max(2sinx, 2cos(2x)-1)^2 * min(sin(2x), cos(3x))^2 = max(2sinx, 2cos(2x)-1)^2 * min(2(sinx)(cosx), (2cos(2x)-1)cosx)^2, where cos(3x) = 4cos^3(x)-3cos(x) = cos(x) (4cos^2(x) - 3) = cos(x)(2cos(2x)-1) [as you found at 21:24 (before being erased)]. You now can leave that cos(x) factor out off the min(-,-), resulting: max(2sinx, 2cos(2x)-1)^2 * min(2sinx, (2cos(2x)-1)^2 * (cosx)^2. Now observe that max(_,_) and min(_,_) have the same expressions, so the max and min are irrelevant. Therefore you have, finally: Integral from 0 to Pi of 4sin^2(x)*(2cos(2x)-1)^2*cos^2(x)dx = Integral from 0 to Pi of 4sin^2(x)*(4cos^2(x)-3)^2*cos^2(x)dx = Integral from 0 to Pi of 4sin^2(x)*((4cos^3(x)-3cos(x))/cos(x))^2*cos^2(x)dx = Integral from 0 to Pi of 4sin^2(x)*((cos(3x)/cos(x))^2*cos^2(x)dx = Integral from 0 to Pi of 4sin^2(x)*((cos(3x))^2*dx = Integral from 0 to Pi of (sin(4x) - sin(2x))^2*dx. Where in the last equality we used the product to sum formula sin(A)cos(B). Now we have an easy integral: Integral from 0 to Pi of (sin(4x) - sin(2x))^2*dx = Integral from 0 to Pi of (sin^2(4x) + sin^2(2x) - 2sin(4x)sin(2x)dx = Pi.
Smart solution for the forth one: Basically it's symmetric around π/2, you divide it then transform to π/2-x, then take sin(x) out of the right side Then you left with sin^2(x) times the product of min and max of the same thing, which is just their product From there it's trivial You get to Sin(2x)*(1+2cos(2x)) all square You transform to 2x Which cancels the 2 factor from the initial separation and gets (sin(x)+sin(2x))^2 from 0 to π You separate, and get (sinx+sin2x)^2+(sinx-sin2x)^2 Which is 2sin^2(x) + 2sin^2(2x) from 0 to π/2 The latter of which we transform then split again to get 2sin^2(x) as well So we have 4*sin^2(x) from 0 to π/2 Transform, add and divide by two, then integrate We get 2*π/2=π
18:10 well consider the following if f and g are functions in x and for any positive number k , k*max(f/k,g/k) =max(f,g) . Now just consider the following : |Cos3x/cosx| = |4cos^3(x)-3|=|2cos2x-1|,|sin2x/cosx| = |2sinx | , irrespective of what you do |cosx | is always positive . So the integrand simplifies to (1/cos^2(x))*max^2(sin2x,|cos3x|*min^2(sin2x,|cos3x|) . So just the max and min vanishes leaving the integrand with ( 2sinxcos3x)^2 . Now it is an easy trig intgral leaving us with pi
it's fascinating to see that the golden ratio has fibonacci-like properties, e.g. the nth power is equal to the (n-1)st plus the (n-2)nd power which is like a recursive property
15:46 isn’t the exponent outside the floor function of the golden ratio, meaning, in every interval it is just 1 raised to the power of 1. So, the answer is just 10 because of 10 intervals?
@@maths_505 Understood. On an unrelated topic, do you do programming? I just started at uni after having a bio major in high school. People say programming is just maths but I struggle with it though I'm good at maths. I just wanted to hear another math person's perspective.
@musicburst2513 yeah I code in python but I've rarely ever thought of it as math. Programming in c++ felt more like a math problem tbh because there was planning and structure so coding often felt like writing a solution development. My work in python however is just alot of libraries, modules and chat gpt 😂
@musicburst2513 What I do while programming in Python is write the algorithm of the code, then write it in syntax. Python is actually very simple as long as you know the syntax
23:50 Since f-g >0 then f>g but f = (2cos2x-1)^2 , but you wrote 4(sinx)^2 and (sin2x)^2 < (cos3x)^2 but you wrote (cos3x)^2 I've check the integral of 0 to Pi of (2cos2x-1)^2*(sin2x)^2 dx and it equals to Pi, Am I wrong or ..........................
Yeah homie I basically missed a negative but kept winging it💀 And no that factorisation doesn't work, just multiply the terms and you'll get a -6n instead of 6n
For max(4sin^2(x), (2cos2x-1)^2), g-f = 16sin^4(x) - 12sin^2(x) +1 varies between -1 and 1 in [0, pi] and so happens for min(sin^2(2x), cos^2(3x)) where g-f = 16cos^6(x)-20cos^4(x)+5cos^2(x) varies between -1 and 1. I don't know what are you doing, sincerelly, at that step.
@@maths_505i don't get it too, if you grafh 16sin⁴x - 12sin²x + 1 you can have negative values, for ex: x = π/4. would be -1, just because they are even exponentials doesnt mean that it's always gonna be positive so i'm still confused about that step
Shouldn't it be int[0,10](floor(phi)^floor(x))dx? So then because phi is ~1.6... floor(phi) is just 1, and then 1^x is just 1, so its int[0,10]dx which is just 10????
NOTE: I made a mistake writing integral 3; the floor(x) is on the golden ratio inside the "outer" floor function.
You are killing those integrals! you've upgraded from integration bee to integration WASP. Amazing work
13:51 the part where he does mit level antiderivatives, solves infinite series but struggles with the most basic arithmetic is so relateble
I'm just glad I made it out alive
My favourite math creator has replied to my comment, now I can live my life in peace
@@bixxuxboxplayz4986 I try to reply to everyone. Terribly sorry in case I missed any of your previous ones.
max(2sinx, 2cos(2x)-1)^2 * min(sin(2x), cos(3x))^2 = max(2sinx, 2cos(2x)-1)^2 * min(2(sinx)(cosx), (2cos(2x)-1)cosx)^2, where cos(3x) = 4cos^3(x)-3cos(x) = cos(x) (4cos^2(x) - 3) = cos(x)(2cos(2x)-1) [as you found at 21:24 (before being erased)].
You now can leave that cos(x) factor out off the min(-,-), resulting: max(2sinx, 2cos(2x)-1)^2 * min(2sinx, (2cos(2x)-1)^2 * (cosx)^2. Now observe that max(_,_) and min(_,_) have the same expressions, so the max and min are irrelevant. Therefore you have, finally:
Integral from 0 to Pi of 4sin^2(x)*(2cos(2x)-1)^2*cos^2(x)dx = Integral from 0 to Pi of 4sin^2(x)*(4cos^2(x)-3)^2*cos^2(x)dx = Integral from 0 to Pi of 4sin^2(x)*((4cos^3(x)-3cos(x))/cos(x))^2*cos^2(x)dx = Integral from 0 to Pi of 4sin^2(x)*((cos(3x)/cos(x))^2*cos^2(x)dx = Integral from 0 to Pi of 4sin^2(x)*((cos(3x))^2*dx = Integral from 0 to Pi of (sin(4x) - sin(2x))^2*dx. Where in the last equality we used the product to sum formula sin(A)cos(B).
Now we have an easy integral: Integral from 0 to Pi of (sin(4x) - sin(2x))^2*dx = Integral from 0 to Pi of (sin^2(4x) + sin^2(2x) - 2sin(4x)sin(2x)dx = Pi.
Awesome
Smart solution for the forth one:
Basically it's symmetric around π/2, you divide it then transform to π/2-x, then take sin(x) out of the right side
Then you left with sin^2(x) times the product of min and max of the same thing, which is just their product
From there it's trivial
You get to
Sin(2x)*(1+2cos(2x)) all square
You transform to 2x
Which cancels the 2 factor from the initial separation and gets
(sin(x)+sin(2x))^2 from 0 to π
You separate, and get (sinx+sin2x)^2+(sinx-sin2x)^2
Which is 2sin^2(x) + 2sin^2(2x) from 0 to π/2
The latter of which we transform then split again to get 2sin^2(x) as well
So we have 4*sin^2(x) from 0 to π/2
Transform, add and divide by two, then integrate
We get 2*π/2=π
can you refer some source for the fact used that min*max is just the product?
18:10 well consider the following if f and g are functions in x and for any positive number k , k*max(f/k,g/k) =max(f,g) . Now just consider the following :
|Cos3x/cosx| = |4cos^3(x)-3|=|2cos2x-1|,|sin2x/cosx| = |2sinx | , irrespective of what you do |cosx | is always positive . So the integrand simplifies to (1/cos^2(x))*max^2(sin2x,|cos3x|*min^2(sin2x,|cos3x|) . So just the max and min vanishes leaving the integrand with ( 2sinxcos3x)^2 . Now it is an easy trig intgral leaving us with pi
Beautiful stuff
it's fascinating to see that the golden ratio has fibonacci-like properties, e.g. the nth power is equal to the (n-1)st plus the (n-2)nd power which is like a recursive property
@@balubaluhehe2002 indeed
Great work, Kamaal; you are insane!
Floor function of phi then raised to the power of floor function n
Yeah I fixed it a few moments later😂
15:46 isn’t the exponent outside the floor function of the golden ratio, meaning, in every interval it is just 1 raised to the power of 1. So, the answer is just 10 because of 10 intervals?
It's actually inside it and I had written the integral in LaTeX incorrectly.
@@maths_505 Understood. On an unrelated topic, do you do programming? I just started at uni after having a bio major in high school. People say programming is just maths but I struggle with it though I'm good at maths. I just wanted to hear another math person's perspective.
@musicburst2513 yeah I code in python but I've rarely ever thought of it as math. Programming in c++ felt more like a math problem tbh because there was planning and structure so coding often felt like writing a solution development. My work in python however is just alot of libraries, modules and chat gpt 😂
@musicburst2513 What I do while programming in Python is write the algorithm of the code, then write it in syntax. Python is actually very simple as long as you know the syntax
14:06 it should be 2/3*Pi^2-4-25/12
Indeed
I'm training fir this exam since kast year but i miss it, see you doing helped me to stat focus , so thanks for work !
ooh my god! Patience really worth it😱😱
Thank you for the support you given me on instagram sir. This year's integration bee had been difficult than that of pervious years.
@@dp_maths_world I honestly think 2023 was the toughest one
23:50 Since f-g >0 then f>g but f = (2cos2x-1)^2 , but you wrote 4(sinx)^2
and (sin2x)^2 < (cos3x)^2 but you wrote (cos3x)^2
I've check the integral of 0 to Pi of (2cos2x-1)^2*(sin2x)^2 dx
and it equals to Pi, Am I wrong or ..........................
And the tradition continues
Math
Heritage
For problem 1 wouldnt it be easier to just use arctanh or is that not allowed?
Yes that can be used to skip a couple steps
am i crazy or at the end of #2 is -4 + 25/12 supposed to be -23/12 and not -73/12
It's supposed to be - 4 - 25/12, so you do end up with -73/12.
@nicolasferraro4632 so there's a plus when there should've been a minus?
He messed up the partial fraction decomposition at 10:33 since the correct factors are (n-1) and (n-5) and not (n+1) and (n+5).
Yeah homie I basically missed a negative but kept winging it💀
And no that factorisation doesn't work, just multiply the terms and you'll get a -6n instead of 6n
@@Akamaikai0923 yeah, he just switched them up at 14:02.
You could invoke the Lucas sequence for the golden ratio integral
For max(4sin^2(x), (2cos2x-1)^2), g-f = 16sin^4(x) - 12sin^2(x) +1 varies between -1 and 1 in [0, pi] and so happens for min(sin^2(2x), cos^2(3x)) where g-f = 16cos^6(x)-20cos^4(x)+5cos^2(x) varies between -1 and 1.
I don't know what are you doing, sincerelly, at that step.
Those are all even powers so we don't get the -1
@@maths_505i don't get it too, if you grafh 16sin⁴x - 12sin²x + 1 you can have negative values, for ex: x = π/4. would be -1, just because they are even exponentials doesnt mean that it's always gonna be positive so i'm still confused about that step
Understandable, have a great day
uh oh 28:42 should've been -4y+4 not +1 basic arithmetic lowk feeling harder than the integrals😭😭😭😭
Indeed it is😭😭
You probably already found this out or someone else mentioned it but in problem 2, -48/12 + 25/12 is -23/12
13:36 i love the self encouragement lol
😂😂😂
Shouldn't it be int[0,10](floor(phi)^floor(x))dx? So then because phi is ~1.6... floor(phi) is just 1, and then 1^x is just 1, so its int[0,10]dx which is just 10????
No it turns out I wrote the integral wrong in LaTeX
My favourite person since 2024❤
9:06 am I crazy? But the derivative of 1/(u^2 + 1 + u) is equal to (-2u -1)/(u^2 + 1 + u)^2 and that is not what it was 1 step before…
The second u is not a u, it's a n (a constant).
Wow. Just wow
@@mathscribbles thanks mate
Only the first one was easy.......got the idea easily........the other ones were untouchable
Where's Cleo when you need her?
Hi,
Terribly sorry about that : I lost the timing I did yesterday 🤥I thought I posted it but no 😔.
But I liked the video, as usual.🙂
No problem my friend
ahh shittttt here we go again
X,2x+5=8
I don't understand how anyone can enjoy this and I'm studying physics 😅
офигительно круто!
At 10:33 you messed up the factorisation n²+6n+5=(n-5)(n-1)≠(n+5)(n+1)
@bennetdiesperger4080 aight my math's bad but I do know that what you wrote has a -6n 💀
Okay cool ❤
oookay cool
Lfg!!!