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@phat5340 Сағат бұрын
There is a really cool richard borcherds video on the borwein integral, I'd suggest everyone here takes a look
@onusiddartha1641 2 сағат бұрын
Never heard of this 😂😂😂 U r the best 😂
@zzzluvsk 14 сағат бұрын
Dude thanks sm for these. Im still in calc 2 learning series, but once im done with this semester I cant wait to go thru this advanced playlist
@ninadmunshi2879 15 сағат бұрын
This man is a machine!
@MayankXOR Күн бұрын
Nice content dude keep this series alive please.
@cdkw2 Күн бұрын
This is quality
@oonkderdoonker Күн бұрын
Please correct me if I'm wrong but at 18:26 isn't that I and not 2I? Also just wanna say very nice and good video I like it :)
@Silver-cu5up Күн бұрын
oh yea i got mixed up lol, that integral is actually "I" still.
@enzocoronel354 Күн бұрын
11:12 from this moment you could separate into 2 integrals by using u2+1=(u2+1)+1 and then distributing the denominator. This separation allows you to complexify one of the integrals with u=tanθ and the other one gets easier because is an arctan. After adding the integrals and then multiplying by 4 I got π.
@Silver-cu5up Күн бұрын
How do you complexify dθ/(3tan^2θ+1) nicely???
@enzocoronel354 Күн бұрын
@@Silver-cu5up dθ/(3tan^2θ+1) = dθ/(3(sin^2θ)/(cos^2θ)+1) = cos^2θ.dθ/3(sin^2θ)+(cos^2θ) = complexifying = (2cos(2θ)+2)dθ/(-6cos(2θ)-6)+(2cos(2θ)+2)
@Silver-cu5up Күн бұрын
@@enzocoronel354 ohhhhhhhh, gotcha
@cdkw2 Күн бұрын
21:21 nvm I found something to comment about LOL!
@Silver-cu5up Күн бұрын
lmao, thanks for this timestamp xD
@cdkw2 Күн бұрын
I dont know what to say so I guess I am commenting for the algorithm 🔥🔥🔥🔥🔥
@Silver-cu5up Күн бұрын
does commenting actually cause that?
@cdkw2 17 сағат бұрын
@@Silver-cu5up yeah! Thats my many KZbinrs do question of the day shenanigans
@ninadmunshi2879 2 күн бұрын
You can drop requirements 2 and 3 and replace it with pi periodicity instead. This gives the more general and more useful trick which reduces to half the integral of f from 0 to pi.
@cdkw2 2 күн бұрын
I am hosting a integration bee in my school, these will def come in handy!
@Silver-cu5up 2 күн бұрын
oh shizzz, nice!! i hope it turns out well!!!
@joshpradhan3292 2 күн бұрын
8:38 3+cosx = (cosx - 2) + 5 in the numerator and then you split to get a t sub with a cosx in just the denominator.
@Silver-cu5up 2 күн бұрын
That definitely would be faster!! 0o0
@fxrce6929 3 күн бұрын
In Australia we call this t-formulae
@Silver-cu5up 3 күн бұрын
oh thats right 0o0 In general trig, you guys call this t-formula, but usually for solving complicated algebraic trig-identities. (I only know cuz Eddie Woo lol)
@cdkw2 3 күн бұрын
13:30 did bro pull out a gun 💀
@Silver-cu5up 3 күн бұрын
Lol, that was my pen and tablet xD
@cdkw2 3 күн бұрын
gg
@MayankXOR 3 күн бұрын
Thanks 🙏
@TheOfficialFLP 4 күн бұрын
Bro... yeah, ggs
@joshpradhan3292 4 күн бұрын
Yeah this is definitely the hardest part of section 12. Last q: I got -1/3 - 2(1/5 - 1/7 + 1/9 - 1/11 + ... ) = 1 - π/2 using arctanx Taylor series Is this cooked or fire?
@Silver-cu5up 4 күн бұрын
You got it!! Nice job!!! 🤌🔥
@phat5340 5 күн бұрын
Don't wanna be that guy but at the end were you break up the integral don't you presuppose were it converges ie the reimann summation theorem
@Silver-cu5up 4 күн бұрын
Im not tooooo familiar with the riemann summation theorem. Idk how this can be applied.
@evankniffen 5 күн бұрын
bro is cranking these out; best week ever
@MayankXOR 5 күн бұрын
Great lecture.
@qwertyman123 5 күн бұрын
nice
@Stormisrunner 7 күн бұрын
5:12 in morning but gonna crunch
@Stormisrunner 8 күн бұрын
fav video
@joshpradhan3292 8 күн бұрын
hype
@bra1nwave172 14 күн бұрын
22:29 If we let u = lnx then the integrand becomes the product of an exponential function and cosine.
@Silver-cu5up 14 күн бұрын
Yep, thats another nice method too! Also can follow up with complexifying the integral.
@simoelabbassi390 Ай бұрын
Amazing work 👏👍 .. keep it up ❤
@simoelabbassi390 Ай бұрын
Explain us multi and line integral please ❤❤
@Silver-cu5up Ай бұрын
I could maybe explain multiple integrals in the future but I don't really know too much in multivariable calculus. I definitely don't remember how to do line integrals ;-; since I dont ever work with those problems
@sigmainclination9483 Ай бұрын
Nice ❤
@ismaailmousa1447 Ай бұрын
Are you ever going to have some merch?
@Silver-cu5up Ай бұрын
Very unlikely xD I'm a terrible businessman lol
@vismof Ай бұрын
could you explaine more detailed how you rewrite e^(isinx) at 3:05?
@vismof Ай бұрын
oh, nvm i figured it out, its from the identity e^(i theta)
@simoelabbassi390 Ай бұрын
In the second integration why did you change cos^2(2x) with 1/2 ??
@Silver-cu5up Ай бұрын
The work was done mentally: cos^2(2x) = 1/2+1/2*cos(4x). By periodicity, cos(4x) becomes 0 due to the bounds being 0 to 2pi.
@simoelabbassi390 Ай бұрын
@@Silver-cu5up wow I see now thanks 👍
@4axel3loop33 Ай бұрын
When we do a u sub like u = ix, how do we change the bounds if it’s a definite integral? (like when you did u = x*sqrt(1-i) but the bounds stayed as 0 and infinity?)
@Silver-cu5up Ай бұрын
Oooooooh crap, I didn't think about that. For sure 0 becomes 0, but for infinity is a very amazing question, im super glad u mentioned this. I've honestly never came across a case where the integral bound of infinity become complex infinity, and maybe back to real from using 2 different u-subs. Looking at it now, I'm thinking inf * sqrt(1-i) = inf * 2^(1/4)e^(-i*pi/8) = inf* i. At first I thought it was because we were taking the imaginary part, so the integral works. However, cos(x^2) also converges even though the bound is inf*i, so my idea on that was wrong. Unfortunately I know nothing about complex analysis, but after experimenting with WolframAlpha, it's also treating the bound i*inf as inf.
@simoelabbassi390 Ай бұрын
Keep it up ❤
@MayankXOR Ай бұрын
GOAT
@evankniffen Ай бұрын
i’ve been waiting for floor function for so long
@4axel3loop33 Ай бұрын
the 🐐 is back!
@koreantop Ай бұрын
I've been waiting for you, and I'm so happy to see you again!
@snitchzerboo Ай бұрын
i was waiting for this!
@HarryStoltz Ай бұрын
Great vid!
@sigmainclination9483 Ай бұрын
Nice !! Keep it up bro independent of views !! U will be most watched soon ❤
@joshpradhan3292 Ай бұрын
he is indeed the goat
@Silver-cu5up Ай бұрын
Thank u guys!!
@joshpradhan3292 Ай бұрын
I swear the second integral is easier with just plain IBP? Split the integrand into ∫ e^e^x * e^x * cos(e^e^x) dx + ∫ e^e^x * e^x * cos(e^x) dx. First one is an obvious u sub and for second let u = e^x and then its a quick classic.
@Silver-cu5up Ай бұрын
It gets annoying sometimes ;_;
@4axel3loop33 Ай бұрын
For 22 you don’t even need u-sub! (1+ln(x))^2024*(ln(x)+2026) = (1+ln(x))^2025 + 2025(1+ln(x))^2024 = 1 * (1+ln(x))^2025 + x * 2025(1+ln(x))^2024 * 1/x, which is reverse product rule
@evankniffen Ай бұрын
RETURN OF THE KING!!!!!
@MayankXOR Ай бұрын
Finally, you're back again. This week is going to be good!
@codyriceandothers Ай бұрын
too lit