A surprisingly difficult integral: int 0 to π/2 arctan(2sin(x)) solution using Feynman's trick

[@Maths-Integrals]

It's a very nice integral and the solution is interesting. There are also many very interesting properties of the Dilogarithm function. This simplifies the final result: I= ((pi)^2)/6 -(3/2)*(ln(phi))^2 where "phi" denote the golden ratio

[@ranaranino4731]

at 7:40 multiplie denominator and numerator inside logarithm by alpha and make substitution alpha=sinh(y) leads Quickly to the result

[@Nottherealbegula4]

These just keep getting better and better, thankes for accepting my request!

[@MrWael1970]

I'm proud for attending this video. Best regards.

[@cosimo7770]

A work of art.

[@xanterrx9741]

Absolutle great solution, well spended first 20 min after woken up , thanks for your hard work and waitng for next video

[@Mosux2007]

Have you ever dealt with integrals involving elliptic functions?

[@JakePinedo-ns4yu]

“Terribly sorry about that” it’s fine bruh nobody cares if u accidentally write something weird

[@CM63_France]

Hi, I know that tan^-1 (-2) = 2 tan^-1 (phi) (it's the angle between 2 adjacent faces of the dodecahedron) so tan^-1 (2) = pi - 2 tan^-1 (phi) , so tan^-1 (2) / 2 = pi/2 - tan^-1 (phi) , so tan^-1 (2) / 2 = tan^-1 (1/phi) . 13:43 : sin theta = 2 z / ( 1 - z^2) , minus and not plus, but fortunately this term cancelled out later . "ok, cool" : 1:33 , 10:01 , 14:24 , 19:21 , "terribly sorry about that" : 3:34 , 6:30 , 11:16 , 12:02 , 15:22 , 16:39 , 16:42 , 16:52 , 17:31 , 17:46 .

[@mcalkis5771]

A good day when you upload Kamal.

[@symmetricfivefold]

watching this feels like watching teacher solving a "simple question" and ends up with me not knowing where am i

[@barryfortnite3849]

can you do the integral from 0 to infinity of 1/x!

[@jejnsndn]

Can you do it using double intrgral?

[@Rundas69420]

I think so, because the applicability of the Leibnitz rule is equivalent to transforming the single integral into a double integral by taking a derivative.

[@maths_505]

Yes indeed

[@BoringExtrovert]

I haven’t finished the entire thing yet, but why don’t you jut use inverse hyperbolic tan for the form dx/(a-x2)?

[@threepointone415]

As the saying goes: It's worth it's weight in Golden Ratios

[@daveydd]

Sir I have a question, why or how is ln(2) so related to the trig functions? Especially when their integrals involve natural logarithms. What's its essence? What's its meaning behind it? Is it because of ln(sin(π/4)) and ln(cos(π/4)) being equal to -(1/2)(ln(2)) ? Perhaps that's only one of the bunch of reasons, I've tried googling but I haven't found much.. any ideas?

[@utkarshsaini5650]

its very early for you to start with integration first learn functions, inverse trigno, other topics

[@Tosi31415]

could've used a particular formula at the end to remove the dilogarithms but cool nonetheless

[@Circuito28]

Yo bro do you like algebraic or complex geometry problems/topics? Also algebraic number theory would be awesome, think about it

[@willemesterhuyse2547]

You didn't pick up the error by 8:35!

[@TMH2007]

I was busy for a few months and after coming back i sense a change in style of thumbnails, or is it only me?

[@willemesterhuyse2547]

He picked it up by 10:48.

[@kingzenoiii]

ah yes, the leibniz trick

[@willemesterhuyse2547]

At timestep 6:39 you left out an alpha in the denominator.

[@gowipe-grandcross]

Hello, I would like to send you a result about a generalisation of the dirichlet integral (with the proof ofc) : the integral from 0 to infinity of sin(x^n)/x^n. I don't think you already present this on your channel. Is there a way to do it ?

[@maths_505]

I actually plan on presenting my solution soon

[@gowipe-grandcross]

​@@maths_505what method did you use ? To see if it differ from mine

[@Mario_Altare]

I'm afraid I've got the sub wrong: after considering that sin x = tan x/√(1+tan^2⁡x), I let tan x = u, so I'(a) = ∫_0^∞ u du/{(√(1+u^2 )[1+(a^2+1) u^2])} Then, after letting 1+u^2 = v^2, I'(a) = ∫_0^1 dv/[1+(a^2+1)(1-v^2 )] I got stuck because I obtained the following integral: I(a) = 1/2 ∫_0^2 1/(√(1+a^2 ) √(2+a^2 )) * ln⁡((√(2+a^2 )+√(1+a^2 ))/(√(2+a^2 )-√(1+a^2 )))da which is a bit of a hell... Maybe there's a further sub to go on from here, but maybe there's not

[@alexkaralekas4060]

I mean it feels a little cheated by not taking the definite integral at 7:40 because if you did you wouldn't be able to take the dilogarithm

[@maths_505]

@@alexkaralekas4060 we still can since the only change would be the upper limit of the integral (replacing 2 by, say, x)

[@MathematicFanatic]

ok now in terms of alpha 😳

[@jackkalver4644]

I learned that my calculator can be deceptive.

[@orionspur]

Ohhhkay cruel!

[@Babyshark-co8ks]

No views in 12 seconds? Bro fell off

[@maths_505]

Indeed

[@Bruh-lb5ck]

I’m bored of this kind of comment. Who even started it in the first place

[@dr.hanamantkarande9544]

Hello

[@stuffthings1417]

not surprising. but i've been out of the game for 20 years.