M.I.T. Integration Bee Question

[@rob876]

At 7:40 you could have made the substitution (u + 2)/√3 = w to arrive at the answer quicker using partial fractions.

[@TomRocksMaths]

Yes nice spot!

[@kenroyadams2762]

Absolutely fantastic! Thank you so much for taking the time and effort to do this integral.

[@TomRocksMaths]

you're very welcome :)

[@polychromaa]

Alternate method is hyperbolic weierstrass substitution; t=tanh(x/2)

[@TomRocksMaths]

nice!

[@gurkiratsingh7tha993]

When you are done with the u sub then you can use partial fractions to solve the rest, I know that u^2+4u+1 is not Factorable over rationals but it is factorable over reals, i.e. u^2+4u+1 = (u+2+√3)(u-√3+2)

[@TomRocksMaths]

nice!

[@chenkuanmin]

yes , my thinking is also this

[@gurkiratsingh7tha993]

@@TomRocksMaths thanks

[@giuseppepalazzo6022]

Didn't know Machine Gun Kelly was that good at math

[@TomRocksMaths]

Side hustle

[@a_k__]

I really like the style of your teaching. I always feel I learned something when I watch your videos. I think it would be very useful if you can make a video about different part of Math as a subject. Kind of like a roadmap that shows different parts of math and how it is used in different industries. I think your unique style of teaching is great for it

[@YawnGod]

You NSA? CIA?

[@NazriB]

Lies again? Most Irresistible PES

[@yugiohsc]

I love watching the integration bee! It’s the one sport i enjoy watching

[@TomRocksMaths]

olympics next?

[@itsreeah2663]

Happy New Year! Have a mathematical one

[@hugolindholm4889]

Why not factorize it into (u+2-sqrt(3))*(u+2+sqrt(3))? Then do partial fraction decomposition?

[@timothyaugustine7093]

Exactly. That'll not require any hyperbolic trigonometric functions.

[@Zonnymaka]

I guess that wouldn't have been fancy enough

[@franolich3]

Simpler method similar to what others have suggested: J = Integral [ 0 to infinity : dx / (2 + cosh(x)) ] = Integral [ 0 to infinity : 2e^x dx / (e^2x + 4e^x + 1) ] The roots of (y^2 + 4y + 1) are y = -2 +/- sqrt(3) J = Integral [ 0 to infinity : 2e^x dx / (e^x + 2 - sqrt(3)) (e^x + 2 + sqrt(3)) ] Let u = e^x + 2 => du = e^x dx J = Integral [ 3 to infinity : 2 du / (u - sqrt(3))(u + sqrt(3)) ] By partial fractions: J = (1/sqrt(3)) * Integral [ 3 to infinity : 1/(u - sqrt(3)) - 1/(u + sqrt(3)) du ] = (1/sqrt(3)) * [ 3 to infinity : ln( (u - sqrt(3)) / (u + sqrt(3) ) ] = (1/sqrt(3)) * [ ln(1) - ln( (3 - sqrt(3)) / (3 + sqrt(3) ) ] = (1/sqrt(3)) * ln( (sqrt(3) + 1) / (sqrt(3) - 1) )

[@TomRocksMaths]

nice!

[@ghostek7792]

WTF@!@! that app is insane i've never seen it. i'm about to start integral calculus for software engineering and knowing there are such tools to assist in learning is going to give me peace of mind when i encounter problems that I may struggle on. the break down on the app is so thorough as well i love it! thx for the video

[@TomRocksMaths]

glad you find it useful!

[@IbraHermoso]

Hi Tom, happy new year!!

[@TomRocksMaths]

and to you too!

[@owen7185]

I love the symmetry in the answer

[@ThatWeirdCat]

i didn't like math because i can't get past 70% in every exam i take in my school, but i like this dude. 10/10 🔥

[@smellund]

I don’t even begin to understand integrals but still watched the whole thing, good video

[@TomRocksMaths]

as long as you had fun :)

[@angelodiavolo3915]

Yesterday I studied integrals for the first time. Today I watch this video. I have surprised understood more then I thought I would hahah

[@TomRocksMaths]

Great job :)

[@ranpancake]

yet another banger ! [also, love the shirt :)]

[@TomRocksMaths]

thanks!!

[@ArjunBhanap]

Great question!

[@TomRocksMaths]

Try some other questions from the M.I.T. Integration Bee as selected by Tom in this FREE Maple Learn worksheet: learn.maplesoft.com/?d=LTBSKOBSBSLIESPFETLUHHCMINPMMIGQELHHCNCUGIBTDLEGIFNMNHPHOGBJBMNONGEPGFITMMKJMKKRHQBQCSLTMPALJSBNKULU

[@iTeerRex]

It’s one of those problems with 2 stars at the end of the chapter, and after banging your head against the wall for some time. You get the answer with a negative sign 😂 Happy new year everybody, and may this omen be lifted soon.

[@Carlos-vn4ec]

this is so buck wild. I forgot how cool doing integrals can be

[@TomRocksMaths]

you said it

[@rossg9361]

Beautifully explained, but even as a decent amateur mathematician I would struggle to know where to start. The way in?

[@TomRocksMaths]

I think replacing cosh by it's definition in terms of exponentials was really helpful as then it starts to look like a polynomial

[@adrianh1307]

1:51 "is given here by this function" it's not a function, it's a number

[@chanddd2]

wtf did I just watch...? I'm a lawyer

[@TomRocksMaths]

As long as you had fun :)

[@d314159]

Took over a minute to do it in my head, including some checking. Getting old ...

[@mathunt1130]

Why not directly use t=tanh(x/2)? People are taught a similar trick when you have trig functions, and it works when you have hyperbolic functions as well.

[@TomRocksMaths]

Yes this would also have worked.

[@purim_sakamoto]

おー　お手本のような式変形　こういうの訓練しときたいなあ

[@gujjalamohanmanjunath3156]

Try jee advanced integration questions if you are perfect in them then you are perfect in integration

[@Genus-Homo_Species-Sapiens]

For ur kind info, even JEE Adv AIR 1, 2020 lost in MIT Integration Bee semi finals 2024 MIT ki baat ho rahi h, unko gyan dene ki jarurat nhi h IMO Gold Medalists h waha,,JEE Adv se tougher que solve kr k baithe h wo

[@sergiolucas38]

great video :) just a curiosity, from what i see, most people dont use to rationalize expressions, why is that so? for example, that sqrt3 at the end. thanks.

[@TomRocksMaths]

I should have done this really, but hopefully you can see the two answers are equivalent.

[@kevincardenas6629]

Hi Tom I really enjoy your videos, I recall once seeing a video where you solve an optimization problem that was related to entropy, there was this sum over some probabilities but I've looked over your channel and can't really find it :( Do you remember doing a video like that? If so it would help me a lot you could direct me to the video!

[@TomRocksMaths]

Oooooo this might be one of the questions I did in an exam on Flammable Maths - maybe try here: kzbin.info/www/bejne/mIG0iWyti6l7ibs

[@danielbenton5817]

Why don't you take the minus sign out and sub in tanh so you don't need to look any integrals up?

[@nitayweksler3051]

I learned all those tricka in calc1 but we didnt really speak of hyperbollic functions so the coshx=.... is the hardset part for me lol

[@ninjaneer0]

This man teaching math looks like a punk rock guitarist with killer tattoos? My attention is grabbed immediately.

[@TomRocksMaths]

If only I had any musical talent...

[@drliorsilberman]

Why so complicated? Like any rational function of trig functions, the standard "tangent of half the angle" substitution solves the problem immediately. Let $t = \tanh(x/2)$. Then $dx = \frac{2dt}{1-t^2}$ and $\cosh x = \frac{1+t^2}{1-t^2}$. Making the substitution rewrites our integral as $\int_0^1 {2dt}{3-t^2}$ and this is immediate.

[@ytpanda398]

sorry I understand your first part just fine, but how is cosh(x) equivalent to [1 + tanh(x/2)^2] / [1-tanh(x/2)^2] ? Thank you in advance

[@drliorsilberman]

@@ytpanda398 They aren't "equivalent" -- they are straight up equal. Indeed $t^2 = \frac{e^x+e^{-x}-2}{e^x+e^{-x}+2}$, so $1 + t^2 = 2\frac{e^x+e^{-x}}{e^x+e^{-x}+2}$ and $1-t^2 = \frac{4}{e^x+e^{-x}+2}$. It follows that $\frac{1+t^2}{1-t^2} = \frac{2(e^x+e^{-x})}{4} = \frac{e^x+e^{-x}}{2} = \cosh x$.

[@drliorsilberman]

It's the kind of formulas you should know if you want to compete in an integration bee.

[@Bee-gp3nq]

Approved.

[@TomRocksMaths]

thank you Bee

[@superdan2593]

Hey Tom! Great video! I have special request for you... Would you be able to explain the fluid physics that Adam Savage demonstrated with his gauge blocks on his latest Tested video. You could skip directly to 24:17 for an example.

[@TomRocksMaths]

I'll check it out - thanks for the tip!

[@anawilliams1332]

Hey 👋. First of all i love your videos, they are littteraly amazing. But i didnt really get some of this video. Are functions like cosh and arcoth related to exponential functions or trigonometric ones. Im kind of confuced and feel like a bit stupid. Im in year 12 so i think its something we havent covered yet but im really curious now

[@TomRocksMaths]

They are related to both trigonometric and exponential functions. You should come across them in Further Maths I believe.

[@anawilliams1332]

@@TomRocksMaths just covered them a couple days ago, will need to rewatch this video now :)

[@eashanshenai4980]

6:24 why is the denominator to (u+2)² root 3 and not just 3?

[@eashanshenai4980]

@Khaled Mohammed oh I see, thanks!

[@morehave2417]

hi man, try to use partial fractions method and you gonna end up with ( ln( (e^x+2-root_3) / e^x2+2+root_3 ) ) / root_3 from 0 to infinity, then sub in the result it will be ( ln(2+root_3) ) / root_3

[@jannien4129]

how can you say (u+2)/sqrt3 = coshy ???

[@TomRocksMaths]

This is the substitution I am using to try to simplify the integral

[@KenStarkey]

I feel like expanding out the hyperbolic solution to the ln function is an unnecessary step. Just because Maple did it does not make it a more natural solution.

[@TomRocksMaths]

I agree it isn't needed to get the answer, but the purpose of using it was to show the solutions are equivalent as you say. It also helps if someone does it using partial fractions as then they will get a log also.

[@leif1075]

@@TomRocksMaths Thanks for sharing. O really hooeyou can respond to my other comment in the new year when you get a chance. Thanks very much

[@cxkeisdelusional]

why is xqc teaching math

[@TomRocksMaths]

that's a new one

[@trillionman2105]

Shine x

[@TomRocksMaths]

yep

[@andrewgonzales1359]

If this guy went to Oxford, why does he have to use a calculator?

[@amritkumarpatel5717]

here

[@solarine-7354]

Machine gun Kelly solve math…

[@TomRocksMaths]

side project

[@meiwinspoi5080]

tom is un-neccesaarily complicating the calculations. treat the equation in denominator as a quadratic in e^x and use partial fractions to get a neat expression. the lower bound will be sqrt 3 and upper bound infinity. the integral is standard integral and for upper bound apply limit t tends to infinity of ln (t-1)/(t+1) which is zero and get your final answer. fewer steps and neat. btw the app sucks as an ipad app. why would any one write on a paper and take a picture and compute. so ipad unfriendly. one should be able to directly write on the app and get results.

[@NeverTalkToCops1]

A puzzle to solve while you are in the hospital. KZbin math puzzles have little merit. The value would be in thoroughly understanding what integration is and how to apply it in various endeavors. Otherwise, you're just some 7 year old grinding away on Rubik's Cube, forever.

[@romanemul1]

Let the sin shine.

[@TomRocksMaths]

i stand by shine

[@mohammadalkousa2856]

Great! Recently it was published a book about MIT integration bee, under the title " MIT Integration Bee, Solutions of Qualifying Tests from 2010 to 2023" You can simply find it!

[@Zeusbeer]

??????????????????????????????? Why not just use partial fractions, its not that hard since it is a linear term

[@beachboardfan9544]

Seems like a terrible problem, seems to require a ton of pre-existing knowledge to solve.

[@TomRocksMaths]

You need to know hyperbolic functions, but aside from that it is just a couple of clever substitutions.

[@beachboardfan9544]

@@TomRocksMaths How do you know the substitutions?

[@two697]

Not really. The only part people might not remember is the integral of cosech

[@nitinankareddy5538]

When you are done with the u sub then you can use partial fractions to solve the rest, I know that u^2+4u+1 is not Factorable over rationals but it is factorable over reals, i.e. u^2+4u+1 = (u+2+√3)(u-√3+2)

[@TomRocksMaths]

nice!