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!

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

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

This is another approach: kzfaq.info/get/bejne/msh-m9WjvLvPm5c.html

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

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)

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

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

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

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.

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

Didn't thought bout that amazing technique!

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

Excellent presentation 👌

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.

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

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

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

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

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

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

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???

"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.

Great 👌 video , much helpful!

Thank you!

This video made my day🔥

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

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.

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.

Man, that was beautiful!

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

absolutely fantastic

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 😁

Bruh this was literally a question in my homework today

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

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

Love this!

can you make a video of limit integration?

Brilliant First time seeing this trick

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

Thank you sir

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

Metodo interessante,bravo!!!

Thanks

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

What a nice idea to integreat the seemingly impossible func

explain modulus function intrgals plz

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

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

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

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！

wow my math guy🔥👊💪

So what happens when u have mutiple parameters?

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

That’s wild!

Can't we solve using partial fraction

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

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

thanks bro💖🌺

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

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. 😊

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

Qué genio eres, y Feynman también.

man of action.

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

It's so reminiscent of eigenvalue equations

Why can't we use partial fraction method ??

Sir more good problems 👍👍

I think it pretty good... and as you end with, cool!

Excellent.

can you teach natural logarithms from start to finish...

👍 thank you

damn my teacher took a similar example in class and this gave me more clarity on the concept :D

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 .

Very cool

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

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.

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

That's so cool

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 ?

This seems easier then trig substitution

What is this itegral used for?

damn why dont they teach these bangers at school, this is so coool

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

Magician of parameters

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

That was awesome. How would we find c here?

Can be done IBP.

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

Shouldn’t the last bit just be equal to tan ^-1(x) + c when a = 1?

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

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

why cant the a=-1?

At first, I thought that's a DOTA T-shirt :D

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.

Integration of 1/(1+x⁴) please

Please more often new videos)

Feynman was a crazyyyy dude

Very complex, was this the right answer?

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