THE COOLEST LOG TRIG INTEGRAL ON YOUTUBE!

This is truly ridiculously awesome! Full solution development using some beautiful applications of symmetry and other cool tricks.
The integral I that'll knock your socks off:
• Solving this surprisin...
My complex analysis lectures:
• Complex Analysis Lectures
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[@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

[@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!

[@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

[@dwightswanson3015]

Truly an absolute banger!

[@CM63_France]

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

[@slavinojunepri7648]

Excellent

[@mab9316]

A beauty !

[@iWilburnYou]

Love symmetric integrals

[@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

[@robmaddock8531]

You're awesome

[@Tosi31415]

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

[@Jalina69]

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

[@yoav613]

👏👏👏

[@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.

[@archinsoni1254]

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

[@user-un6ns4nl7t]

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

[@Ben-wv7ht]

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

[@theIndicLearner]

Sir, please can I solve the same equation with the same method on my channel.In my native 'Hindi Language ' Please let me do it for some videos.I will get a good starting.

[@maths_505]

You have permission to take all integrals from my channel. Good luck my friend, I understand Hindi because I'm Pakistani and speak Urdu so I might watch some of your videos.

[@theIndicLearner]

@maths_505 OH, is it, you are Pakistani ? I am really sorry for the embarrassment of us both, but I can't do it now. We both give taxes to our governments, and when they fight, it's the public who suffer. If you want to support us, bring a government in your country which is friendly with India. Ours, already is such. Thanks a lot for your support as a common man.

[@maths_505]

@@theIndicLearner 🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️🤦🏾‍♂️

[@theIndicLearner]

@maths_505 does that mean you belong to Pakistan but you are currently a citizen of some other country. If this is so, then I am sorry again we suffered a blow very recently at 9th this month from your country's side as the circumstances are saying it loudly so even the name hurts, right now. Really sorry again.

[@agrimmittal]

​@@theIndicLearner you talk a lot

[@giuseppemalaguti435]

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

[@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?