how Richard Feynman would integrate 1/(1+x^2)^2

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We can use trig substitution (letting x=tanθ ) to do a typical calculus 2 integral, the integral of 1/(1+x^2)^2. However, we will use Feynman's technique of integration (aka Leibniz's integral rule, aka differentiation under the integral sign) to come up with a very nice integral formula.
Here's the integral of 1/(1+x^2) by using complex numbers 👉 • integral of 1/(x^2+1) ...
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[@AndrewDotsonvideos]

He has another famous technique (at least for people who calculate Feynman diagrams) called using Feynman parameters. It’s a way of re-casting an integral you’re solving into a form with temporarily more integrals that make the original integral easier to evaluate. Of course this is only helpful if the remaining Feynman parameter integrals can be solved analytically or are at least less expensive to solve numerically (it’s usually the latter). Not sure if you’ve ever made a video on it, but in the same spirit of Feynman integration tricks!

[@maalikserebryakov]

Dude you’re still interested in symbolic calculus?

[@AndrewDotsonvideos]

@@maalikserebryakov never know when it’ll help with an integral I’m trying to solve for research 🤷🏻‍♂️

[@alozin5339]

yo whens the next upload

[@nicolasmendoza6183]

@@maalikserebryakov buddy, you're watching the wrong channel if you're not interested!

[@abrarazad1571]

You also watch him????

[@plainbreadmike]

Can't get enough of these integrals with Feynman's technique videos, they're just so satisfying!

[@SydneyWeidman]

P

[@aurelio3532]

yes!!! keep making more please!!!!

[@hfgfgnnfgng5562]

Fr ...

[@gagadaddy8713]

This kind of Integration trick is really OUT OF THE BOX, only from the brain of those genius ..... not the ordinary maths student 😆

[@epikherolol8189]

Usually that's the case but we use these techniques after their inventions to carry on their legacy. That's how new discoveries are made​@@gagadaddy8713

[@flowingafterglow629]

That answer is really slick, because if you look at it, the first term just looks like the integral of 1/u^2, where u = 1+x^2 and then the second term is just some version of the integral of 1/(1+x^2) Random factors of 2, I agree, but the form is pretty cool

[@davidalexander4505]

For definite integrals, I now see that there is actually no differentiating under an integral sign (requiring something like dominated convergence theorem) it's actually much prettier, we can write it as follows: d/dx (1/a arctan(x/a)) = 1/(a^2+x^2) Hence, d/da d/dx (1/a arctan(x/a)) = d/da 1/(a^2+x^2). By commutativity of partial derivatives, d/dx d/da (1/a arctan(x/a)) = d/da 1/(a^2+x^2). Thus, an anti derivative for d/da (1/a arctan(x/a)) is d/da 1/(a^2+x^2). (then work out these partial derivatives)

[@NarutoSSj6]

Rip Chen Leu. Although maybe never uttered by name again, you have a special place in all our hearts.

[@vishalmishra3046]

This video has an innovative new method of solving such integrals. Here is the old boring way for the same - set x = tanT which changes the problem to INT { cos^2 T = (1+cos2T)/2 } dT = T/2 + sin2T/4 = T/2 + 2tanT / 4sec^2 T = [ arctan x + x / (1+x^2) ] / 2

[@AyushGupta-cj3sy]

We indians flooded everywhere 🤣🤣

[@UnknownGhost97]

@@AyushGupta-cj3sy This Equation says about feyman technique

[@AyushGupta-cj3sy]

@@UnknownGhost97 buddy i mean to says indian 🇮🇳could easily solve these

[@AyushGupta-cj3sy]

@@UnknownGhost97 I understand Bhai But it's mostly in their higher studies But we have in 12

[@UnknownGhost97]

@@AyushGupta-cj3sy Hey im an IT professional i can solve these problems easily just exploring at these logics here

[@jellowz3556]

Thank po sir! I hope you will also teach this topic "definition of exp z for imaginary z" under the linear equations with constant coefficient

[@SlipperyTeeth]

"Why do we add the +C at the end?" It depends on what you consider integration to be. Normally we just think of integration as the opposite of differentiation. But then, what is differentiation? If you think of differentiation as a function from functions to functions, then integration should be its inverse function. But there isn't in general a left inverse for differentiation, because it's not one-to-one - and there are multiple right inverses. So you might consider "integration" to be the entire set/class of right inverses of differentiation - such that whenever you compose "integration"/differentiation, you pull back this abstract layer of set/class and compose them with every instance of an integration function. So differentiation after "integration" is just the set/class of differentiation after right inverses of differentiation - which all collapse to the identity. And there's the added bonus that with just a little more information (such as a single point on the curve) you'll be able to choose one of those integration functions to "act" as a left inverse for a specific input - so the whole set/class of integration functions can act as a left/right inverse for differentiation. For single variable calculus, that's about all you need to consider, and this is a perfectly fine way to define the integration notation. For multivariable calculus, there's a new wrinkle. You can have a function that's constant in one variable, but not another (Let f(x,y) = y, then d/dx (f(x,y)) = 0). So if you integrate a function in the variable x, then you pick up a constant in the variable x. And then if you differentiate that by the variable a, it doesn't always become 0, because "constant in the variable x" doesn't imply "constant in the variable a". Sure, some functions are constant in both x and a, but not all. So if we compose differentiation with "integration", some of those compositions will collapse the constant, but not all. We didn't add +C at the end, it never should have been removed.

[@Francesco-bf8cb]

I've tried a differents (but much longer) method You know when you differentiate f/g you get (f'g-g'f)/g², so know it becomes a differential equation

[@MathNerd1729]

I recall seeing you do this in October 2018. Still a very neat video! :)

[@chanduudarapu7906]

Thank you so much, lots of love from India 🇮🇳

[@adamlopez2339]

wow, what a nice way to solve this integral. Thank you for the video

[@pirnessa]

One mistake: C is a constant in terms of x not in a. Hence, the partial derivative d/da C is not zero in general, it is just another constant in terms of x (which you added back in the end). Nice video! (PLEASE SEE EDITS BELLOW BEFORE YOU COMMENT) Edit: C may depend on parameter a however it wants. Thereby, C can be any function of parameter/variable a, and so, it might not be differentiable with respect to a. So in the end, the best way is just to find one antiderivative of ∫1/(a^2+x^2)^2dx. Then when we set a=1, we now know one antiderivative of ∫1/(1+x^2)^2dx. But we know that all the other antiderivatives of ∫1/(1+x^2)^2dx are obtained by adding a real constant to the antiderivative we already know, since g'(x)=0 for all real x if and only if g is a constant function. This is basically what @blackpenredpen did, but the reasoning that C is a constant with respect to a is not right although it is irrelevant mistake for the main point of the video. Edit 2: First of all my claim is that ∫1/(a^2+x^2)dx=(1/a)arctan(x/a)+C(a) where C:R->R is any function and R is the set of real numbers. If you think that I'm wrong and that ∫1/(a^2+x^2)dx=(1/a)arctan(x/a)+c, only when c is just any real number, i.e. constant with respect to both a and x. Then your claim against my claim is that if C:R->R is not a constant function, then (1/a)arctan(x/a)+C(a) is not an antiderivative of 1/(a^2+x^2) with respect to x. To help you, I will go through what you are trying to prove and why it is not true. So you need to take a function C:R->R which is not constant, for example you can think C(a)=a. Then you need to prove that (1/a)arctan(x/a)+C(a) is not an antiderivative of 1/(a^2+x^2) with respect to x. But you know the definition of antiderivative for multi variable functions, so you know that by the definition you need to prove that the partial derivative d/dx ( (1/a)arctan(x/a)+C(a) ) is not equal to 1/(a^2+x^2). But we know the following partial derivatives: d/dx (1/a)arctan(x/a)=1/(a^2+x^2) and d/dx C(a) = 0. So by the linearity of partial derivative you have d/dx ( (1/a)arctan(x/a)+C(a) )=1/(a^2+x^2). Thus, your claim is wrong and we have ended up proving that (1/a)arctan(x/a)+C(a) is an antiderivative of 1/(a^2+x^2) with respect to x if C:R->R is any function. In the comments you can find also different reasonings and how other people realized this. If you still disagree, please read the 50+ other comments in detail, read my arguments, read others arguments, read why in the end they realized that C can be a function of a. Our comments are not the best source so I also recommend studying or recalling multivariable calculus and before that one variable calculus. Even better is to go to talk people in some university's math department. If after this you still feel that I'm wrong, then G I M M E A V A L I D P R O O F of the direction you are claiming and cite to my previous comments and show where I went wrong so the conversation is easier and faster.

[@tobechukwublessed4274]

A constant is a constant, independent of any variable... That's what I think. So it's pretty much staright, no mistake

[@pirnessa]

​@@tobechukwublessed4274 Yes you are correct, but here C is only a constant in terms of x. So first we were just in the "x-world" where C is just a constant. But when we introduce the parameter aka new real variable a we are not anymore in the "x-world", we are in the "xa-world" where C could depend on a while it does not depend on x. It might sound nit picking but this is really important in multivariable calculus. One real variable: Here when we talk about integration in one variable we mean antiderivative aka inverse derivative aka indefinite integral, i.e. that if F'(x)=f(x), then ∫ f(x)dx=F(x)+C, where C is just a constant aka a real number. So the antiderivative ∫ f(x)dx gives the set of all functions F whose derivative is f. Two real variables: In this video to do the Feynman's trick our function depends on two variables, namely x and a. We want to have the same property as in one variable, that is, ∫f(x,a)dx gives the set of all functions F whose partial derivative with respect to x is f. Thereby, if d/dx F(x,a)=f(x,a), then ∫f(x,a)dx=F(x,a)+C(a), where C is now a real function, which could be a constant. Suppose that we allow C to be only a constant, this is a valid definition but not very useful which I try to clarify by the next example. Let F(x,a)=x+a. Then d/dx F(x,a)=1, and so, ∫d/dx F(x,a)dx = ∫1dx=x+C. If we don't allow C to be a function of a we don't have the nice antiderivative property mentioned above, i.e. F(x,a) do not belong to the set of functions obtained from the antiderivative ∫d/dx F(x,a)dx. More practical example why we want that the antiderivative property is satisfied is that we want to have working tool to solve partial differential equations. Also @Phoenix Fire has really nice comment also in this comment section which clarifies this thing. Hopefully this clarifies.

[@tobechukwublessed4274]

@@pirnessa it's a cool observation, it really reminds me of solving ODE's by method of Exact equations where some constant may pertain to some variable after integration. But... If you examine carefully what he did, he differntiated both sides partially with respect to a... Now in partial differentiation, the only thing that is permitted to stand is the variable which we are differentiating with respect to, all other variables and/or constants will be assumed as constants for the time being, and what happens when we differntiate constants?... They vanish!

[@tobechukwublessed4274]

But When we integrate partially, a constant function of the other variables apart from the one we integrate with respect to comes in place.... For instance.... integrate an f'(x,y,z) partially with respect to x will yield f(x,y,z) + h(y) + g(z) where h(y) and g(z) is any function of y and z respectively, be it a constant function or any other type.... But when we differntiate partially, all constants what so ever must vanish. That's my point...

[@tobechukwublessed4274]

Finally, he was integrating, yes, but he differentiated while integrating, that's the beauty of that method. It's stainless

[@neutron417]

Didn't thought bout that amazing technique!

[@rjc3343]

Bro, you're so talented, I have my undergrad as a mech E and still come to your page for fun! Please don't stop the videos lol

[@maalikserebryakov]

How’s mechanical engineering treating you 8 months later? Symbolic calculus isn’t going to help you as an engineer you know. I bet you’re thinking of dropping out.

[@jmz_50]

How are you currently doing? Just curious, good luck btw

[@rjc3343]

Man that's a very toxic outlook on math and its integration (no pun intended) into engineering. I use the skills taught in diffEQ almost daily. And yeah, it was the best decision of my life to leave active duty in the military and pursue that degree. I have a solid job that keeps me entertained daily. And yeah, I'm not a quitter, hence already having the undergrad when I wrote this originally, but thanks for your concern. @@maalikserebryakov

[@rjc3343]

I'm good, love the degree. Work with a lot of EE and Physicists that treat me and my ideas with a lot of respect. @@jmz_50

[@lumina_]

​@@maalikserebryakov why are you being so negative? You seem like a miserable person to be around

[@haaansolo8568]

I learned about Feynman's trick when it came up in Howard and Sheldon's fight on tbbt, and have been stunned by it ever since.

[@SahajOp]

What is that

[@a_beats5529]

@@SahajOp tbbt is the big bang theory, a famous american sitcom

[@6612770]

What is the episode name?

[@think_logically_]

Is it legitimate to use Feyman's trick with indefinite integrals? An indefinite integral ∫ 1/(a²+x²) dx is in fact 1/a arctan(x/a) + C(a), where C(a) is not just a constant, but any function depending on 'a', but not on 'x'. Then the derivative d/da has an additional term C'(a) which is impossible to find ! In this case it's a mere coincidence that result is correct (if it really is). Take another example F(a)= ∫(x^a) dx =x^(a+1)/(a+1)+C(a). If you ignore C'(a), differentiate F'(a)=∫(x^a) ln(x) dx = x^a - x^(a+1)/(a+1)², then let a=0, you get ∫lnx dx = 1-x, while the correct answer is ∫ lnx dx = xlnx - x + C. BTW, I don't believe Feyman himself ever used his trick with indefinite integrals, so the title of the video looks misleading to me. 😊

[@jirisykora9926]

I don't know if anyone wrote it before but if you plug in -1 for a it's going to be the same as for a=1 because of the nature of the formula, where a^3 and tan^-1 cancel the negative sign of each other.

[@CliffSedge-nu5fv]

7:20

[@orenfivel6247]

i thought U gonna do IBP w/ DI method😁. By the way 4:57 the constant C, it does not depend on x, but might be depend on a (Like in an Exact ODE solving procedure). Thus technically, when differentiate WRT a, we should have C'(a) which is another constant that des not depend on x, and eventually U will rename the last integral constant as C or c or whatvever u wanna 😁

[@BetaKeja]

Yeah, I had to pause at 7:04 when he added the C back. Nope, C should not have been removed. It is constant w.r.t. x not a.

[@epikherolol8189]

​@@BetaKejaC is technically the arbitrary constant of integration which is just a variable number. But it is a number after all and thus differentiation of any number with respect to anything is 0

[@maciejkubera1536]

Great video as usual! On 0:49 You forgot to square the constant c ;) ;) ;) ;)

[@frencyii5370]

You can also solve it by substituting x=tan(u), it allows u to simplify until coming to intregral(cos^2(x)), which is easily solvable with some goniometric formulas.

[@dictetord12]

Correct

[@helloworld2024-h8i]

Did the Same thing :)

[@CliffSedge-nu5fv]

Any time I see a sum or difference of squares, I immediately reach for trig sub (assuming an inverse-chain rule "u-sub" wouldn't work).

[@joykukreja4270]

Cool method. I did it by putting x = tan theta in 5 steps.

[@blackpenredpen]

Learn more problem-solving techniques on Brilliant: 👉 brilliant.org/blackpenredpen/ (20% off with this link!)

[@와우-m1y]

asnwer=1 isit

[@NurHadi-qf9kl]

Itu jenis intgral pecah rasionak. Maka dimisalkan 1/(1+x^2)= {(Ax+B)/(1+x^2)}+{(Cx+D)/(1+x^2)^2} Dari asumsi tsb, konstanta2 ABCA dapat ditemukan 1=(Ax+B)(1+x^2)+(Cx+D) 1=Ax+B+Ax^3+Bx^2+Cx+D 1=(B+D)+(A+C)x+Bx^2+Ax^3 Yg berarti A=B=0; D=1; C=0 Int menjadi

[@ivanegorov3425]

一个小小的建议，您可以在视频的演算完成后留一两秒左右方便截图，视频很棒，感谢您的付出！

[@blackpenredpen]

好, 謝謝! btw, 這部也有中文版的 (自己幫自己打廣告 哈哈) kzbin.info/www/bejne/qZmYlYRnj92Zirs

[@pashaw8380]

​@@blackpenredpen You need to take a read at the above comments made by @Sakari Pirnes . Some of those here are professional grad students who have gone through real analysis and functional analysis. In your video you made a mistake by considering C as a normal constant treated the same as that in single variable calculus. But the C here is not an ordinary C; it is a function of 'a' not necessarily zero after being differentiated with respect to the parameter 'a'.

[@arkadipray1210]

What if we substitute x = tan a , Then , dx = sec² a da Substituting in above 1/(1+x²)² dx → sec²a/(1+tan²a)² da → sec²a/(sec²a)² da → 1/sec²a da → cos²a da → 1/2 × 2cos²a da → 1/2 × (1+cos 2a) da Integrating we get , 1/2 (a + (1/2) sin 2a) + c Now, since x = tan a And we know sin 2a = 2tan a/(1+tan²a) Therefore, sin 2a = 2x/(1+x²) → 1/2 ( tan^-1 x + (x/1+x²)) + c That's the answer....

[@CliffSedge-nu5fv]

That's not the point of the video. Yes, you can do that to verify the result, but the point of the demonstration was not to evaluate the integral, but rather to show how Feynman's technique can be used. It's like using trig sub to evaluate int(x/(1+x²))dx. Of course, you don't need to use trig sub, just do u=1+x², but that's not the point. It is to demonstrate how trig sub works using an integrand that is easily handled some other way to check the result.

[@manjumanl5279]

Just one question here ; Who gaves you the right to deffereciate partially ????? please explain !

[@blackpenredpen]

Leibniz

[@manjumanl5279]

@@blackpenredpen Hhhhh , simple ,hhhhhh

[@DavyCDiamondback]

So does it wind up not mattering that the value of c, for any constant boundary conditions, is variable with respect to changing the value of a???

[@CliffSedge-nu5fv]

Yikes. Reading through the comments, there are a lot of people confused about the purpose of the video. This is not a video on how to evaluate int(1/(1+x²)²)dx. It is a video about Feynman's technique using that integral as a simple example. The point is not to evaluate that integral. Anyone with basic calculus knowledge can do it by parts, partial fractions, trig substitution, and probably several other strategies. The fact it can be done easily some other way is why it is a good example. You can verify the result yourself using your favorite method - or, better, try doing it two or three different ways to make sure you can get the same result each time.

[@dodokgp]

Nice! The caption of the video could have been "Integrating a function without integration"

[@nowhereman000]

If Bro can make an entire playlist on feynman's technique : I one over zero percent sure I will watch it completly.

[@kokainum]

How about the integral of 1/((1+x^2)(1+x^y)) over (0,inf)? I've heard it doesn't depend on y and it's always pi over 4, but idk how to prove it. I guess it's using Feyman's trick.

[@CliffSedge-nu5fv]

If you are integrating wrt x, then y is treated as a constant.

[@abhishekchoudhary4689]

Just substitute x=tan theta you will get theta/2 + sin2theta / 4 where tantheta =x

[@Zuhair_Sadman_Mahir]

Good

[@Zuhair_Sadman_Mahir]

Best !!! Thanks

[@CliffSedge-nu5fv]

The purpose of this video is not to evaluate the integral. You're missing the point. Yes, you can do that to verify the result, but the point of the demonstration was not to evaluate the integral, but rather to show how Feynman's technique can be used. It's like using trig sub to evaluate int(x/(1+x²))dx. Of course, you don't need to use trig sub, just do u=1+x², but that's not the point. It is to demonstrate how trig sub works using an integrand that is easily handled some other way to check the result.

[@gio5969]

Calculus, the most powerful mathematics in the world and it would blow your mind clean off, you've gotta ask yourself one question: "Can I integrate? Well, can ya, punk?"

[@CliffSedge-nu5fv]

In general, there are infinitely more functions that cannot be integrated than ones that can, so the ratio of integrable functions to all functions is zero. Therefore, the answer to your question is "no, I a punk, cannot integrate (in general)."

[@MahfuzRaian]

What is the rule of differentiating the integrand inside the integral?

[@MochaE44449]

I have no idea what any of this is, but it’s fun to watch

[@8bitenginedayo]

Bruh this was literally a question in my homework today

[@rafikyeghoyan6634]

why are u allowed to swap differential and integral?

[@blackpenredpen]

Leibniz’s rule

[@nishantmiglani1952]

wait what about when a = i, you get the integral of a real function as a complex function ??

[@abrahammekonnen]

I guess I'm just confused on the point of the video: was it to solve the integral by using the technique, or just show off the technique? Both are good just thought it was the 1st not the second.

[@CliffSedge-nu5fv]

It is to show off the technique. This integral is used as an example, because it is easily evaluated several different ways - by parts, partial fractions, trigonometric substitution, etc. - so anyone can verify the result using their preferred method. It's not like this is some crazy-hard integral, and this is the only technique that works; anyone with basic calculus knowledge can do this integral easily using standard calc 1 or calc 2 techniques. It's like if I used int((1+2x)²)dx as a beginner's example to show how to use u-substitution. Of course you don't need u-sub, but that's not the point of the demonstration.

[@PapiCiencias]

i love you man, you are very carismatic even not trying it

[@VijayMishra-uz5dc]

Integrate 1/(x²+a)^n+1 from 0 to infinity . Can you integrate it

[@PREMSINGH-bu2kf]

Why can't we use partial fraction method ??

[@CliffSedge-nu5fv]

You may use whatever method you want. That's not the point. This video is demonstrating how to get the correct result using Feynman's method. You can get the same result using trig sub or partial fractions, or ...

[@veerdabas5578]

You could just solve it by substitution taking x= tanθ and dx = sec^2θ now in the denominator (1+tan^2θ) = sec^4θ now in the end we get cos^2θ and using cos 2θ formula we get θ/2 +cos2θ/4 now by subst. in the end we get tan-1x/2 + 0.5((x^2)/(1+x^2)) simplest method i could have thought about .Takes about 2 minutes to solve .

[@CliffSedge-nu5fv]

You're missing the point. Yes, you can do that to verify the result, but the point of the demonstration was not to evaluate the integral, but rather to show how Feynman's technique can be used. It's like using trig sub to evaluate int(x/(1+x²))dx. Of course, you don't need to use trig sub, just do u=1+x², but that's not the point. It is to demonstrate how trig sub works using an integrand that is easily handled some other way to check the result.

[@XKhanBTC]

This video made my day🔥

[@venkybabu8140]

Write all the numbers on the number line with powers and find what is left. 1 square. 2 3 gap 5 6 7 gap 10 11 12 13 14 15 gap 17 18 19 20 21 22 23 24 gap and so on. Highest gap is somewhere around 300 and after that no gaps. Higher integrals merge. That's why the universe has a huge black hole rubber band around it.

[@onbushshifting4000]

So what happens when u have mutiple parameters?

[@IitBombay-o3z]

Hey, why we can not substitute x = tan(Q) i am getting different answer Reply my comment

[@joluju2375]

What is the key idea behind this technique, I mean how could one think of doing this, and hope it could work ? It looks like if instead of playing with x, we're playing with the 1. It is as if instead of taking the direct route, we find it easier to take a detour. I would like to understand the reasoning, at least in broad terms.

[@gcewing]

How to think of things like this: Step 1: Be Richard Feynman. Step 2: Think really hard. Step 3: Write down the answer.

[@trueriver1950]

The hint is that Feynman was a physicist not a mathematician

[@maalikserebryakov]

@@trueriver1950 feynman didn’t even invent this technique, he just learned it from a book titled “Advanced Calculus” by Woods. So you can remove his ballsack from your mouth now.

[@trueriver1950]

@Maalik Serebryakov is true that RF did not invent this technique, despite it now being commonly named after him. However, unless you can tell me differently, Woods never got a Nobel prize in either Maths or Physics. Feynman did, for his ground breaking work on QED. He also invented a method of summing a set of path integrals where there are several different paths to consider -- the Feynman diagram (which is more than a pretty picture, as each component of the diagram can be calculated using separate integrals). Are we expected to change the name we use for RSA asymmetric cyphers now we know that British mathematicians beat those three Americans to that technique? I don't think so. My guess is that this technique we now often referred to as Feynman's technique became associated with the Feynman diagram because it is often needed when calculating the separate integrals. Whether my guess is right or not, Feynman did not claim the idea of differentiating under the integral as being original to himself, the name got put on the technique by others. Wood's technique (if you want to insist on calling it that) is also probably inspired by perturbation theory, which is used in both classical and quantum mechanics: certainly Wood's work (or that of previous mathematicians) makes perturbation theory rigorous, which it wasn't at the time physicists started using it. If you read his description of Babylonian vs Greek mathematics, or watch the video of the address where he describes that, you will see the point of my comment: what a mathematician wants from maths is different on a philosophical level from what a mathematician needs, and a lot of Feynman's work is not rigorous as would be required in a mathematics course. And I don't feel it's appropriate to be gratuitously offensive in a non family-friendly way to make the point that this technique has became known after the person who popularised it. If you have ever referred to Pythagoras's theorem then the same comment could be made of your interaction with Pythagoras the theorem having been previously invented by both the Babylonians and the Chinese; and Pythagoras would certainly have known of the work of the Babylonian mathematicians. Theorems and methods become associated with the people who made them famous, not always the same as the people who originated the idea. Get over it, and don't be an offensive troll. And for his work on QED and his later work on QCD I do happen to think, as a physicist, that he was the greatest physicist of the 20th century. (Yes I rate him above Einstein). That does not mean I would go to bed with him, not even if he were still alive (which sadly he's not).

[@trueriver1950]

@@maalikserebryakov PS the technique was not invented by Woods either -- it's sometimes named after Leibniz who clearly know it. I don't know if Leibniz invented it either.

[@bottom-up1981]

物理里这种操作真的多，代数求和或者积分结构加偏导，真的是很漂亮的做法

[@pashaw8380]

然而他在影片裏展示了錯誤的訊息，他將C看成了一個普通的常數，但C并不是一個普通的常數而是一個C(a)的實變函數，他在影片裏把它看作了C(x)因此直接偏導得零，這是一個數學裏很大的錯誤（即使他很幸運地得到了正確的結果），他思考的不夠嚴謹，沒有往多變數微積分的方向去思考。如果他瞭解汎函分析和實變函數論，他就不會犯這個錯誤。倘若你用已知 ∫ 1/√(a² + x²) dx = ln ∣x+√(a² + x²)∣ + C 的訊息去用他上述的方式解 ∫ 1/∛(a² + x²) dx 你就會知道我的意思了，不會得到正確的答案。費曼的方法其實就是萊布尼兹的積分法則，但萊布尼兹法是建立在有界的積分上，并且要遵守收斂定理，不可隨意亂用因爲那是有局限性的。這裏的作者拿它來解不定積分就是一個錯誤的做法了。這個如果不去做深入的講解會誤人子弟的。數學是一個非常嚴謹的學科，不能有漏洞的。

[@bottom-up1981]

@@pashaw8380 多谢老哥指点，我并非数学和物理出身，做的也是一些无关紧要的脏活，很多东西确实不太了解。

[@bottom-up1981]

@@pashaw8380 对于泛函，我就在分析力学，qm以及工程数学中有浅薄的了解，推导和证明是很快乐的事情，但更多拿来算。

[@user-wu8yq1rb9t]

Feynman is here! ... Cool I love Feynman (he's my favorite scientist ever).

[@harshbansal7524]

sir please of this: integral of sqrt(x^3+1) w.r.t dx

[@dr.rahulgupta7573]

Excellent presentation 👌

[@chivoronco4853]

The constant C may not dissapear by taking partial wrt a since it may depends on a. With this he dont need to add C at the end 8:52

[@박찬혁-u1m]

This is what I really wondered about.. Could you explain why it shouldn’t disappear?

[@박찬혁-u1m]

I didn’t understand why there should be constant at the end. I thought that the fact that lhs is an indefinite integral doesn’t fully substantiates the suddenly appeared arbitrary constant, but the constant must be there anyway.. So there must be some reason the constant remains there, and I think your explanation help me understand it much better

[@CliffSedge-nu5fv]

In this case, +C is +C(a) for a=1, since a=1 was the condition given at the beginning.

[@dorol6375]

What do sin(cos(sin(cos(sin(cos..(x) and cos(sin(cos(sin(cos..(x) converge to?

[@SuperYoonHo]

Thank you!

[@WizardAmbrose]

I have a question for you. If Sam has 2 apples, Rick has 10 apples and Sandra has 3 apples, Calculate the distance between Earth and Sun.

[@trueriver1950]

This is not an appled math class

[@jhondoe1618]

No hay duda de que Feynman era un genio. Gran video, saludos desde Santa Marta, Colombia

[@yusufshaharyar8528]

Isn't there a much traditional method We can add and subtract x² And then separate the two terms from it Like (1+x²)/(1+x²)² - x²/(1+x²)² The first term here can be integrated easily And for the second term, integration by parts can be done by taking just x as the first function and the rest as second function

[@The-Devils-Advocate]

Yes, but this is another way that can be more interesting

[@dyhsehehb6232]

Substituting x=tan(theta) is better

[@maalikserebryakov]

Cant we do partial fraction decomp ?

[@The-Devils-Advocate]

@@maalikserebryakov but would that be the Feynman trick?

[@maalikserebryakov]

@@The-Devils-Advocate there is no feynman method

[@vietdungle1237]

What a nice idea to integreat the seemingly impossible func

[@fyrerayne8882]

Thanks

[@awarecitizen3732]

I might be really dumb to ask this but the answers don't match so how does this help us..?

[@user09832]

This happens a lot in indefinite integration where two methods gives right answer but look different. To verify your answer, you have to subtract both answers and if you end up with a constant, your answer is correct

[@ritikraj26_]

Wow. Why did I never think of this?!?

[@madhavharish2575]

Bro!! you could have done this by just substituting "x" as "tan theta", which will give you integration of Cosine square theta w.r.t theta and that's very easy to solve without using Feynman's Technique!! Btw the Feynman's technique is excellent!!😁

[@CliffSedge-nu5fv]

Yes, you can do that to verify the result, but the point if the demonstration was not to evaluate the integral, but rather to show how Feynman's technique can be used. It's like using trig sub to evaluate int(x/(1+x²))dx. Of course, you don't need to use trig sub, just do u=1+x², but that's not the point. It is to demonstrate how trig sub works using an integrand that is easily handled some other way to check the result.

[@illumexhisoka6181]

Brilliant First time seeing this trick

[@adityasharma5692]

Can we do this by x=tantheta and then diffrentiatte that functin w.r.t to x then putting the value of dx through this and then final it become (cos)²theta dtheta ??? Can we do this

[@CliffSedge-nu5fv]

You can do whatever you want. There are at least three different ways to evaluate this integral. But that's not the point. The purpose of the video is to show Feynman's technique in action. It was never claimed that you must do it that way, only that it works.

[@Trizzer89]

Wha I dont understan about Feynmans trick is how do we know which function to start with. Just guess?

[@averagegamer9513]

Feynman technique of integration is OP!!

[@holyshit922]

known for more than two centuries before Feynman

[@mathematicalguru2602]

Put x=tan theta you will see a magic 😀😀

[@nimmubhai4779]

can we even find the integral of 1/(1+x^2)3 ???

[@CliffSedge-nu5fv]

Could try partial fractions.

[@xenoxander9274]

I have better soln. What if we add and subtract x² in numerator, and then solve ...

[@smith_7347]

Hey, I got a problem which boggles my mind. Let's assume we have a limit: Lim An/Bn as "n" approaches infinity ("An" and "Bn" are some sequences). I was wondering if we can substitute the "An" with another sequence (Cn) which satisfy the condition: Lim Cn/An = 1, as "n" approaches infinity. The condition says that, when "n" is enormous", the Cn and An are basically the same (with great approximation), so this is when the question comes - Will the substitution change the value of the limit?

[@konchady1]

No, the substitution wont change the limit. This follows from products of limits rule and noticing that as the Cn/ An limit is finite, we dont run into indeterminate forms. In fact, An/Bn is allowed to diverge to +/- infinity and this would hold.

[@juhakivekas2175]

Man, that was beautiful!

[@oraz.]

This seems easier then trig substitution

[@aniketde357]

what if i took x=tantheta? wont it be shorter

[@BABIN_2020]

did not understand why partial derivative undet integral

[@coder6783]

Simply put x = tan theta and then solve the given integral. Single line solution 😅

[@prakashgupta8342]

Integration of 1/(1+x⁴) please

[@necro5379]

Can be done IBP.

[@gregotieno5457]

wow my math guy🔥👊💪

[@shone7064]

It's so reminiscent of eigenvalue equations

[@zhelyo_physics]

Love this!

[@cwater9795]

Can you use same technique to integrate 1/(x^2+x+1)^2 ?

[@CliffSedge-nu5fv]

I don't see why not. You can complete the square to put denominator in the same form as (a²+u²)²

[@arkasaha2156]

Sir, how do we integrate when dx is in denominator... I mean : integral something/dx

[@kavi5990]

U can't, integral basically means sum of area of infinite rectangles which u get by f(x)dx F(x) /dx is meaning less

[@arkasaha2156]

@@kavi5990 yeah, but I had encountered a problem, don't exactly remember what it was, where there was this scenario when the dx was in the denominator

[@kavi5990]

@@arkasaha2156 would love to know the problem Actually I have seen a similar problem it is integral of (e^dx)-1 Thats it there's no dx multiplied with (e^dx)-1 So how do we integrate it I don't really know if this is legit the answer I came up is thatt since dx is an infinitesimal( very small there fore very near to zero)(e^dx)-1 from Taylor mclauren series could be aproximated to dx Basically lima--->0 (e^a-1) /a =1 Then we can get it to be the integral of dx which is x

[@Ashish-li5yw]

How do one make sure that the end result is obtained from what specific standard result. I mean had there been a cube ... The how to proceed ?

[@harshitpal4144]

a interesting problem for you----- Consider a natural number N=Divide[2020!,Power[2,1010]*1010!]. If n is the remainder when N is divided ny 1000, then find the sum of prime numbers which divide n

[@saravanan3833]

Can't we solve using partial fraction

[@saranshkushwaha7040]

Its a very easy again put x= tan∅ You very get cos²∅ Then the Integral shorts to ∅/2+1/4sin2∅ Putt values to get I = 1/2tan-1 x + 1/2(x/1+x²)😊

[@nada88882]

Yo can you make a video on how to get better at math? Ive been struggling for so long

[@arpanmukherjee4625]

I came here after Howard mentioned about it in TBBT to an answer to Sheldon's question.

[@almazchati4178]

Probably this technique was known long before Feynman. I used to use it not knowing that it was his technique.

[@Shivam-ee2pd]

How do you integrate 1/(x^a+a^x)?

[@aserioussalamander9475]

Me: (10 years since doing calc during first 10 seconds of the video): *slaps head* "of course!"

[@i3moryy797]

yo please if i want to ask a question can i ask you and how to ask thanks

[@General12th]

Hi BPRP!

[@blackpenredpen]

Hiii

[@s.m.m99203]

Thank you sir

[@gregorysadofyev3981]

This is not correct solution. You can use Feynman technique only in integrals with limits (from 0 to + infinity for example) and only if you proof that can take derivative under the integral sign. Right solution is using integrate by parts: du=dx & v=1/(1+x^2)

[@samsaraAI2025]

What is this itegral used for?

[@tapankumardas3292]

man of action.

[@aurelio3532]

absolutely fantastic