Before watching the video, the substitution I made was pi/2*x = u, then used the double angle formula for sin in the denominator of the integrand. In the numerator, I used the fact that sin^2 u + cos^2 u = 1, then integrated by parts.
sin^2 u + cos^2 u = 1 is a basic trigonometric identity. In general, it's a strategy to cancel out a product of sines and cosines in the denominator by setting 1 = sin^2 u + cos^2 u in the numerator.@@monikaherath7505
Using Euler reflexion formula you can also show that 7ζ(3)/pi^2 = integral(gamma(3/2-x)*gamma(3/2+x)) from -1/2 to 1/2. It doesn't help to integrate, but it's beautiful ;-)
Nice one ! I now challenge you to solve the integral from 0 to pi of x * (sinx)^n dx. It took me around an hour or 2 to solve and is absoluetly gorgeous
@@grigoriefimovitchrasputin5442 Tbh it's not too difficult but you need to be creative, normal methods don't work. Or at least I couldn't get very far using them
I think by applying property of integral, it breaks into pi * integral 0 to pi/2 (sin x) ^ n dx. Should be easy from there to express in terms of gamma function.
Should be doable by integrating by x = sqrt(x) then u = x/2 and then letting u be sin x which leads to a structure very similar to the beta function can’t rlly go further because I have no pen rn
Corretto, ❤ho usato sinx, formula esponenziale, poi la serie geometrica... Anche se a me risulta un segno -... Ah ah, al primo colpo non mi viene mai... Ok, il segno - è scomparso... Is correct
It looks like it can be generalised. If n=3, then ans can be written as (2n+1)zeta(n)/pi^{n}. May be exist some types of integrals which are generalised that way?
Yes I’m also curious if there are analogous formulas for zeta of odd (or even all) n. One can go try to go back through the calculation and find a point where the integrand can be tweaked in such a way to yield this kind of formula. I’m too lazy to do this right now though
Oh cool....I used to tutor A level students..... Thanks bro....you'll understand them all pretty soon cuz you'll study the basics in your A levels....then you'll solve these integrals and DEs on your own.
@Jacques-kc4qy it’s amazing that people like yourself are incredibly smart when it comes to this stuff I don’t understand any of it, But yet when it comes to political leanings regarding logic and common sense we have some of the stupidest people on the planet
![⭐️( ลอยกระทง 2024 ) Ver. แดนซ์ Remix BY [ ดีเจกิต รีมิกซ์ ]](http://i.ytimg.com/vi/rMWOLhgXUHc/mqdefault.jpg)
Before watching the video, the substitution I made was pi/2*x = u, then used the double angle formula for sin in the denominator of the integrand. In the numerator, I used the fact that sin^2 u + cos^2 u = 1, then integrated by parts.
Could you write it out please I don't understand how you got to sin^2 u + cos*2 u
sin^2 u + cos^2 u = 1 is a basic trigonometric identity. In general, it's a strategy to cancel out a product of sines and cosines in the denominator by setting 1 = sin^2 u + cos^2 u in the numerator.@@monikaherath7505
Using Euler reflexion formula you can also show that 7ζ(3)/pi^2 = integral(gamma(3/2-x)*gamma(3/2+x)) from -1/2 to 1/2. It doesn't help to integrate, but it's beautiful ;-)
GREAT!! An interesting class of integrals: integrate from 0 to 1 [ x^n (1-x)^m / sin(πx)].
Bro was like LOL do it by parts and stuff but really didn't bother by going the distance, great video as always brother.
Thanks mate. I've been looking for an integral to use that series on
Thank you for this innovative integration. Smart solution.
damn, maths in 4k is fancy
Hi, use the properties of the gamma function like is relation to 1/sin(x.pi) and x(x-1)
Nice one ! I now challenge you to solve the integral from 0 to pi of x * (sinx)^n dx. It took me around an hour or 2 to solve and is absoluetly gorgeous
Thanks ! It looks interesting to solve
@@grigoriefimovitchrasputin5442 Tbh it's not too difficult but you need to be creative, normal methods don't work. Or at least I couldn't get very far using them
@@quentinrenon9876 it looks like Wallis integrals, except that there is a x. Anyway, i'll give it a try
I think by applying property of integral, it breaks into pi * integral 0 to pi/2 (sin x) ^ n dx. Should be easy from there to express in terms of gamma function.
Should be doable by integrating by x = sqrt(x) then u = x/2 and then letting u be sin x which leads to a structure very similar to the beta function can’t rlly go further because I have no pen rn
Corretto, ❤ho usato sinx, formula esponenziale, poi la serie geometrica... Anche se a me risulta un segno -... Ah ah, al primo colpo non mi viene mai... Ok, il segno - è scomparso... Is correct
Exhilaratingly beautiful!
Awesome integral!
It looks like it can be generalised. If n=3, then ans can be written as (2n+1)zeta(n)/pi^{n}. May be exist some types of integrals which are generalised that way?
Yes I’m also curious if there are analogous formulas for zeta of odd (or even all) n. One can go try to go back through the calculation and find a point where the integrand can be tweaked in such a way to yield this kind of formula. I’m too lazy to do this right now though
Isn't the essence of the proof the following Fourier Expansion for csc(x) = 2 SUM_odd (sin(nx))?
cr7 fan here too xD, loved that one 👍
SUIIIIIIIIIIIIIIIIIIIII
I love your solution though I don't understand one of your solution. I'm an A level student.😊
Oh cool....I used to tutor A level students.....
Thanks bro....you'll understand them all pretty soon cuz you'll study the basics in your A levels....then you'll solve these integrals and DEs on your own.
Bring back dark thumbnail 1/4
Okay tomorrow
Didn't go the way I thought
How is this applied to every day life?
Integrals are everyday life
Integrals are life.
@@daddy_myers example please
@@damrgee8279 Your question alone indicates you could use more integrals in your daily life.
@Jacques-kc4qy it’s amazing that people like yourself are incredibly smart when it comes to this stuff I don’t understand any of it, But yet when it comes to political leanings regarding logic and common sense we have some of the stupidest people on the planet