A very unfriendly integral problem!

[@Rajsaday1]

That was the cutest offspring sighting 🥰

[@stephensu4371]

3:36 yes she is the cutest offspring sighting

[@goodplacetostop2973]

I know, right?

[@neilgerace355]

But not unfriendly :)

[@Sanguinium_Light]

@@stephensu4371 Haha, that was good.

[@shubhamsnehil6145]

you take my words :)

[@rishavgupta2117]

Subsituting x wit sin2u will make it lot simplfied as we can easily write sin2u+1 as (sinu +cosu)^2 and 1-sin2u as (cosu-sinu)^2 and in interval 0 to pi/4 we will get the denominator as 2 cos u since cosu>sinu in the interval 0 to pi/4 .and now we will simply have cos2u in num and cosu in den then it will be easy to integrate

[@Swybryd-Nation]

Seriously Brilliant

[@SingaporeSkaterSam]

Love it. Had to look up some standard results but still, very cool sub.

[@paolomorseletto3030]

Very well done

[@bharatpakkha7787]

@@Swybryd-Nation alternatively, if one choses x=cos u, then it reduces to simple integral of -(cos (u/2)-sin(u/2)) in 0 and π/2.

[@leif1075]

Why on earth would anyone thinknof subbing x with sin2u though at all??

[@Hooeylewissukz]

A marathon on ODEs and PDEs would be very interesting to this very jaded graduate!

[@henrymarkson3758]

This channel already has a first class 97 video playlist on ODE's which is quite comprehensive. For this reason I would prefer a PDE marathon or a Linear Algebra marathon .

[@ethanbartiromo2888]

I love PDEs, I took a grad PDEs class last semester, it was super fun, I second that

[@skylardeslypere9909]

@@henrymarkson3758 I am not sure if Michael teaches PDE's though. It would be hard for him to "teach" an entire semester worth or PDE theory for a video if he doesn't teach it normally.

[@felixbelair9251]

Here is ODE one kzbin.info/www/bejne/r5_GgZiOnreXmc0

[@kevinmartin7760]

If you rationalize the denominator of the interior of the ln in the final result (multiply numerator and denominator by √2-1), the denominator becomes 1, and the numerator is (√2-1)^2. Roll the 1/2 outside the ln into this and the result simplifies to: √2+ln(√2-1)

[@nidhiagrawal3354]

You're absolutely right ☺️👍

[@violintegral]

I got √2-ln(1+√2), which is equivalent

[@nidhiagrawal3354]

@@violintegral You're absolutely right as well☺️👍

[@philippelepilote7946]

Sure. I was waiting for this simplification with √1+-x sooner in the process. I tried, but unfortunately this doesn't prevent from solving another limit when x->0+

[@violintegral]

@@philippelepilote7946 there is a simplified form that I reached through a trig and hyperbolic substitution, but I have no idea how someone could reach it otherwise. Yet I did prove that the two were equal, and an integration calculator agreed with me. Anyway, the form for the logarithm I found was ln((1+√(1-x))/(1+√(1+x)))=1/2ln(((√(1+x)-1)(√(1-x)+1))/(√(1+x)+1)(√(1-x)-1))). The graphs are the same on Desmos, too. This equivalency implies that (1-√(1+x))/(√(1-x)-1)=((1+√(1-x))/(1+√(1+x)), which can easily be proven if you cross multiply, giving abs(1-x)+abs(1+x)=2, which is only true on the interval (-1, 1), which so happens to be the domain of our logarithm. I would love to hear if someone has a clever way to reach this simplified form that I have found, starting with the logarithm in Michael's answer.

[@rennanchagas6174]

The child was really unexpected for a video with a mad emoji thumbnail 😂😂

[@habermasnyc]

Either livestream would be great, but my first choice is Abstract Algebra.

[@LUrzidil-rn7nd]

why?

[@hellkr]

...with offspring sighting :D

[@iwalk9234]

I started by letting x=cos(u) and then used double angle formulae to rewrite the denominator in terms of cos and sin of u/2. Don’t have to deal with the discontinuity this way.

[@parthsingh3057]

that's precisely how i did it ! but i may have messed up constants somewhere since i got sqrt(2) + ln(sqrt(2)-1). Also probably because i initially tried to rationalize and then substitute cos2x. EDIT: rationalizing the expression under ln will give the same answer, and thankfully I wasn't wrong :)

[@Forge_1]

Man......I did the same thing but I thought it wouldn't work :cc

[@jaymorf7374]

Nice. Would make a good reply video!

[@samsyet-0074]

Either of Linear Algebra or Abstract algebra would be great! I'm a HS student, and I can definitely use those for uni and also coz I would love to study them.

[@martinnyberg9295]

3:37 ♥️🥰 Cutest mathematician ever. 😊

[@pardeepgarg2640]

Imagine if she likes science instead of mathematics

[@ConManAU]

Since I accidentally skipped the group theory unit when I took maths in university, I’d love to get a crash course in it via a marathon stream!

[@ericthegreat7805]

Did he make one?

[@nyki8056]

I would say an abstract algebra stream would be a nice pick!

[@silversky216]

Wow!! I am loving the new setup. The video quality is awesome, and the audio is really pleasing to the ears. Well done sir!!

[@tinycatzilla]

I'd love to see some first-year analysis content! This level of difficulty ranged to some more difficult calculus would be perfect for me right now.

[@afseraph]

Linear algebra marathon would be great! Abstract algebra also seems nice, but you already have a great group theory playlist in your channel.

[@alnitaka]

I noted that (sqrt(2)-1)/(sqrt(2)+1) = (sqrt(2)-1)^2, because sqrt(2) + 1 is the reciprocal of sqrt(2) - 1. so you finally get sqrt(2)+(1/2)ln((sqrt(2)-1)^2) = sqrt(2)+ln(sqrt(2)-1).

[@weonlygoupfromhere7369]

Lol I love how put the "Offspring Sighted" caption before they came in

[@andreben6224]

Abstract Algebra sounds fun! I'm biased though :)

[@Peter_1986]

I like to try solving integrals in my head when I try to fall asleep during the nights. It is usually something that allows me to do a full u-substitution in the denominator, and then a few tricks with derivatives and with rewriting those equations until I have got rid of the original variable.

[@camrouxbg]

This is a really great problem! Thanks so much :) I love how it covers so many different techniques, plus putting in the limit at the end, then taking care of that limit carefully. Really shows how you have to just be meticulous to not miss any parts, but also just breaking things down into smaller problems we already know how to solve is a really great strategy.

[@insignia201]

Linear algebra would be really cool! Especially if you get into diagonalization and determinants.

[@andreivila7607]

Imagine a complex analysis course with Michael Penn. Heaven on earth :)) P.S. 1/2 ln((sqrt(2)-1)/(sqrt(2)+1)) can be simplified to ln(sqrt(2)-1)

[@MarcusCactus]

I would rather write the solution as root2 - arctanh(root2). Short, thus elegant.

[@asklar]

@@MarcusCactus root2 o ( I - arctanh)

[@-rahul-2908]

Rationalize then substitute x=cos2u then we get a simple integral with 1/2 outside and sqrt2 in numerator and cosu + sin u in denominator just bring root 2 to the denominator we get sin (pi/4 + u) in denominator, its just integral half cosec (u+pi/4) which equals ln [cosec(u+pi/4)-cot (")] from pi/2 to zero by changing limits

[@houseflyer4014]

A number theory Livestream would be great!

[@noumanegaou3227]

kzbin.info/www/bejne/a5a2np2EaKihbNU

[@enejidjsi5939]

yeeeeepppp!!!!

[@martinoreilly8313]

Sweet to follow. I’ll have a go with trig substitutions to see how I get on.

[@skylardeslypere9909]

I really love both linear algebra and abstract algebra. I guess I would prefer seeing a linear algebra livestream, but either is definitely fine.

[@pimpomresolution5202]

Given that you also have an offspring, how about a marathon course on how to get my 11 year old daughter to do her math homework. :-)

[@pianochannel100]

This feels like the integral equivalent of rubbing two wet sticks together to make fire.

[@segmentsAndCurves]

Indeed.

[@b.b4229]

Yes yes yes; I would love a Livestream on Linear and/or abstract algebra. Yes please, as soon as convenient.

[@alexwolffe7805]

Gosh, I love this man DOing (HARD) MATH. A linear algebra marathon would be perfect for a first-year undergraduate, along with the calculus you did.

[@marcinbednara3825]

It would be much easier if we notice that the consider function is even. So it is enough to calculate the integral from -1 to 1 and multiply the result by 0.5

[@henrikholst7490]

Brutal calculation. Reminded me of the calculus class in undergrad where I just had to go home and sleep for a bit just taking notes and following along burned all circuits. ;)

[@goodplacetostop2973]

3:36 Hi 17:11 Posting the video 15 minutes after the expected time DansGame

[@bprpfast]

Hiiiii!

[@goodplacetostop2973]

@@bprpfast Hiiiiii 👋

[@CM63_France]

Hi, My suggestion for the next live stream : geometric algebra. For fun: 0:46 : "our goal is to", 5:37 : "ok, great.", 10:52 : "ok, great.".

[@schrodingerbracat2927]

using substitution u=sqrt(1+x) for the first integral (J1), and v=sqrt(1-x) for the second integral (J2), J1 = integral u²/(u²-1) du from 1 to sqrt(2), J2 = integral v²/(v²-1) du from 0 to 1, they join together nicely as integral v²/(v²-1) du from 0 to sqrt(2) which is easy to evaluate.

[@Nikolas_Davis]

12:25 Here's how a physicist "gets rid of that badness": * sqrt(1+x) ~ 1 + x/2 for small x ==> sqrt(1+x) - 1 ~ x/2 [1] * sqrt(1-x) ~ 1 - x/2 for small x ==> sqrt(1-x) - 1 ~ -x/2 [2] *Divide [1] by [2], you get -1, lose the minus since we're inside the absolute value, then you get 1 * The rest of it is ( sqrt(1-x)+1 ) / ( sqrt(1+x)+1 ), plug in x=0, you get 2/2=1 * all in all, you get ln|1| = ln1 = 0 Done.

[@henrymarkson3758]

Let's go with Plan A, a Linear Algebra marathon.

[@joaopedrobmenezes2977]

Yes

[@patricklynch7574]

I only tuned in to parts of the Calc 2 marathon, but I'd do my best to be there for a full Lin Alg marathon

[@rylanbuck1332]

MIT open courses already has like 60 45minute videos up on linear

[@GroundThing]

One simplification at the end would be to see the 1/2*ln((sqrt(2)-1)/(sqrt(2)+1)) as 1/2*ln((sqrt(2)-1)^2), by rationalizing the denominator inside the ln, then you can bring in the 1/2 inside the ln to cancel out the squaring to get ln(sqrt(2)-1) for a final answer of sqrt(2)+ln(sqrt(2)-1), or alternatively if you want your terms to be positive, as I tend to, you could bring out the negative and get sqrt(2)-ln(sqrt(2)+1)

[@TheJulijiji]

Abstrac algebra seems really nice, Linear algebra videos and marathons are much more easy to find inside KZbin while an abstrac algebra marathon is not.

[@paultoutounji3582]

Hello Michael, I have been watching e few of your demos...You are amazing ! Cogratulations to you ! Pleasure to learn from you ...

[@matthiasbergner8911]

My suggestion would be a marathon about geometric analysis. Looking forward to hearing some new results about Willmore surfaces.

[@hasanjakir360]

The answer should be √2 +ln(√2-1)

[@hendrixgryspeerdt2085]

I solved this integral by: - first rationalizing the denominator - splitting the integral by linearity - substitution: u = sqrt(1+x), w = sqrt(1 - x) (for the respective split integrals) - combining integrals by linearity - then you're left with the integral from 0 to sqrt(2) of u^2/(u^2 - 1)du - finish with partial fraction decomposition answer: sqrt(2) + ln(sqrt(2) - 1)

[@nathanisbored]

yes, please do linear algebra. i struggled with that class and now i have a shaky foundation with abstract vector stuff

[@astonishingzeta4864]

You can use trigonometrical Substitution to tackle the lengthy process to solve that one, If You use x=cos(2z) then it could be most simple to solve than ever... 👍👍👍👍

[@ЛевЯрков-е1ж]

Hello. I found this integral in similar way. First note that in our integral Sf(x)dx f(x) - even function, then we can change boundaries of integration from [0,1] to [-1,1] by multipling integral by one half. Then Sf(x)dx=Sf(x)+g(x)dx, where g(x) - odd function. If f(x)+g(x)=h(x), then f(x)=1/2(h(x)+h(-x)). If we multiply and divide f(x) on (sqrt(1+x)-sqrt(1-x)) we get f(x)=1/2*(sqrt(1+x)/x+sqrt(1-x)/(-x)) -> h(x)=sqrt(1+x)/x. Last part: to find integral from -1 to 1 1/2*sqrt(1+x)/x.

[@kilian8250]

Would be nice to get such a livestream about real analysis or topology :)

[@lavneetjanagal]

The integral can be written simply as the Principal Value integral \sqrt{1+x}/x from -1 to 1. Maybe some contour integral tricks can be used after that.

[@cah220]

Linear Algebra would definitely be interesting as a live stream. Even 20 years out of undergrad, the undergrad computational (say 200 level) linear algebra class makes little sense to me. The senior elective version (as in Axler) made much more sense. Another live stream idea? Big Rudin.

[@thisguyispeculiar]

Here's what I did. Rationalize given form to get this: {1/2 integral of √(1+x) / x} - {1/2 integral of √(1-x) / x} Then in first half, put x = tan²A → dx = 2tanAsec²A And in second half, put x = sin²A → dx = 2sinAcosA Resolve and add up both the terms, you get the same answer.

[@tgx3529]

I have seen very nice solution for these problems.Substitution x=sin2t. But this idea is also nice.

[@shadrana1]

Linear algebra and Matrices,a good subject to start.

[@scipionedelferro]

Definitely an abstract algebra or group theory marathon !! With maybe an advanced topic at the end like the Galois theory and the theorem of non-solvability with radicals of polynomials with degree > 4

[@ElivinMendez]

I would love to see more on Number Theory... more specifically Additive Number Theory

[@calebsyler9515]

Either a Dif Eq livestream of ODEs and PDEs or Linear Algebra is always nice. I think if you do a couple problems with a matrix exponentials would satisfy fans of both maths (if you go the Diff Eq route)

[@CTJ2619]

“all this heinous stuff up here” great video

[@simonzarka-arthus7999]

you can simplify it in sqrt(2) - tanh^-1(sqrt(2))

[@User-gt1lu]

Abstract algebra would be amazing!

[@beanshadow008]

Definitely, abstract algebra is kingue

[@manucitomx]

What a change from yesterday! I’d love the Linear Algebra Marathon. Let’s have more offspring sightings, that was adorable. Thank you, professor!

[@alihamad5246]

I would love it if you do a proper full number theory course because I'm trying to teach that to myself as a high school student, but I need a kickstart at it. Your playlist regarding the subject is cool, but it seems to be significantly incomplete when compared with the contents of number theory textbooks. Like seeing you present case-wise examples as you build up in complexity without presenting examples haphazardly would be amazing. Linear algebra would be amazing as well, but... anything Michael Penn is cool. That offspring sighting.... keep those up!

[@Swybryd-Nation]

A difference in the integrand? Maybe FRULLANI’S Integral is applicable? Or does frullani’s thm. require the integral to go from 0 to inf? Thoughts Prof. Penn?

[@casoheloa7407]

Integral along time axis from 16:05 to 16:10 and find the golden volumn of Mike.

[@sp_danger1729]

I would love a Linear Algebra stream

[@paounn9444]

I would love a differential equations marathon. Loved the topic back at uni, forgot most of the finest details on the topic

[@pseudopathicus2524]

Don’t have time to watch the full vid but maybe try trig substitution when u reached the point at 4:36

[@chilling00000]

How about a live stream on Lie theory

[@mathfincoding]

Very niche topic but would LOOOOOVE this

[@Harshit_Pro]

Please make more videos on Integrals

[@MarcusCactus]

At 1:04 you say "rationalise the denominator". But since x goes from 0 to 1 along the reals, it takes an infinity of irrational values. So 2x is definitely NOT a rationalisation... Should use a word like "unroot" or "derootisation" ?

[@jeromemalenfant6622]

Not to be too picky, but the 2nd term in the final answer, (1/2) ln [ (sqrt(2) - 1)/(sqrt(2) + 1) ] can be simplified to ln [ sqrt(2) - 1 ] by multiplying and dividing the quantity inside the [ ]'s by sqrt(2) -1.

[@rickenbackerlover7386]

I would love to see a livestream about stochastic calculus! Specially about the Itô Calculus thing

[@SAMAmUrl]

Lovely break. Blessing family

[@vishalmishra3046]

*The Magic of Trigonometry to simplify complex integrals* Look at the integral closely to simplify the denominator by setting x = cos 2T and noting that 1+cos2T and 1-cos2T are nothing but 2cos^2 T and 2sin^T respectively (you need perfect squares under square-root). √(1+x) = √2 cosT and √(1-x) = √2 sinT which simplifies denominator [ √(1+x + √(1-x) ] to 2 (cosT + sinT) / √2 = 2 cos (pi/4-T) dx = d cos2T = -2 sin2T dT. Range of integral goes from x=0=cos2T (so T=pi/4) to x = 1 = cos2T (so T=0). Remove negative and reverse the integral range. Therefore, integrate from T=0 to T=pi/4 ( 2 sin2T dT) / (2 cos(pi/4-T) then replace T with pi/4-T to get (sin(pi/2-2T)dT / cos T) = (cos2T/cosT) dT *a super simplification* *In a nutshell, the integral is simply cos 2T / cos T from 0 to pi/4* ( cos 2T = 2cos^T - 1) / cosT = 2cosT- secT which integrates to 2sinT - log(secT +TanT) = 2 sin(pi/4) - log [ sec (pi/4) + tan(pi/4) ] + [2x0 - log(1+0) = 0 - 0 = 0] = √2 - log(√2 + 1) = √2 + log(√2 - 1) *Super simple. Right ?*

[@yoav613]

There is avay to avoid this limits and get the answer quickly and this is by combining the 2 integrals to 1 integral between 0 to sqrt 2

[@johnsalkeld1088]

THERE IS A NICE SIMPLIFICATION WHEN YPUI RATIONALISE THE DENOMINATOR inside the Ln you get (root(2)-1) ^2 over (2 - 1) so the 2 can come out of the ln and cancel the 1/2 so root2 +ln(root2 - 1)

[@jkid1134]

Linear or abstract algebra are great ideas. Also multivariable calculus, analytical geometry, maybe some probability course would also be good.

[@hasanjakir360]

We can simplify a little more. Notice that ½ln((√2-1)/(√2+1))=ln(√2-1). So the final answer can be √2+ln(√2-1)

[@chrisb1047]

I’d love to watch some lessons on solving coupled linear differential equations in state space. Specifically related to mathematical modeling of say harmonic systems or heat transfer would be really interesting !

[@noam.1527]

Hi! As a french student I always find it weird using l’hopital’s rule 😅 We always use taylor expansion and asymptotic analysis !

[@HeavyMetalMouse]

Could do a little simplifying on the final result. (sq(2)-1)/(sq(2)+1), rationalizes to (sq(2)-1)^2/(2-1) or just (sq(2)-1)^2. Then the power of 2 comes out to the front of the log and cancels the 1/2 This leaves you with sq(2) + ln(sq(2) - 1) as the result, nicely getting rid of all the fractions. :)

[@indeedhid380]

we could've used sqrt(1+ax) ~ 1+(a/2)x for the limit.

[@mohan153doshi]

Dear Michael, I feel that no place is a good place to stop as far as your videos are concerned. I would just like them to go on and on. I know that is not possible, but I feel that way. That's because the math is presented here in a way unparalleled anywhere else on YT.

[@GoldenAgeMath]

A group theory marathon would go crazy

[@erics749]

Linear algebra or some sort of introductory abstract algebra marathon would be great!

[@natepolidoro4565]

Literally any class you teach on a live stream I will watch.

[@mathbyprofc8791]

Hey Michael. Can you make a separate video explaining why you draw radical symbols in 2 painful to watch steps? My OCD thanks you. Abstract Alg or Real Analysis, although I'm not sure about the demand for the latter. Keep mathing. You're my fav!

[@ethanbartiromo2888]

I will be taking abstract algebra this upcoming semester so PLEAAAASE do something on it!

[@Pedritox0953]

Wonderful explanation

[@a_llama]

3:39 that 'hi...' though 😅

[@violintegral]

I think the easiest approach is to rationalize the denominator by multiplying by its conjugate, then substitute x=u², 2udu=dx and then use a u=sinh(t) substitution for the first term and a u=sin(theta) substitution for the second term. That gave me ln((1+√(1-x))/(1+√(1+x)))+√(1+x)-√(1-x)+C as my final answer, which nicely condenses the logarithm. Only problem is that the back substitution for the sinh sub is a little more difficult, but not too bad. And I had √2-ln(1+√2) as my answer for the definite integral from 0 to 1. This method also allows you to avoid the tedious process of finding the limit as x approaches 0 of the logarithm. But to be honest, I cannot figure out how my logarithm and his are equivalent, but I know they are because I checked it with an online calculator. If anyone wants to show their equivalency, that would be appreciated.

[@faresberarma3349]

Can be done easily using trigo substitution x=cos(2y) then y=pi/4-y we get integral cos(2t)/cost from 0 to pi/4 the rest is very easy

[@aswinibanerjee6261]

Modular algebra and its relation with the elliptic curve Please 🙏🙏🙏

[@abhishekanand5112]

Substituting X=cos2a, will be easier approach.

[@emperornortoni2871]

At 7:50 I mumbled to myself "plus C." This dude is way smarter than me, my professors just drilled that into my head. I wish I had these videos before the integration bee at my college, I could have placed higher than third, haha.

[@procerpat9223]

how about a complex calculus marathon ?

[@calcul8er205]

Using the substitution x=sin2a gets to the result in a few lines

[@vbcool83]

Not so unfriendly, you have covered some much more unfriendly integrals!

[@methatis3013]

I would really want to see differentiation marathon. I dont know much about differentiating trig functions and more complex functions

[@cpiantes]

I was worried that when you multiplied the integrand by 1 you introduced a 0/0 singularity at x=1. I got burned by things like that many a time in my calculus class.

[@Smrda1312]

Abstract algebra sounds great!

[@tonyhaddad1394]

3:38 potential mathmatician

[@fasihullisan3066]

natural log of minus one is zero or undefined ?