THE COOLEST LOG TRIG INTEGRAL ON YOUTUBE!

[@SuperSilver316]

Mans trying to win an Oscar right near the end there 😂

[@maths_505]

😂😂😂

[@stefanalecu9532]

We love him for moments like this and more

[@Aditya_196]

Fr 😂 I was like bro bro the acting is lit fire just say it out

[@CM63_France]

Hi, "terribly sorry about that" : 1:32 , 6:10 , 11:45 , 12:40 , 14:12 , "ok, cool" : 12:50 .

[@kingzenoiii]

13:05 EXOTIC FUNCTIONS.... ASSEMBLE!

[@Calcprof]

Mathematica does get the integral from 0 to pi/2 of log^2 cos(x) dx = 1/24 \[Pi] (\[Pi]^2+3 Log[4]^2), agreeing with you

[@MrWael1970]

Very interesting video. Thank you.

[@dwightswanson3015]

Truly an absolute banger!

[@iWilburnYou]

Love symmetric integrals

[@manstuckinabox3679]

Ahh... the good ol massacring of integrals that make me wonder if they appear in any real life application. It's been a while since I came here, and so glad it still feels like home. always nice to see myself grow more mathematically mature along side your channel, keep it up big G!

[@night9587]

0:54 En utilisant la methode de feyman on a integral de 0 à π/2 de ln^2(bcosx) en derivant par rapport à b on obtient inegral de 0 à π/2 de 2/b.ln(bcosx) puis on a π/2ln(b)+2/b integral de 0 à π/2 ln(cosx) changement de variable x=π/2-t on integral in(sint) qui donne 2/b×πln2/2 +k en assemblant le tout on a π/bln(b) +π/bln2 qui donne π/bln(2b) d'où M(b) =π/b×ln(2b) en integrant on a un changement de variable 2b=t la solution de l'integrale est π/2×ln^2(t)+ C en remplaçant t par 2b et b par 1 selon l'integrale originelle on obtient bien π/2×ln^2(2) + C c'est la determination du C qui pose un peu problème voilà.merci

[@archinsoni1254]

Is integral 0 to infinity x^(-x) possible ?

[@Туканчик1337]

Yes, Summ( 1 to inf) n^-n

[@GeoPeron]

BriTheMathGuy has a video on it, check it out. It's called "The integral of your dreams (or nightmares)" if I'm not mistaken

[@maths_505]

@GeoPeron f**k him I have a video on it 😎😎 jk check out both videos 😂

[@GeoPeron]

@@maths_505 You have a bunch of videos on the Gaussian, Fresnel and Dirichlet integrals, leave some to other math youtubers, damnit!

[@maths_505]

@@GeoPeron hell nah I'm coming for em all!!!

[@chancia8990]

dude you're so chaotic xd

[@mab9316]

A beauty !

[@slavinojunepri7648]

Excellent

[@robmaddock8531]

You're awesome

[@Jalina69]

It is very easy! You open a bracket, then you close a bracket 😇

[@Tosi31415]

solved last video's homework and got I=π/(2sqrt2*e^(sqrt2))

[@xleph2525]

Solved the "HW" problem from the last video: pi * cosh( sqrt(2) ) / sqrt(2) This look alright, boss?

[@maths_505]

I think you made a mistake somewhere bro

[@ambiguousheadline8263]

I believe the answer should be pi/(sqrt(2)e^sqrt(2)) if I did everything correct

[@Tosi31415]

​@@ambiguousheadline8263you probably forgot to divide by two at the end but it's correct

[@dan-florinchereches4892]

I would like to try something else but not aure if my level of maths is jigh enough as a humble engineer :) I would like to transform 1 into sin(π/2) so 1+sin(x)=sin(π/2)+sin(x)=2sin(x/2+π/4)cos(π/4-x/2)=(using cos(a)=sin(π/2-a)) 2*cos^2(π/4-x/2) 1+sin(-x)=2sin(-x/2+π/4)cos(π/4+x/2)= 2 sin^2(π/4-x/2) All the reig functions are positive from -π/2 to π/2 so using properties of logs makes sense Function=ln(2 cos^2( π/4-x/2))ln(2 sin^2(π/4-x/2))= (ln2)^2+2ln2(ln(cos(π/4-x/2))+ln(sin(π/4-x/2)))+ln(sin^2(π/4-x/2))ln(cos^2(π/4-x/2))=(double angle sine) (ln2)^2+2ln2*ln(sin(π/2-x)/2)+ln(sin^2(...))ln(1-sin^2(...))=ln2+2(ln(cos(x))-ln2)+ln(sin^2(...))ln(1-sin^2(...)) Feels like i am going in circles as i got a factor similar to what i started with and integrating ln(cos(x)) i am not sure will evaluate nicely with Weierstrass either

[@giuseppemalaguti435]

sinx>cosx....formule di bisezione...4 integrali risolvibili...in verità 3/4 sono semplici,ma int(lncos(x/2)*lnsin(x/2))????

[@Ayush-yj5qv]

Thats what i call "AN IIT ADVANCE LEVEL" Question i tried my best couldn't solve fully

[@sandyjr5225]

This is beyond JEE Advanced.

[@Ayush-yj5qv]

@@sandyjr5225 no this is the 1℅ question in jee advance that no one can solve 🥲

[@julianwang7987]

You dropped a factor of 1/2 somewhere. The answer should have been pi/2*ln^2(2) - pi^3/12

[@danielespinosa869]

I agree

[@michaelguenther7105]

At about 1:17 he redefined I to be the integral from -pi/2 to pi/2 instead of from 0 to pi/2 as at the start of the video, so for the redefined I his result is correct.

[@julianwang7987]

@@michaelguenther7105 I think this is a bad practice. Not every viewer just wants to watch the solution right away (or at all). He could have carried a factor of 1/2 through the rest of calculation, like a presenter should do. Dropping constant factors is something I do while calculating for my own private amusement, not when I need to present the result publicly.

[@Ben-wv7ht]

I beg of you , change the bounds to become 0 - pi/3 , this will be MUCH MORE challenge

[@yoav613]

👏👏👏

[@thomasolson7447]

I don't know what this stuff is. I have grade 12. But I'll give it a go. I have Maple and kind of understand how to use it. r[1]=cos(θ) + i*sin(θ) r[2]=-cos(θ) + i*sin(θ) f:= t -> (r[1]^t - r[2]^t)/(r[1] - r[2]) g:= t -> r[1]^t + r[2]^t int_0^n f(t) d t = ((cos(θ) + i*sin(θ))^n*ln(-cos(θ) +i*sin(θ)) - (-cos(θ) + i*sin(θ))^n*ln(cos(θ) +i*sin(θ)) + ln(cos(θ) + i*sin(θ)) - ln(-cos(θ) +i*sin(θ)))/(2*cos(θ)*ln(cos(θ) +i* sin(θ))*ln(-cos(θ) + i*sin(θ))) int_0^n g(t) = ((-cos(θ) +i*sin(θ))^n*ln(cos(θ) + i*sin(θ)) + (cos(θ) + i*sin(θ))^n*ln(-cos(θ) + i*sin(θ)) - ln(cos(θ) +i*sin(θ)) - ln(-cos(θ) + i*sin(θ)))/(ln(-cos(θ) +i* sin(θ))*ln(cos(θ) + i*sin(θ))) I graphed some of them. pi/5 looks like a 3-leaf clover. -i/2 is a sink. I suspect integrals are not meant to be used like this. *maybe a little bit of a ninja edit int_0^n f(t) d t = -(r[2]^n*ln(r[1]) - r[1]^n*ln(r[2]) - ln(r[1]) + ln(r[2]))/((r[1] - r[2])*ln(r[1])*ln(r[2])) int_0^n g(t) = (r[2]^n*ln(r[1]) + r[1]^n*ln(r[2]) - ln(r[1]) - ln(r[2]))/(ln(r[2])*ln(r[1])) Did I just do a general solution to something in polynomials?