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!
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
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
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.
@@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.
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?
Mans trying to win an Oscar right near the end there 😂
😂😂😂
We love him for moments like this and more
Fr 😂 I was like bro bro the acting is lit fire just say it out
Hi,
"terribly sorry about that" : 1:32 , 6:10 , 11:45 , 12:40 , 14:12 ,
"ok, cool" : 12:50 .
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
13:05 EXOTIC FUNCTIONS.... ASSEMBLE!
Truly an absolute banger!
Very interesting video. Thank you.
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!
Love symmetric integrals
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
A beauty !
Excellent
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
dude you're so chaotic xd
You're awesome
Solved the "HW" problem from the last video:
pi * cosh( sqrt(2) ) / sqrt(2)
This look alright, boss?
I think you made a mistake somewhere bro
I believe the answer should be pi/(sqrt(2)e^sqrt(2)) if I did everything correct
@@ambiguousheadline8263you probably forgot to divide by two at the end but it's correct
Thats what i call "AN IIT ADVANCE LEVEL" Question i tried my best couldn't solve fully
This is beyond JEE Advanced.
@@sandyjr5225 no this is the 1℅ question in jee advance that no one can solve 🥲
solved last video's homework and got I=π/(2sqrt2*e^(sqrt2))
It is very easy! You open a bracket, then you close a bracket 😇
I beg of you , change the bounds to become 0 - pi/3 , this will be MUCH MORE challenge
You dropped a factor of 1/2 somewhere. The answer should have been pi/2*ln^2(2) - pi^3/12
I agree
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.
@@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.
sinx>cosx....formule di bisezione...4 integrali risolvibili...in verità 3/4 sono semplici,ma int(lncos(x/2)*lnsin(x/2))????
Is integral 0 to infinity x^(-x) possible ?
Yes, Summ( 1 to inf) n^-n
BriTheMathGuy has a video on it, check it out. It's called "The integral of your dreams (or nightmares)" if I'm not mistaken
@GeoPeron f**k him
I have a video on it 😎😎 jk check out both videos 😂
@@maths_505 You have a bunch of videos on the Gaussian, Fresnel and Dirichlet integrals, leave some to other math youtubers, damnit!
@@GeoPeron hell nah I'm coming for em all!!!
👏👏👏
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?