad Zuhair Khan ln in this situation is not friendlier than ln, since the inside of ln would be a complicated expression. In fact, coth is expressible in terms of ln, so that makes your point moot.
blackpenredpen Tbf, I prefer it because you can see how you get the answer, whereas the tanh is just a standard formula.
Did You make already any video with non-elementary integrals like eliptic ones?
For #8 Integral of sqrt(tan(x)). I think I found a much simpler method and simpler though presumably equivalent answer, I would love someone to confirm or refute if possible.
I did the same sub: u = sqrt(tan(x)) so u^2 = tan x
Differentiate so: 2u du = (sec x)^2 dx
Use tan x to create a right triangle of sides 1, u^2 and sqrt(u^2+1)
Thus sec(x) = sqrt(u^2+1) and (sec x)^2 = (u^2+1)
Now dx = (2u / (u^2+1)) du
The substitution means we have to integrate 2 * (u^2 / (u^2+1) du
Partial fraction gives integral of 2 * (1 - 1/(u^2+1)) du
Integrating gives 2 * (u - arctan u) + c
Back substitute to get final answer of 2 * (sqrt(tan x) - arctan(sqrt(tan x))) +c
What do you think?
You made a mistake, you said: "Use tan x to create a right triangle of sides 1, u^2 and sqrt(u^2+1)"
But in the triangle a,b,c (where a=1 b=u^2) the hypothenuse is equal sqrt(a^2+b^2)=sqrt(1+(u^2)^2) = sqrt(1+u^4)
Then you have (sec x)^2 = u^4 + 1 and you have to integrate 2 * (u^2 / (u^4+1) du
salut monsieur svp j'aimerais avoir un pdf des 100 integrale ou un pdf d'çntegrale pour licence de mathematiues svp
On question number (8). Suppose you let integral equal to Q, then square both sides and integrate twice then take the sqr,, can it work?
Do you think that Isaac newton would have been able to derive all of these integral solutions back in his day
Why IS integral of tan (sqrt x ) impossible to solve
I genuinely don't understand
I hate that solution for ∫ √tanx dx and prefer this one -
∫ (√tanx + √cotx)/2 dx +
∫ (√tanx - √cotx)/2 dx
Use the common denominator, √sinxcosx and split 2 into √2*√2, and rearrange -
√½∫ (sinx+cosx)/√(2sinxcosx) dx +
√½∫ (sinx-cosx)/√(2sinxcosx) dx
Wouldn't it be nice if we had a way to use sin²x + cos²x = *1* on the bottom?
*2sinxcosx*
= 1 - ( *1* - 2sinxcosx)
*= 1 - (sinx - cosx)²*
= ( *1* + 2sinxcosx) - 1
*= (sinx + cosx)² - 1*
Substitute each one -
√½∫ (sinx+cosx)/√(1-(sinx-cosx)²) dx +
√½∫ (sinx-cosx)/√((sinx+cosx)²-1) dx
Substitute
t = sinx - cosx
u = sinx + cosx
√½∫ 1/√(1-t²) dt *-* √½∫ 1/(√u²-1) du
= √½sin¯¹(t) - √½cosh¯¹(u) + C
*Substitute back for t and u and you're done,* unless you prefer to use ln|u+√(u²-1)| in place of cosh¯¹(u). If so remember that √(u²-1) = √(2sinxcosx) and everyone converts that to √sin(2x) for no useful reason I can see but there it is if you want it.
Note that √½ is really (1/√2) but I don't have all day to type that and you don't have all day parsing parentheses in a CZcams comment.
I'm sure it doesn't matter and it's probably just me but I find that solution a whole lot cleaner, easier to follow, and easier to remember with fewer chances of making an algebra mistake.
The long, drawn out version is called _Trigonometric Twins_ (not my video) at czcams.com/video/dT8b8wAjTKM/video.html and watch out for the typo near the end.
You probably need to learn the method bprp showed to pass a test though. I don't know.
I also find the similarity of the intermediate form compared to the algebraic answer in the video pretty interesting.
√½∫ (√tanx + √cotx)/√2 dx +
√½∫ (√tanx - √cotx)/√2 dx
Ni999 Wow, well, this is just extremely pedantic as a comment. Let me address a few things:
1. This is not that much simpler to what is on the video, contrary to what you claim. And the answer he gave was not in its simplest terms, so disputing elegance there is futile.
2. 2sin(x)cos(x) is simplified to sin(2x) because it is, well, *simpler.* Individual trigonometric functions are always preferable to products thereof.
3. BPRP's method is generalizable to higher order roots of tan(x), whereas yours is not. And considering the precedent this has on the channel, it makes perfect sense he explained it the way he did it.
@@angelmendez-rivera351 Ok.
1. It's simpler *for me.* At the end of his solution, bprp had to look more than once to make sure of the substitutions at the end. I hit that same thing every time using the algebraic method for this particular problem. If you say it's not more elegant, fine. It's certainly easier *for me* to finish the substitutions.
2. If I'm using it to solve for a definite integral, and I've already pulled up (or coded) the solutions for sinx and cosx and stored them, then it's easier to multiply the two stored values than to pull up a third trig function.
3. Bprp has a video showing 4 ways to solve ∫ secx dx. The other three ways provide beneficial exercise and food for thought. He even has an alternative video (can't remember it off hand) where he shows integration using this same method - ∫ f(x) dx = ½∫ f(x)+g(x) dx + ½∫ f(x)-g(x) dx. So it's not just "my way" and he didn't avoid showing the overall method elsewhere because it couldn't be generalized.
I never said that one ought not learn what he taught. I even said that you'd probably need to know his way for a test - that instrument to exhibit long-term learning.
I thought that others who had missed the method would find it interesting. I'm not going to apologize because you found my comment pendantic - especially given that you felt the need to resort to numbered paragraphs.
I thought this was also math for fun and anyone is free to agree and laugh with me or disagree and laugh at me. Either way, it's all good.
Clearly I was dead wrong. Let me know if you want me to delete the comment (and therefore the thread), it makes no difference to me. Everyone who knows this channel knows who you are and respects you. I won't be bothering you again.
hello brother. I get a different answer for number 2 intergral ln(1-x^2)dx instead of 1-x i get x-1 and 1+x is same as x+1
Lol, I speak Irish but I don't know if that helps in the slightest
Almost an f-bomb at 27:35!
Hey Im a Calculus amateur. Just wondering what method did bprp used at 38:50. Thx in advance!
Hi! Four years later, are you still a calculus amateur?
Could you solve this integral? Integral of (secx)^(3/2). I wish you did it. Thanks for giving a lot of support
Angel Mendes This integral is non-elementary, so there is no solution anyone can give you that you would be satisfied with.
Thank you so much, bro
Sir, why don't you make a video about proving that the ramanujan formula
Only between you and me!😁
pAHsitive integral
m8 im in high school learning quadratics XD
could u do a video where u explain calculus and why it works sorry i just kinda don't get what ur doing and just don't get calculus - but i still sub
Its all about analyzing a graph of the function. Integral is giving u a surface area under a function. Derivative is the gradient of a line tangent to the function
@@LeeSeungrhee yes i got the practical part but the theory is really confusing (actual formulas etc)
Technically they all have answers, just not elementary ones
How did he found out that we can't do the other one?
How we can know what is elementary and what is not?
can you know if the integration is possible or not just by looking at it ? , and if yes how do you know ?