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

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)

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

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.

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

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

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

"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.

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

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.

Didn't thought bout that amazing technique!

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

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

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.

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.

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 😁

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 ;) ;) ;) ;)

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. 😊

Thanks

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

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.

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

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！

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

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

Bruh this was literally a question in my homework today

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

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

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?"

Why can't we use partial fraction method ??

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 .

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....

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

Thank you!

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

What a nice idea to integreat the seemingly impossible func

This video made my day🔥

So what happens when u have mutiple parameters?

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

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???

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

Excellent presentation 👌

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

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.

This seems easier then trig substitution

Man, that was beautiful!

why are u allowed to swap differential and integral?

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.

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

Feynman technique of integration is OP!!

wow my math guy🔥👊💪

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

Brilliant First time seeing this trick

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.

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

Thank you sir

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

Love this!

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

absolutely fantastic

I am in 12th STD,.. this question came in my unit test .. not joking seriously 😳

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

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)😊

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

That’s wild!

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!!😁

Integration of 1/(1+x⁴) please

It's so reminiscent of eigenvalue equations

Can be done IBP.

just solved this question in my JEE online course.....

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

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

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

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

Feynman was a crazyyyy dude

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

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

I am left stunned with Feymann's genius. This whole year from January till now, that I am learning calculus, I have not seen such a technique! By the way I am 14, so I learn calculus while being a school student, strange but true.

man of action.

Excellent.

I used regular trig sub and got the same answer...

why cant the a=-1?

can you make a video of limit integration?

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

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

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

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

Very complex, was this the right answer?

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

I love Leibniz rule

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

Feynman was the man!

I personally would just factor it and use integration by parts.

What is calc 2?

Watching from Bangladesh 💙