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That will not work no

@@Silver-cu5up ohhhh my bad, I forgot it's -1/x^2 and not -1/x^(1/2), thanks

@@Ex-ws2vb I competed in Fresno State Central Valley Integration Bee and Reedley College. First time at Fresno State I got eliminated at semifinals. 2nd time at Reedley College and won 1st. (I didnt score perfectly tho ;_;) 3rd and 4th time at Fresno State and won both as 1st place with perfect scores

AAAAHHHHHHHHHHH!!!!! I swear my brain just feels lazier and lazier when it comes to speed bashy integrals ;_;

I caught that too

The |x| is mainly for logs. The sqrt(x)'s can actually be left alone, but for logs we need to put |x| on it.

@@Silver-cu5up This isn't true in this case. The domain of this function is already x>0 because of the square root, you don't need the absolute value in the log

@@ninadmunshi2879 Ooohhhhhhhhh, I see what you mean now. Gotcha, yea becuz of the domain of sqrt(x) stops it from passing through negative side, so the |x| is not needed ever in this integral.

@@ben_adel3437 just be very careful with that sqrt(e^u) when doing IBP very fast

its just not that common lol, but for fractional parts its super useful and quick

oooof but i do wish some classmates were more "human" minded rather than having this "smart boi" mask over us and have expectations. Sorry u had to go through that, but absolutely check your work after finishing

very slick way for those who are more comfortable with forced u-sub and memorized the integral of ln(x)

OMG UR RIGHT LMAO

UK University Integration Bee has a lot of integrals that utilizes Feynman Technique and Taylor Series. You could also search on MathStackExchange, but other than that there's not much resources that has a convienent list.

I second this question!

Majority was self-taught and archiving integration techniques from other people's solutions. Unfortunately there was no resources for any of these tricks, so I decided to kind of popularize them and give it a name to help memorize and practice such techniques. The piecewise functions like floor(x), {x}, and max(x), I was kind of trained by my friends on how to solve those types of integrals, and then Brilliant.org helped me generalize a method on how to speedsolve them (unfortunately the page doesn't exist anymore ;_;)

Become a better me xD

@@SCBA-if4wl LMFAO

1+sqrt(2) is the height of a regular octagon with side length 1. Assuming the points are labeled A-H going counterclockwise, construct triangle ABE. The height of this triangle is sqrt(2)+1, and the width is 1. Since line AE bisects angle HAB, which measures 3pi/4 because it's a regular octagon, angle EAB measures 3pi/8, except that's exactly the angle we want. Thus, arctan(sqrt(2)+1)=3pi/8.

@@maxvangulik1988 OMFG, THATS CRAZY!!!

@@maxvangulik1988 I feel like this must be some part of their intention when creating this problem!!! So the actual answer to that original integral would've been 3pi/4!!!

@@adityavsx sheeeeeeeeeeesh, very nice!

@ would u be knowing any resources for algebra like you know so many for calc? I’m prepping for JEE Advanced and math is kind of??my strongest point and hence I need to sharpen it more so…

@adityavsx oh dang, unfortunately idk much resources for JEE Algebra ;_; Id probably look into those JEE videos on TH-cam where coaches do oneshots or problem marathon like Vedantu or other coaches. But if those dont help, sorry idk much about it ;_; You could also look into a book called JEE Algebra Booster (i forgot the author name but its a blue book). Thats as much as i know sadly. Sry about that.

@ no problems uff, chillax 😂thnx doe

@@fxrce6929 self-taught ;_; There werent that much resources back then sadly...so i started writing a journal to collect integration bee techniques and archive common sneaky integrals during high school.

CMM says it should be csc(2)-pi/4 ;_;

@Silver-cu5up That's the same thing as well 🤩 Well I guess they wanted the simplest possible form...

@@eggcorpprod really? Huh, last time i put it in wolframalpha, it was a different numerical answer

@@calcul8er205 shouldve speedsolved passively 😭

If you go to my channel playlist, I have an Integration Bee Training Series in all 3 levels!! 😃 I believe the Gamma function is in Intermediate level, but the Infinite Series is in Advanced level. Thank you, I really appreciate it!!!

​@@Silver-cu5upThe pleasure is mine, man- You're doing a great job with the content quality! 100% eager to continue watching, and I hope you use your passion to bring new ideas for future videos. 💯💜

lmfao

Surprisingly yea xD

I started doing that a lot more now that I'm more advanced xD you'll even see me do this more often in advanced level integrals lol

oh shizz!! I hope it goes well!! This integral is one of the infamous integrals known for eliminating a lot of competitors. There's not exactly an application for this integral, it just reoccurs a lot in intermediate-advanced level integration bees.

@@Silver-cu5up I see, thank you!

it's used in physics a lot but I only used it in physics so far, I don't know about others besides maths

@khanggamr7454 sounds about right lol

that works too!!!

@@cdkw2 LMAO, someone has told me that before XD

the upper bounds are from 1/2 to 3/2 because w = u+1/2. Our bounds were from 0 to 1 before, so with our subsitution we get our new bounds 0+1/2 = 1/2 and 1+1/2 = 3/2.

@@TM-ln5mr Most are already fully explained in my integration bee training series!! 😁

@@Jolobee-x7c oh dayum, good luck!!

This really made me so mad ;_;

@@jonathanalvarado8007 its just a wacky integral using a different variable. Its pretty much an integral of x^2 dx from 0 to 1. But instead they used d' to confuse the competitors.

I was genuinely astonished and devastated xD

@@esp-elagogisch-sozialepart9701 That works yea!!

oh I just realized TH-cam doesnt do the email setting on my end, but u can just go ahead and send it here: silveralchemist544@gmail.com

@@ninadmunshi2879 idk if it would be the simplest. The quadratic is too tricky to deal with for some others (although very satisfying), and the negative can be easily missed. And then the sqrt(x^2+1/4) theyd have to deal with after.

@Silver-cu5up I can understand the hesitation behind quadratic formula, but there's no way around this integral without some kind of "trick" and quadratic formula is a one and done instead of chugging through hordes of identities. Second, the integral of sqrt(x^2+a^2) is the area under a hyperbola. Either you can quickly do the cosh^2t integral or it's just muscle memory because this integral shows up all the time. But that means computation wise there's only two things to do that require work instead of multiple things.

@@cdkw2 Im a problem writer for the BMT Integration Bee xD. This year 2024 tho i was assisted by a lot more people, so idk the official integrals and I dont have the problem sheet yet ;_;

@@Silver-cu5up ooooh npnp

@@fxrce6929 I dont think so no. I dont think theres an exact closed form for it.

@@abdulllllahhh OMG AINT NO WAY!! DID U WON YOUR FIRST INTEGRATION BEE????

@ 5-0 in the finals (shoutout frullani) 💪💪

@@abdulllllahhh SHEEEEEEEEEESH!!! CONGRATS!!!

@@abdulllllahhh you mind sharing some integrals? 🥺

@@Silver-cu5up they were all lowkey relatively easy, like one of them was literally a simple kings rule so I instantly knew it was 𝛑/4, the other was a frulani with Arctan, another was 0 to 1 of (-logx)^-1/2 which just becomes Γ(1/2), another one evaluated to ζ(2) but I can’t remember what the integral was, 5th one was actually hard 0 to 1 of (loglog1/x)/(1+x)². Idk how they expected us to evaluate it, but I recognized the integral from Maths505 so I had a general idea of what to do. u = log1/x. Then take I(α) as 0 to inf (x^α-1)(e^-x)/(1+e^-x)² and notice that I = I’(1). Then use geometric series and rearrange to see that I(α) = Γ(α)η(α-1), differentiate and plug in 1 to get I = 1/2 log (𝛑/2e^γ) where γ is the euler mascharoni constant. Really wide variety of difficulty with these integrals lol

oh shizz xD thanks and welcome!!

Timing myself: 1st Attempt: (2 minutes, 12.72 seconds) = 2:12.72 2nd Attempt (after knowing the process): 1:32.08 3rd Attempt (knowing the process better): 55.60 4th Attempt (memorized some answers of the process): 38.24 sec. Now if I have wait long, i should be slow again forgetting some answers in the process. Then remember it back and repeat until I fully memorized the shortcut to this integral. **SPOLIER: ** I'll end up remembering that: integral of sqrt(x+sqrt(x)) dx from 0 to 1 = 4sqrt(2)/3 - 1/8 [integral of cosh(2x)-1 dx from 0 to arccosh(3)].