An Integration Conundrum - Numberphile
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- เผยแพร่เมื่อ 28 ก.ย. 2024
- Featuring Ben Sparks doing some calculus. See brilliant.org/... for Brilliant and get 20% off their premium service (episode sponsor)
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Ben is the unsung hero of this channel. Dude is incredible at opening your mind about math we're already familiar with
He is brilliant. A super communicator and always such enthusiasm 😊
I actually went through the third ring of Dijarmha and reached the eternal light of Nurrzhenzhis by watching this on LSD. opening my mind is an understatement.
@@aceman0000099 I’ve watched a lot of these on nutmeg; so, I can *_KINDA_* feel, where you’re coming from 😵💫.
@Zach Gates True and beautifully motivated.
Yes, Love love Ben!
Fantastic video. I’ve actually been struggling with this integral for an upcoming test. Very timely!
Glad to be of service!
Destin doing a level maths???
well i could have used it a few month back when i had my calculus exam!
??.
integration is an interesting area
ha
ha
Nc hahaha and derivative is a pretty interesting slope
That about sums it up well
Galois theory is an interesting field
I think it would have been informative to also discuss that since cos(2x) = cos²(x) - sin²(x) and cos²(x) + sin²(x) = 1, then you can work out that cos(2x) = 2cos²(x) - 1 = 1 - 2sin²(x), so that the 3 answers are indeed identical up to a constant [ (-1/4)cos(2x) is (-1/2)cos²(x) plus a constant and is (1/2)sin²(x) plus some other constant, both absorbed in the undefined constants in the answers].
It may also have helped to label the constants c1, c2, c3 and c4, to explicitly say they are not necessarily the same (they are not) though that would have given away the punchline.
Yeah I know it's "mathy" but we shouldn't skip the trig identities to show that all three are indeed the same.
Yeah, I was doing that in my head a bit during the video and was waiting for this reveal. Felt incomplete without it! But the graphing was nice too. :)
Right, and it might also lead to a confusion in definite integration, "there's no C so which method is right!" But since the functions really are the same up to a constant, any will work with a definite integral where the constant sort of cancels out.
Yeah when I show this to calc 1 students I stress that they are all equal up to a constant.
That's really helpful, thank you!
i have given this "conundrum" in every calculus class i taught. Most students tended not to have a problem with it, but rather they were entertained. The few who struggled afterwards seemed like something finally clicked. By this i don't mean my students were all aces in calculus, just that they were a little more clear on the meaning of indefinite integrals.
happy to see this "conundrum" appear on Numberphile.
What I like about this particular example is that the meaning of the constant becomes obvious. (Replace (cos x)² with 1 - (sin x)² and expand out, and you get the other integral with a +1 term - the (c + 1) terms group into a new constant.) It's something I wish I had been taught instead of having to figure it out, although figuring out the meaning of the +c was interesting in itself.
I love Ben Sparks' videos
Ben, is this your burner account?
@@numberphile If it is, it's so well disguised that I don't recognise it :)
I am not Ben Sparks, and so far have never claimed to be.
@@OneTrueBadShoe ...which is exactly what Ben Sparks would say if you were him, or he were you.
@@OneTrueBadShoe bro, he's just joking.
Numberphile going back to its roots! This is outstanding! You should have one of these a week, your audience would learn so much! Thank you for doing this one.
It wouldn't be a Ben sparks numberphile video without geogebra somewhere in it
What I want to know is why Ben didn't show that the answers were equivalent by more than one method?
Between the double-angle formula for cos and the Pythagorean identity, you can show that the three different expressions only differ by their constants (which is easier than sketching the three graphs to do on paper).
Because that wasn't the point of the video
I like a video that challenges my intuition. I know there’s multiple ways to integrate, but it hadn’t occurred to me that the functional expression might be different (though in effect equivalent). Very cool!
Oh nice you can do this by parts, substitution, trig identity and recognition, I'm definitely going to have to use this question in future
I really love this guy because he also uses visualization in a great way which can sometimes really clarify why something is the case.
Method 1 and Method 3 are equivalent because *cos 2x = 1 - 2 sin² x* so *-1/4 cos 2x + c = 1/2 sin² x + (c - 1/4) = 1/2 sin² x + c'*
Method 2 and Method 3 are equivalent because *sin² x + cos² x = 1* so *-1/2 cos² x + c = 1/2 sin² x + (c - 1/2) = 1/2 sin² x + c''*
That's what I missed in the video: you can use the trig identities again to actually show how the functions are equivalent
Yeah. I was also confused as to why did not use trig identities after his integrals to simplify, just like he did in one of his methods. The video title and mood, initially made the different answers seem paradoxical.
You need to make your next video shedding light on the trig identities that explain how to convert between the 3 or 4 very different looking formulas, given specific C values.
Agreed. It would also be great to see the double angled formula proved by using the unit circle.
Ben Sparks has got to be my favorite Numberphile regular.
this is why I fell in love with integrations, finding new and tricky functions to integrate and differentiate to check the answer! Introduction to calculus made me appreciate Mathematics big time :)
It doesn't matter how many times I see it, the fact that a trig function squared is jsut another trig function (plus some shifts and frequency change) never ceases to be unintutitive
There is actually another method, and it’s the easiest in my opinion
You can simply consider the cos(x)dx as a differential of a trigonometric function which is sin(x) so, cos(x)dx=d(sin(x))
Integral(sin(x)d(sin(x))) = (sin^2(x)/2) + C
And you get the same result as in the last methods
That's method 3 in the video, just phrased slightly differently. "cos(x)dx = d(sin(x))" is the same as saying "cos(x) = d/dx(sin(x))". Usually you'd just refer to this as u substitution, using u = sin(x), du = cos(x)dx. It's all the same thing, just the chain rule in reverse.
That's literally what he did for method 3. He just went through all the steps for why it becomes half the square.
@@stanleydodds9 Exactly
I think method 2 and 3 are both "the best" since it's basically a substitution, one of the more useful methods for solving integrals.
That being said, I went for partial integration and I'm glad it appeared in the video!
Method 5
Put sinx = t (now diffrentiate both side wrt x)
Cosx =dt/dx
Cosxdx=dt
Now just put the value of sinx = t & cosxdx = dt
To get integral of
t dt
Which is on integrating
(t^2)/2 + c .......(1)
Now put back the value of t (= sinx) into the eqn (1)
So the answer is
((Sinx)^2)/2 + c
Same ans as in method 4. But simpler than integration by part....
That's method 3.. :)
Interesting wrap-up at the end. Despite having never learned calculus, that part was insightful.
I think the average calc student would be more likely to use integration by parts because that stuff would be fresh in their mind. Trig identities not so much unless they're especially studious!
My Calculus professor used to say "differentiare humanum est, integrare diabolicum"
and never forget the trig unity, sin² x + cos² x = 1 which can be used to show that methods 2 and 3 are the same, just off by a constant
Method 3 & 4 show constant as the same value c. But methods 1 and 2 have different solutions so the constant can't be c, choose other values e.g. a and b. That makes it easier to explain and understand
This was a cool video! Reminds me of how 1/(x+1) and x/(x+1) has the same derivative, for the same reason. And it freaked me out the first time I saw it
An excellent example, but do you need to use -1/(x+1) and x/(x+1) (or vice versa), for them to have the same derivative?
1/(x+1) and x(x+1) don't have the same derivative. 1/(x+1) and -x/(x+1) do, and the reason (I know you know, but just to complete your idea for other readers) is that:
1/(x+1) = (1+x-x)/(x+1) = [-x + (x+1)] / (1+x) = -x/(1+x) + (x+1)/(1+x) = -x/(x+1) + 1,
which means that they are almost the same function except shifted one unit vertically one from the other, and hence they have the same slope for any value of x.
"Do you remember some calculus from your school days?"
"No, I don't."
Me, "OK cool, maybe I'll be able to follow this." (instantly totally lost)
“Do you remember calculus from school days?”
Me who didnt even lear it:”Yes.”
The notation of putting a number after fuction name normaly means apply the function that many times, but it is customary to use it with trigonometric functions to raise that function to the power, because those are not usually applied one after other. But of course they are in navigation and in survey where you can see sin(sin(x)) and others.
We can slightly change 3rd method:
try y=sinx
dy/dx=cosx
dy=cosxdx
Sinx*cosx*dx=y*dy
Integral (ydy) =y^2/2 + C = 0.5*(sinx)^2+C
That's why you should write the areafunction as definite integral: A(x) = ∫ _0^t f(t) dt
Because then, the constant 'C' is vanishing away, and you get the correct answer.
If f(x) = eˣ, then the ∫₀ˣ eᵗ dt = eˣ-1. By comparison, c=-1, so the lower limit should be the inverse antiderivative evaluated at 0 (i.e. F⁻¹(0)) in general.
@@adiaphoros6842
You're right. What I wrote is not always correct.
Because there are functions you can't evaluate at x=0.
F(x)= 1/x, F(x)=1/x², F(x)=1/(x+x²+..), F(x)=e^(1/x), etc...
There are better ways to handle this constant.
sin²x is poor notation because it suggests sin(sin(x))
Yeah I learned this back in calculus when I got a different answer than the answer sheet, so I graphed them on Geogebra and noticed exactly what this video explains.
use laplace transforms (laplacian of derivative) to more directly get the proper integral of any function, just have your laplace tables completed
I'm not sure whether this is already well known, but the double angle identities are trivially easy to rederive if you know Euler's Formula:
exp(iθ) = cos(θ) + i sin(θ)
exp(2iθ) = cos(2θ) + i sin(2θ)
exp(2iθ) = exp(iθ) · exp(iθ)
= (cos(θ) + i sin(θ))^2
= cos(θ)^2 + 2i sin(θ) cos(θ) - sin(θ)^2
Now equate the two sides:
cos(2θ) + i sin(2θ) = cos(θ)^2 + 2i sin(θ) cos(θ) - sin(θ)^2
Hence:
cos(2θ) = cos(θ)^2 - sin(θ)^2
sin(2θ) = 2 sin(θ) cos(θ)
I'm sure it varies from person to person, but I find it much easier to remember Euler's Formula than the angle identities so this is how I remember them!
You can easily extrapolate for the angle sum identities as well.
You know you're a fresh green physics student when you start with integration by parts.
I bet the engineers are laughing with glee as they jump straight into their trig identities
I did A level maths 53 years ago no maths since and I immediately saw it as integral y dy/dx dx and got y^2/2 where y= sinx.
Both method 2 and method 3 are going to save you at some point (3 works like a charm if you have both exponential and trigonometric functions at once). My favorite way of writing it would be
∫sinxcosxdx=
=(1/2)∫2sinx(sin)'(x)dx
=(1/2)∫((sinx)^2)'dx
=(1/2)(sinx)^2+c
though, because it happens in an uninterrupted line.
I learned 2 and 3 as u-substitution. You can write sin x as u. Then you take du/dx = cos x => du = cos x dx, which gives integral of u du = 1/2u^2 = 1/2 sin^2 x.
I insta-changed variables, sin(x)=s, ds=cos(x), answer is s²/2 ie answer 3 and 4. I found his explanation of "methods" 2 and 3 a bit confusing/shortcut-ish, it's nice to change variables explicitly and show what's happening.
Ben is the best math demonstrator on this channel. Period.
Could we get Ben to talk about the terrible sin squared notation on Numberphile or Numberphile2?
I think the main reason is that usually, putting a power on a function should apply to the function, not to the result of the function. Like sin⁻¹x=y is the same as sin y=x. It reverses the function. It doesn't mean 1/(sin x). If we were consistent about this, sin²x would mean sin(sin x), not (sin x)². So basically sin⁻¹ and sin² are two completely incompatible notations, and yet you see them next to each other all the time.
I really liked how using different approaches leads to different looking answers that all correct. The easiest method, for me, was u-substitution (let u=Sin(x) then du=Cos(x)dx). Using Trigonometric identities, the results can be shown as equal to each other. Now, showing that would be a great trig review for the students, though they might not think so. 😁
Yeah to me too. Although the half angle tangent substitution would have been fun too lol.
What I might have added on as an explanation is a more general interpretation of the relationship between derivatives and integrals. By performing an indefinite integral, you're essentially starting with the rate of change and asking what function has that (the original) function as its rate of change. Since a rate of change doesn't care about where you start, there has to be a degree of uncertainty (i.e. the constant at the end).
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I’ve always used U sub for these types of integrals. u= sinx du = cosxdx dx = du/cosx so the integral comes out to be u du.
Just another technique to add to the ones presented: since sinx cosx dx = sinx d(sinx), we can make a variable substitution y=sinx and integrate for y: Int(ydy) becomes 1/2y^2+C and finally, replacing back y, 1/2 (sinx)^2 + C. And of course, we can also go the other way around with -d(cosx)...
Game of integration constant and trigonometry. Expansion of Cos(2x)=2cos²x-1=1-2sin²x
Numbers concerted into integration constant.
The course I'm teaching recommends Geogebra, and I've found it fairly useful. My students hate graphing though, so they don't appreciate it as much. I just might show them this video (or assign it as homework). I know that the fact that there is no "the answer" and only "an answer" will drive them up the wall. :)
2:46 - The constant c is already missing in the row with the squared brackets.
In beam theory, when integrating from rotation to displacement, then twice to shear and finally to bending, the need to carefully establish what the constant is at each integration.
This is a brilliant observation. I learnt it when I was doing integration problems but getting the different answers for same problem.
Wow - so far over my head! This is the first time I've ever had to play a video at 3/4 speed, and I still don't have a clue. Guess I need a remedial math class 🙃 but even without understanding this video I enjoyed it, and I always love the Numberphile and Sixty Symbols channels.
Integration simply gives you an area under a curve. This is akin to saying you can find the area of a square by doing a*a, or a^2, or if you are more inclined, by splitting the square into two triangles, etc.
There are multiple ways of finding the same answer, all valid. In these examples, all the functions express the area under the curve, which is why all the results look the same with an offset.
@@neonglowmusic Thank you! I'll watch the video a few more times and see if I can get it.
Don't worry too much about it. Some of this is university level math, specifically the integration by parts.
If you're curious on how that works, it's a rearrangement of the division formula of differentiation:
f(x) = uv -> f'(x) = vu' + v'u.
Integrate both sides: f(x) = ∫vu' + ∫v'u
f(x) = uv -> uv = ∫vu' + ∫v'u
Therefore, ∫vu' = uv - ∫v'u
QED.
@@denny141196 Thank you! I I appreciate your taking the time to break it down further.
I don't think integrals are something you can sit down and watch slowly and suddenly understand. I mean no disrespect, like once someone defines an integral and sits you down in a calculus course, it becomes clear. Otherwise it's just a funky symbol.
Hey can Brady actually take an A-Level exam (a mock maybe?) I'd be really interested in how much maths has actually rubbed off on Brady while doing these videos. Is he actually learning anything or does it only stay active while his filming and editing and once it's on the web the knowledge goes flying out of his brain?
I worthy experiment if there ever was one.
That is my ACTUAL anxiety dream come true - that I am re-sitting maths exams in high school but don't know any of the material!
This channel is about exploring math concepts and ideas in way a non math person can get a grasp of. So his math ignorance is an integral, pun intended, part of the proces. Solving A--Level problems is kinda boring at least for me.
Thanks for tossing a brick at cos^2(2)
I doubt that's anyone's biggest reason to hate maths, but it can't be helping.
Yet another version, very similar to 2&3 (maybe you'd say the same only different) that I was also taught.
substitute y = sin(x)
Then dy = cos(x)dx
I = (sin(x))(cos(x)dx) = y dy = y^2/2.
Substituting sin(x) back in for y
I=(sin(x))^2/2
And identically for cos(x) to get the other result.
Ben Sparks is one of my favorate Numberphile presenters. Truly enjoyable!
Every one of these videos is another part of forgotten maths integrated into my head once again.
I wanna talk about sin²x being terrible notation because I AGREE SO MUCH.
Actually my most hated notation bs that I can remember :/
that's interesting that he said he thought students should use the double-angle or the integration by parts methods, at the school where I work there is a huge emphasis on u-substitution and I would expect almost all our students to use that method
@12:00 "What method would a school student have used?" I would always have checked integration by parts first, because it's fairly brainless and nearly always works, and you don't have to try to remember specific trig identities in a time-pressured exam context.
I have the worst kind of confusion now...the one where: "I probably actually knew this at one time, but NOW this is as confusing at the Infinte Sum of 1+2+3... is -1/12!!!" Still...I come back for more
That trick to make the sum of positive integers equal -1/12 isn't valid, though :P
The sum of positive integers is a divergent series, and cannot be manipulated by the same tricks that a convergent series can. The proof for that treats a divergent series like a convergent one in a few steps - which while it looks ok, it's actually violating some pretty fundamental rules and leads to a nonsensical answer.
7:27 finally someone said it
Seen this many times before.
All are equivalent, with a different value for the constant, that is determined by trig identities (like sin^2 + cos^2 = 1 or cos2x = cos^2(x) - sin^2(x) )
By parts is where I started. I didn’t know the other methods.
This reminds me of a thought I've had for some time, that calculus would not have the reputation it has if calculus teachers didn't have such an obsession with tricky trig identities. Calculus itself is, conceptually, not that hard, but as I recall Calculus I was more about banging your head against the wall with brain teaser trig identities that it was dealing with the concepts of slopes and areas under curves.
2:30 Sadly; there’s no Chain Rule, for integration. At least, not an elementary Chain Rule.
solve this with a substitution and the chain rule (u-sub). u = sinx du = cosxdx, integral of u du is u^2/2, bada bing bada boom you get your 1/2sin^2x + C
Nice challenge!
u=sinx du=cosxdx. Integral is u^2/2 + c. (sinx)^2 /2 + C
Ah, great example of why the constant term isn’t just some triviality teachers like to take off points for omitting.
I would have used a 5th method - u substitution. u=sinx, du=cosxdx, then just integrate udu.
The "d" in "dx" should be in an upright font.
I like half sine squared the most because c=0 when starting from x=0
I think that might be why geogebra chose that one, because it goes through the origin
The way he draws the letter "X" is so weird! It's like a backwards "c" and a forward "c" back to back! Is that something you learn to avoid mixing up multiplication symbols and the letter "x?"
Presumably you could algebraically show the answers are equivalent by subtracting one from another and seeing if you can simplify the result down to a constant. Not that that is necessarily easy to do.
Before you say "it's different," plug the two formulas into Wolfram Alpha and have it graph them...
(spoiler)
They _are_ different, but only if you ignore the *c* term.
It might be easier to change the identity. Sin(x)cos(x) = 0.5sin(2x). After changing it, just work from there.
@7:50 the notation is terrible for the reason that (sin^2)(x) == s * i * n * n * x or alternatively (sin^2)(x) == sin(sin(x)) depending on what you take juxtaposition to mean. And lots of people use sin(x)^2 -- computer programmers.
Very nice video! Especially liked the graphs in the end. I'd pick the 2nd/3rd methods (they are essentially the same).
Feel embarrassed that I didn’t know any of these fundamental lessons on why the constant c is so important
(Sin(x))^2 is by definition the same as sin^2(x)
Double angle is what we SHOULD use, but probably won't notice. U substitution is what we probably will use. Integration by parts is what we will avoid.
I used Leibniz method on this one and got (-1/4)*cos(2x).
Defined F(t) = integral( sin(x)*sin(x + t*Pi/2) )dx, then found that F'' + (Pi/2)^2*F(t) =0, solved for F(t) and evaluated F(1) = integral( sin(x)*sin(x + Pi) ) dx = integral( sin(x)*cos(x)) the desired integral
Integrating by u-substitution was my 1st thought
I went with method 3
Shouldn't method 4 have had c/2, not just c? Because he didn't divide by 2 until after the integral was resolved and c was present.
I realize it doesn't change that method 3 and 4 are the same if you forget to put c though, so the rest of the video is correct either way.
If possible it would be nice to see a video with Dr Holly Krieger. It's been quite a while.
I have never heard anyone say “cuz” rather than “cosine” when reading cos(x).
Also, from Day 1, I tell my students to always use parentheses around the argument. Clarity is paramount.
Also, one should type
(1/2)cos(x)
rather than
1/2cos(x)
because
1) it’s clear and
2) some machines are programmed so that implicit multiplication takes precedence over explicit multiplication and division.
Never type
1/2cos(x).
Type
(1/2)cos(x)
or
1/(2cos(x))
depending on which you mean.
It’s always better to have more parentheses than might be required than to not have a set that is necessary.
Avoid any potential ambiguity or miscommunication.
Can't say I could follow the techniques used for the solves (I don't know Calc at all), but I _did_ understand the underlying point! Cool stuff :)
BEN SPARKS!
I'm gonna have to send this video to my calculus teacher from 20 years ago. I bet they marked some of my answers wrong that weren't!
How would a student have done it when I was in secondary school (let's say 45 years ago)? They would have moved the cos(x) behind the "d" to make it "d sin(x)". Then you end up with "∫ sin(x) d sin(x)" so that's "1/2 * sin(x)^2+c"
Missing the c once let to me 'proving' 0=1 and then spending a lot of time finding the error. Using integration by parts on ∫cot(x)dx = ∫cos(x)/sin(x)dx with u=1/sin(x) and v'=cos(x), so v=sin(x) and u'=-cos(x)/sin²(x). Then it follows:
I = ∫cot(x)dx = ∫cos(x)/sin(x)dx = sin(x)/sin(x) - ∫sin(x)*(-cos(x)/sin²(x))dx = 1 + ∫cos(x)/sin(x)dx = 1 + I
Missing the c and subtracting the I from both sides then leads to 0=1.
I see Ben, I click.
I would have used precisely none of these methods!
Method 5: u substitution!
Let u = sin(x). Then du = cos(x) dx. This then becomes ∫ u du = 1/2 u² + C = 1/2 (sin(x))² + C.
You can think of the four answers all having the same slopes at every point. Thus they have the same derivative which is the first function being integrated.
12:30 I immediately went for integration by parts simply because I can never remember my trig. identities.
Can integrate using substitution too.
Use graphing might clear all the confusion
omg this would've been such a convincing reminder to never forget the "+c" for integration...
back when i studied this 20 years ago -___-