I’m pretty sure what you’ve been calling “periodicity” is what is meant by orthogonality. In the context of the 4th integral, it just means that int_0^{2pi} cos(mx)cos(nx)dx=0 for integers m=/=n. In linear algebra, an inner product space is a vector space equipped with an abstraction of the dot product called an inner product, usually denoted as . The inner product space that is useful to consider for this integral is V= real linear combinations of B={1,sinx,cosx,sin(2x),cos(2x),…} with inner product =int_0^pi f(x)g(x)dx. We call B an orthogonal basis of V because for any distinct f,g in B, =0. Properties of this particular inner product space are also incredibly useful in understanding Fourier series and also allows for generalisations of Fourier series.
OHHHHHHH, that connection makes a lot of sense now!! Thank you for the explanation. Now the real question is, how on earth do you apply this knowledge with this integral in speed?
@@Silver-cu5up I think that complexifying then either residue theorem or a different orthogonality result is the quickest approach to this integral. By writing cos(5x) as Re(e^{5ix}), the integrand can be rewritten as (1/32)Re(2e^{ix}-1-e^{2ix})^5 which is nicely set up for the residue theorem. Alternatively, you could consider expanding out this trinomial expansion and using the fact that int_0^{2pi} e^{nix} dx =0 for non-zero integer n, so only the constant term in the expansion will actually matter.
first one is not bs, atleast, its quite a standard technique, like when you have to integrate e^xf(x) you look for a way to write f(x) as f(x) + f'(x), if you havent seen it before, fair enough for considering it to be difficult, but in my jee coaching classes its a standard technique and most integrals where you have e^x times something you solve like that. So i imagine the contestants were well versed in it, that said its not trivial to spot how to split f(x) into f(x) + f'(x) here, its certainly challenging, but much more doable when you're 90% sure youre trying the right approach
ooh i got the 12:38 one in less than 3 minutes, atleast the method for it, like you combine the sin20x and sin22x and with the 2sin(a+b)/2cos(a-b)/2 identity, do the same for the other bracket but with the cos identity, you get a common term to factor, cancel the squares and its trivial from there. Its quite doable without complex right? I dont understand why you think its not doable otherwise. Hell im surprised this is a finals integral i struggled way more on the others, now im wondering if i made a mistake in my calculations, let me write everything out here sin22x + sin20x = 2sin(21x)cosx the first bracket becomes ((sin21x(2cosx+3))^2 cos22x + cos20x = 2cos(21x)cosx and the second bracket becomes ((cos(21x)(2cosx+3))^2 factor the common term you get (2cosx+3)sqrt(sin^2(21x) + cos^2(21x)) now its just integrating 2cosx + 3 im not sure where i went wrong, it seems too easy
"I don't know what that means" this man cannot be more relatable I have to say I agree with some of these but for other ones I think they're pretty good problems.
I LOVE THIS DUDE!!! HE IS BACK AT IT AGAIN fresh off the goat series!!!
I’m pretty sure what you’ve been calling “periodicity” is what is meant by orthogonality. In the context of the 4th integral, it just means that int_0^{2pi} cos(mx)cos(nx)dx=0 for integers m=/=n. In linear algebra, an inner product space is a vector space equipped with an abstraction of the dot product called an inner product, usually denoted as . The inner product space that is useful to consider for this integral is V= real linear combinations of B={1,sinx,cosx,sin(2x),cos(2x),…} with inner product =int_0^pi f(x)g(x)dx. We call B an orthogonal basis of V because for any distinct f,g in B, =0. Properties of this particular inner product space are also incredibly useful in understanding Fourier series and also allows for generalisations of Fourier series.
OHHHHHHH, that connection makes a lot of sense now!! Thank you for the explanation. Now the real question is, how on earth do you apply this knowledge with this integral in speed?
@@Silver-cu5up I think that complexifying then either residue theorem or a different orthogonality result is the quickest approach to this integral. By writing cos(5x) as Re(e^{5ix}), the integrand can be rewritten as (1/32)Re(2e^{ix}-1-e^{2ix})^5 which is nicely set up for the residue theorem. Alternatively, you could consider expanding out this trinomial expansion and using the fact that int_0^{2pi} e^{nix} dx =0 for non-zero integer n, so only the constant term in the expansion will actually matter.
@@calcul8er205 0o0 Oh dayum!! Thanks for sharing this idea!
first one is not bs, atleast, its quite a standard technique, like when you have to integrate e^xf(x) you look for a way to write f(x) as f(x) + f'(x), if you havent seen it before, fair enough for considering it to be difficult, but in my jee coaching classes its a standard technique and most integrals where you have e^x times something you solve like that. So i imagine the contestants were well versed in it, that said its not trivial to spot how to split f(x) into f(x) + f'(x) here, its certainly challenging, but much more doable when you're 90% sure youre trying the right approach
Yeahhh agreed
ooh i got the 12:38 one in less than 3 minutes, atleast the method for it, like you combine the sin20x and sin22x and with the 2sin(a+b)/2cos(a-b)/2 identity, do the same for the other bracket but with the cos identity, you get a common term to factor, cancel the squares and its trivial from there. Its quite doable without complex right? I dont understand why you think its not doable otherwise. Hell im surprised this is a finals integral i struggled way more on the others, now im wondering if i made a mistake in my calculations, let me write everything out here
sin22x + sin20x = 2sin(21x)cosx
the first bracket becomes ((sin21x(2cosx+3))^2
cos22x + cos20x = 2cos(21x)cosx
and the second bracket becomes ((cos(21x)(2cosx+3))^2
factor the common term you get (2cosx+3)sqrt(sin^2(21x) + cos^2(21x))
now its just integrating 2cosx + 3
im not sure where i went wrong, it seems too easy
Integration bee is soposed to be challenging and just for fun.its not a competition if it's not challenging
Does anyone know if there are solutions to all the integrals somewhere? I'd like to know if I'm doing it right.
@@Atomcodes_16 unfortunately theres no official solution for all integrals ;_;
"I don't know what that means" this man cannot be more relatable
I have to say I agree with some of these but for other ones I think they're pretty good problems.
@@danielrosado3213 yeee lol
12:38 dont doubt the JEE aspirants man, they pull out random shit that just works
@@cdkw2 LMAO
3:40 please tell me how to solve this using reverse quotient rule
The answer is actually reverse product rule. But im assuming its one of those guess n check. Other than that, idk... ;_;
@@Silver-cu5up damn...
@@Silver-cu5up how do u know it's reverse product rule?
@@Kishblockpro the answer it gave shows that it uses reverse product rule ._.
but without knowing the answer idk
Hi @@Kishblockpro
What is the glassers master corollary?
"Let f be a decreasing function. Then integral of f(x+1/x) dx from 1/2 to 2 = integral of f(sqrt(x^2+4))dx from 0 to 3/2."
@@Silver-cu5up you mean increasing, right? 🤓
@tommy4808 The corollary says decreasing 0_0
@ log is increasing though?
@@tommy4808 its from here:
artofproblemsolving.com/community/q2h3012296p27059643
2022 was definitely the most braindead year
Absolutely ._.