2:39 This integral form has become something that has piqued my curiosity for a while now. I remember in last time's case, that is when alfa was equal to zero, I wondered if invoking summation of 1/1+sin^4(x) (cos(x) as well by king's property) would work, based of the equation we can prove using contour integration or the beta function. However, it seems we relied on the same method as last time, which still is mind-blowing to me! 5:01 oh yeah LOL! multi-valued functions go brr! Amazing vid as always my man! keep pumping them up!
Same integral was given in my class under feynmann tech just instead of cos^2(x) there was x^2. So no complex was needed and its not in our syllabus also . BTW the golden ratio at the end was too goood!! 🤯🤯
Hey! Nice videos, I was wondering if you wanted to do also some cool probability stuff at some point. I guess there are a lot of really nice problems and tricks to do there too.
Terrible solution. I mean the calculation of (√(1+2i)-√(1-2i))/i Definitely the result should be the real number, so squaring the expression , then simplifying it and taking the square root gives us the result (1+2i+1-2i-2√(1+2i)•√(1-2i)))/i²= -2+2√5= 4•(√5-1)/2=4•2/(√5+1)=4/phi Where phi is the golden ratio The answer is π/2•√(4/phi)= =π/√phi
Bro i noticed u reaaaaallly love Feynman's technique
@SayedHamidFatimi 😂😂😂
Feynman's technique is literally never boring to watch.
2:39 This integral form has become something that has piqued my curiosity for a while now. I remember in last time's case, that is when alfa was equal to zero, I wondered if invoking summation of 1/1+sin^4(x) (cos(x) as well by king's property) would work, based of the equation we can prove using contour integration or the beta function.
However, it seems we relied on the same method as last time, which still is mind-blowing to me!
5:01 oh yeah LOL! multi-valued functions go brr!
Amazing vid as always my man! keep pumping them up!
Same integral was given in my class under feynmann tech just instead of cos^2(x) there was x^2. So no complex was needed and its not in our syllabus also .
BTW the golden ratio at the end was too goood!! 🤯🤯
so cos^2(x^2) or just x^2
@@uggupuggujust x2
@ 3:30 The cos x terms in the denominators should both be squared.
Could you do something about Maxwell’s equations? Both integral and differential form
Hell yeah 🔥
I think this is the first time I have heard swearing on this channel, and the answer was so satisfying and beautiful ❤
Very interesting integral. Thank you for your smart plan to solve this type of integrals.
I love how you used phi in the process, and then the golden ratio popped out
I avoid θ at all costs 😂
Only issue for me was that you used phi as an angle and then used phi as golden ratio. Confused me for a second or 2
14:47:me memorise that
(sqrt5-1)/2=1/y (y=golfen ratio)
At 3:28, you are missing squares on the cosines in the partial fraction decomposition. This confused the heck out of me, so just a heads up.
Oh sorry about that mate. My bad.
Hi, incredible youtuber!! Could you tell me what blackboard are you using in your videos?
Hey! Nice videos, I was wondering if you wanted to do also some cool probability stuff at some point. I guess there are a lot of really nice problems and tricks to do there too.
Elegant!
Sweet maths dude
I'm curious what app do you use for your videos? I wanna take my notes in that.
It's Samsung notes. I use an S6 tab.
@@maths_505 Thanks man!
all in all, very cool!
Thanks bro
why is it necessary to take the limit for the tangent structure? isnt arctan infinity just pi/2
1. Those are "limits" of integration.
2. Infinity isn't a number.
@@maths_505 thanks but i meant why turn arctan into the log form when the variable is just being divided by a constant
@@aryaghahremani9304 its a good teaching practice to show things like how such results farrt forward into the complex realm.
brother where do you get these integrals???? you even create them???
This one's from Micheal Penn and Rizzy Math (Instagram)
The one I'm about to upload is one I made up.
sir i got a nice way to evaluate this monster , it took me 2 min.
I=π/√2(√(√5-1))
why did you add +C when it's a definite integral? does that have something to do eith fynman's technique?
He calculated the *indefinite* integral of I'(alpha) - this is separate from the fact that I(alpha) is defined as a definite integral.
Amaze Balls
This integral is complex enough for Feynman's technique
Terrible solution. I mean the calculation of
(√(1+2i)-√(1-2i))/i
Definitely the result should be the real number, so squaring the expression , then simplifying it and taking the square root gives us the result
(1+2i+1-2i-2√(1+2i)•√(1-2i)))/i²=
-2+2√5= 4•(√5-1)/2=4•2/(√5+1)=4/phi
Where phi is the golden ratio
The answer is π/2•√(4/phi)=
=π/√phi