This trick is new to me!

[@paul21353]

This is an integral I have never seen before either even though I have a uni degree in math and have taught math all my life. Such a neat trick! I love this video. Also, it begs the question whether this result can be generalized to a denominator of degree>2. My guess is that the degree has to be even. It needs a look into.

[@julienmallet8989]

yeah intuitively with an odd degree i'd say you'd lose the sort of symmetry making it equal to zero

[@justacutepotato2945]

The coefficients of the polynomial would also need to be symmetric, i think.

[@scollyer.tuition]

With the same substitution (x=1/t), then \int_0^\infty ln(x)/(ax^4+bx^2+c) dx = -\int_0^\infty (x^2 ln(x))/(a + bx^2 +cx^4) dx so the generalisations to higher powers don't look as cute as the one in the video, at least with the even-power choice of integrand that I've made. The odd powers don't look much better either.

[@joesweeney8200]

Pretty sure it'd work for any degree (just the trick of reducing it) as long as the polynomials coefficients are symmetric i.e. f(x)=a_n*x^n+...+a_1*x+a_0 with a_i=a_(n-i). Whether you'd be left with a solvable integral is another story.

[@leif1075]

HOW AND WHY did someone deduce this fact--there must be a reason--from experimenting with different u substitutions I guess--I would think no one would get this right away no matter how smart..

[@johnkane9080]

Excellent, but you change your variables more often than I change my duvet cover!

[@shruggzdastr8-facedclown]

I don't change my duvet covers, bc I don't have any duvet covers 😏

[@drhubblebubble7]

My colleague who always loved showing off his math skills showed me a problem very similar to this right after I learned integrals 5 years ago. I was so obsessed with solving it and I simply could not find a way, then after a month of trying and wasting my spring break he showed me and for the first time I thought he was definitely wrong, he did not use symmetry, and I was very sure his substitutions were wrong. I searched the answer on Symbolab and I got a very long complex result for the indefinite form and I moved on feeling unsatisfied. Now after 5 YEARS you have sparked my interest and I'm eating the shit out of your solution, I love what you did!

[@57thorns]

11:51 is an interesting cut. The sound goes "like that! so let's see". And there are some changes to the deleted parts of the blackboard.

[@MichaelPennMath]

My college is having its annual science festival this weekend! It is completely remote and has some nice events. Our keynote speaker is tonight: author of the Physics of Superheroes: · 7:30 pm: Keynote Speaker James Kakalios, The Uncanny Physics of Superhero Comic Books Here is a link: randolphscience.org/scifest/

[@leif1075]

Thanks for sharing. Can you please share why anyone would think of doing those substitutions at all especially the first one? I really would appreciate an answer.

[@josephgrossenbacher7642]

@@leif1075 , sorry , i'm not M. P. ; but i often tried "such substitions" in case there was some "log-term" in the integrand , since [ ln ( x ) ] ' = 1 / x

[@josephgrossenbacher7642]

thanks for sharing Michael ( i still like your clips though there's "some lack of time" on my side to watch more of them ... , a fact that makes YOU kind of speciel , so thanks for that one too : & quite "unusual" : some weeks ago , a student of mine "saw" the "a = c thing" by herself & she also noticed that i was impressed ... !!!

[@leif1075]

@@josephgrossenbacher7642 wjat do you mean? I don't see and I don't think that would lead anyone to the trick he uses in the video..i dont see why amyone would..its randkm..

[@mokkapatisiddharth5793]

For those in a hurry, it's sufficient if you watch the video till 6:29 because everything after is pretty standard calculation. But yeah, this is definitely a very neat trick to take out that not so easy ln(x) out of the integral.

[@jackhandma1011]

Basically another good place to stop 😂

[@mokkapatisiddharth5793]

@@jackhandma1011 Yeah 😂

[@Bayerwaldler]

Yay - so many baby-steps...🙄

[@leif1075]

I dont see why anyone would think to replace x with 1 over t..random.

[@Wksfr]

“And we are just going to do one more substitution...” 🤣 this guy is just flowing the maths

[@davidbrisbane7206]

I thought the substitutions would never end even although it was clear that we were going to get arctan as the anti-derivitive on the second last board. But I do like a good overkill 🤣😂🤣😂.

[@roarprawn]

Me too, I was concerned Michael would run out of letters :)

[@davidbrisbane7206]

@@roarprawn 🤣😂🤣😂👍👍👍

[@mtaur4113]

Yeah, arctan comes up often enough that the last two or three lines are worth memorizing or looking up.

[@mrflibble5717]

But, that does make a mistakes less likely and the final result simpler. Michaels approach is very thorough, not many are anywhere near as clear.

[@sergeipetrov5572]

Thanks a lot for the detailed explanation! I have never seen such a trick before too. Symmetry is power.

[@goodplacetostop2973]

HOMEWORK : We have a calculator with two buttons that displays an integer x. Pressing the first button replaces x by ⌊x/2⌋, and pressing the second button replaces x by 4x+1. Initially, the calculator displays 0. How many integers less than or equal to 2014 can be achieved through a sequence of arbitrary button presses? (It is permitted for the number displayed to exceed 2014 during the sequence.) SOURCE : Exercise 6 from Combinatorics of 2014 Harvard-MIT Mathematics Tournament

[@reshmikuntichandra4535]

This looks SO SIMILAR to the Collatz Conjecture

[@eduard9929]

@@reshmikuntichandra4535 Not quite because we can press the first button even if n is odd. For example we can press only the second button and we get an infinite serie, tho it isn't what the problem asks for.I'll give it a try later

[@reshmikuntichandra4535]

@@eduard9929 IK Just saying

[@eduard9929]

Well I found something amazing! I used a base convertor and test in base 4 what happens ! First we have 1 ,which is 1 in base 4. Then we go to 5 , which is 11 ; then to 21 which is 111 and so on, if we press the second button. For the first button, we get number 2. For example, when we are at 21 and press the first button , we go to 10, which is 22 in base 4. If we have a number made only of 1 there are 2 cases: If we press 4x+1 we add another 1 at the number If we press [x/2] we get a number only made of 2 and with one less digit I must study more cases, but it is very nice !

[@goodplacetostop2973]

SOLUTION *233* We consider the integers from this process written in binary. The first operation truncates the rightmost digit, while the second operation appends 01 to the right. We cannot have a number with a substring 11. For simplicity, call a string valid if it has no consecutive 1′s. Note that any number generated by this process is valid, as truncating the rightmost digit and appending 01 to the right of the digits clearly preserve validity. Since we can effectively append a zero by applying the second operation and then the first operation,we see that we can achieve all valid strings. Note that 2014 has eleven digits when written in binary, and any valid binary string with eleven digits is at most 10111111111 = 1535. Therefore, our problem reduces to finding the number of eleven-digit valid strings. Let F_n denote the number of valid strings of length n. For any valid string of length n, we can create a valid string of length n+1 by appending a 0, or we can create a valid string of length n+2 by appending 01. This process is clearly reversible, so our recursion is given by F_n = F_(n−1) + F_(n−2), with F_1 = 2, F_2 = 3. This yields a sequence of Fibonacci numbers starting from 2, and some computation shows that our answer is F_11 = 233

[@VibratorDefibrilator]

At first I was thinking for Laplace transform, but after I clicked on thumbnail... it was new to me either. Nice trick!

[@randomdyfbrn]

nice name coaie

[@georgemissailidis3160]

There was no need to divide by 9 in the fractions part. Just note that x^2 + 2x + 9 = (x+1)^2 + 8, then substitute x=3t to yield (3t+1)^2+8. Much quicker way to derive the expression for the denominator in the integrand.

[@PubicGore]

This video is so satisfying.

[@The1RandomFool]

That was a good trick. The method I was aware of to evaluate this type of logarithm problem is a keyhole contour with complex analysis, which is more complicated.

[@paulkohl9267]

Another cool substitution trick on the channel! Generalizations: n-th order polynomial in denominator and odd function of ln(x) in the numerator.

[@goodplacetostop2973]

12:38 Alright so I’m coming from the future. I just have enough time to say that is a good place to stop and that the Riemann hypothesis is-

[@reshmikuntichandra4535]

Can we have a face reveal anytime soon?

[@goodplacetostop2973]

@@reshmikuntichandra4535 I doubt I ever will do a face reveal on KZbin or any social media. Why? - Despite all my activity on the internet, I’m a very private person and I want to keep some degree of privacy, especially for a meme account. - I don't want people to have an opinion of my content and who I am based on what I look like. People would try to invalidate my account or harass me if I said I was a woman or a black or a trans person or an Asian or a disabled person or too old or an Israeli or too young, etc...

[@pardeepgarg2640]

Is time travel is costly Pls reply

[@Oleg-mv6cx]

GUY WHAT THE FUUUUUUUUUCK

[@goodplacetostop2973]

@@Oleg-mv6cx 🤷

[@MrRyanroberson1]

it's amazingly satisfying when you just substitute every factor out of the integral at the end like that, even though it's probably not the best content long-term due to its time-consuming nature

[@AhmedDZUSA]

I really enjoy watching this video. It shows step by step how to solve this kind of problem.

[@egillandersson1780]

Looking for the usual little mistake, which is not there this time, makes us more attentive. Clever ! I didn't know this initial trick. Thank you, Michael !

[@ksatyasirisha2013]

I don't get how x=t? 3:35

[@egillandersson1780]

​@@ksatyasirisha2013 As the bounds of integration don't change, x or t are just here dummy variables, as in undefined integrals. So you can use any name indistinctlty.

[@johnnath4137]

@@ksatyasirisha2013 The variable in a definite integral is a "dummy" variable - you may replace it by any letter of the alphabet, and this wouldn't change the integral value.

[@MichaelRothwell1]

Well, there was missing differential at the end of the board just before he wiped it clean...

[@TheoH54]

@@johnnath4137 even using the runic or Etruscan alphabet :-)

[@funkygawy]

2:34 through 3:17 sorry didn't hear you was trying to figure out what I missed with that negative infinity :)

[@noahtaul]

Very interesting! The trick can certainly be used in a more general form to evaluate that integral without the first u-sub: you try x->3/t instead of x->1/t, and the resulting denominators are the same again. The 3 inside the ln(3/t) gives you the same integral as you got with the u-sub.

[@leif1075]

Why would anyone think to replace x with 1 over t..I don't see why anyone would..surely you can a solve it some other way without that..

[@ZipplyZane]

@@leif1075 I's one of the more common substitutions, as it turns +inf into 0+ (or -inf into 0-) for the bounds of integration, making it finite instead of infinite.

[@justacutepotato2945]

@@ZipplyZane i mean, it also changes 0 into inf, so i'd argue it's not so much for making things finite but more so to keep the bounds from 0 to inf (or something like that) which can help in definite integration. (Like keeping the bounds similar/same, helped us to prove that said integral was 0.)

[@whatthehelliswrongwithyou]

Hey! This integral can be done pretty easily with another nice trick and complex analysis, just look at the same integral but with ln squared and make a cut from 0 to infinity, from there it's pretty straight forward

[@musicman9023]

5:28 "six t plus nine dt". Nice.

[@mahmoudqaddoumi28]

Symmetry is the important idea in nature...... Odd Symmetry.... Inversion.......thanks alot my respected teacher

[@pbj4184]

So basically the essence is noticing that we needed the same coefficient in both x^2 and C^2, the constant and that x=Ct does that while also fitting well with the log due to a basic log property. Now we have reduced the problem to the integral of 1/quadratic which is a solved problem. Cool!

[@mathewyaski]

I really liked how you presented the video. That's a fine trick. I wish this video was around when I was taking my Math Methods in Physics course... alas I had to learn the hard way. Nice work though. I'll be sure to share this!

[@michaelschmitt2427]

I have not seen this trick before, either. It is very nice and saves us from the ln(x). After that, just straight forward substitutions.

[@willyh.r.1216]

A good refresher for me, thx Michael.

[@arielganot1423]

At 3:38, how can you make the substitution that X=t if the entire way to get there is based on the fact that X=1/t ?

[@yuseifudo6075]

It's just a variable

[@Hacker118]

7:37 Alright Ima take the 1/9 out of the integral 8:46 Ok forget about it let's take that 1/9 back in Some question: I think we can also complete the square even the coefficient of x² term is not 1, isn't it?

[@axelperezmachado3500]

yep, just substitue u^2=9x^2

[@Hacker118]

@@axelperezmachado3500 great, thanks!

[@yatharthgupta6468]

Awesome method

[@tgx3529]

I think,mabye Tric for zero integral Is from residue theory. (ax^2+bx+a)'=2ax+b, when we take x=-b±sqrt(b^2-4a^2)/2a, we have ±sqrt(), then there Are residums (Log(x1))^2/sqrt; (Log(x2))^2/-sqrt. We are interesting for Imaginar part 2pí*i*( res1+res2). I have seen this 30 years ago

[@humester]

I could not understand why he put minus infinity. Thank goodness he corrected it!

[@christopherrice4360]

I love all your Math videos Michael Penn, keep up all the great content on your channel👏👏👏👏!!!

[@mrflibble5717]

Beautiful!

[@violintegral]

Also, the integral from 0 to infinity of f(x)dx is equal to the integral from 0 to infinity of f(1/x)/x²dx, which you basically just showed here. This means that the integral from 0 to infinity of f(x)dx is also equal to the integral from 0 to infinity of 1/2[f(x)+f(1/x)/x²]dx.

[@mathematics_and_energetics]

Extremely Nice! Thank you! 😊

[@richardheiville937]

You could factor 8 sooner and thus avoid two changes of variable.

[@shahinjahanlu2199]

I was afraid of integrals Now because of you ,I love them.thx

[@juanixzx]

The trick was super cool, but that change of variables was really odd, from integral with denominator (t+1/3)²+(sqrt(8)/3)² just doing a trigonometric substitution t+1/3=sqrt(8)*tan(theta)/3, calculating the undefined integral and then putting the limits I think would work better.

[@juanixzx]

I guess you do it for get integral of 1/(x²+1), which is a very known, but I don't know, doing every substitution can carry an error easily.

[@flikkie72]

Why is it that you can substitute x=t at 3:39? Since x was substituted for 1/t before, shouldn't you sub for 1/x now?

[@pandumulyamuhammadsyah]

I'v a litle confused, why in part of subtitution back t --> x use t = x instead of t =1/x?

[@lilellia]

When we do definite integrals, the actual variable doesn't matter ("dummy variables"). The two Xs aren't the same in this case, and the second substitution just replaces all the Ts with Xs.

[@pandumulyamuhammadsyah]

@@lilellia well, i get it. So as long as the integral is definite one the dummy variable's behavior could be applied?

[@lilellia]

@@pandumulyamuhammadsyah Yep. If the bounds are the same, then the integral of f(x) dx = integral of f(t) dt = integral of f(u) du = integral of f(θ) dθ = ...

[@jonathanweihing9650]

Just mind blowing trick

[@keksauraisks]

That's a really nice trick

[@tomgreg2008]

Fun problem! Thanks!

[@Bjowolf2]

Cool - and very impressive 😉 Can you say something about where the boundaries lie, as to which types of integrals are solvable by our current methods? For instance I saw recently - much to my surprise - that there actually doesn't exist any exact formula for the circumference of an ellipse ( and probably not for super-ellipses either? ), even though you can calculate its area - only more or less accurate approximations that work fairly well for the not too extreme cases. Are there any reasons for this? What is stopping "us" ? 😅

[@BikeArea]

I think there is a video out there covering that exact topic. Have a search! 😊

[@elrichardo1337]

oh yes, the classic coefficient-reversing trick

[@subashkc7674]

Wow amazing trick❤. Thank you.

[@mathisnotforthefaintofheart]

Typically when you have an improper integral with two "improperties" (being zero and infinity) one has to split the integral. That gets conveniently avoided here, thanks to the 1/t sub. It works out nice indeed. There is a problem like this in Stewart Calculus (with a=c) so the problem is not unique...but cool nonetheless

[@faismasterx]

Great video. Nasty integral.

[@alcodark]

u factored a 9, only to multiply it again..

[@Djentstructer]

Hey it happens. Not a concern.

[@erwinfinder9182]

you're my new teacher and figurative father figure.

[@sihasnesh2016]

Do we not have to prove that the integral with coefficients a=c, is equal to zero before directly applying it to the sum

[@procerpat9223]

nicely done 👍🏻

[@tahasami3409]

Thank you

[@carlosdiezdesollano5918]

Quite interesting. Thanks!

[@MrGoofy42]

Good thing I did not pause the video at 2:37 for 15 minutes to recap the substitution rule, trying to find out where that minus came from.. :D

[@antonior9991]

The trick is complex analysis

[@antonior9991]

Really cool anyway

[@janami-dharmam]

It took me some time to appreciate the beauty!!

[@kantanlabs3859]

Was just watching a video on the Wow signal, I know where it came from now !

[@vedants.vispute77]

I have no money for time travel

[@pardeepgarg2640]

Me also 😭😭

[@JCCyC]

When I get to the point that the integrand is 1/(x²+a²) I just jump to (1/a)arctan(x/a) and call it a day. Ditto for 1/(x²-a²) and others.

[@kamcreative6529]

How i don't understand,first you change it x=1/t and change t=x...how..it is consistent?

[@peaudest]

Wow this trick is powerful.

[@GustavoMerchan79]

11:20 something incomplete there in that 8 simplification

[@davidgould9431]

I thought so too, but it turns out that, because du = -1/y² dy, the y² distributes over the denominator, turning (1/y² + 8) into (1 + 8y²). Got me confused for a while.

[@ahmednesartahsinchoudhury2628]

In your videos, you tend to show the steps required to prove/show something but you don't explain how that thought came into your head, or how to approach a similar problem. I'd like to know how exactly your head processes the problems and how you start attacking it. For example, in the first step in this this video, you used the substitution x = 1/t. Did that come from trial and error? What else did you try that failed and lead you to this substitution?

[@nguyentrantinh1483]

Great thank you so much

[@damianbla4469]

Please Sir do a video about checking if a given function is continuous at given point.

[@YitzharVered]

He has a series on real analysis. He does that there.

[@jkid1134]

Cool video!

[@TaladrisKpop]

Shouldn't the consequence of the trick when a=c be that the integral is 0 or divergent? (It is convergent but it should be proved separately). In the same way that the improper integral of an odd function from -infinity to infinity is not always 0.

[@jmjoebar]

If a=c=0 the integral is divergent I think

[@TaladrisKpop]

@@jmjoebar Right. Because in that case, lnx/bx > 1/bx>0 for x>3 so the integral diverges on [3,infinity), hence on (0,infinity).

[@JohnDoe-oo9ll]

Helps to have wiki open as you listen

[@nahuelcaruso]

Nice trick!!! However, is it possible (in the integral trick) one the improper integrals diverges whereas the other converges?? I did some calculus and I bet it is not possible, what do you think??

[@manucitomx]

Wow. Just that wow.

[@クッキー-m8g]

It is very interesting!

[@mihaipuiu6231]

M.Penn...this integral is very interesting. Nice TRICK...!!!

[@benjamingross3384]

To three or not to three, that is the question.

[@U9191-e6s]

how at the beginning do you assume x=1/t and then you assumed t=x !?

[@koenth2359]

6:08 ... of chocolates, you never know what you gonna get

[@leif1075]

@MICHAEL PENN QUESTION HOW AND WHY did someone deduce this fact--there must be a reason--from experimenting with different u substitutions I guess--I would think no one would get this right away no matter how smart..

[@criskity]

How many substitutions were made here? I don't think I've ever seen that many!

[@kleinbogen]

To me the key to this problem is in the 1st 6 minutes of this video. The last 6 minutes should be easy for any competent Cal 2 student.

[@francescovecchione3106]

Now, just a question. If I find something like this in an official test, do I have to demonstrate everything or can I apply directly by writing the general integral with "if a = c, then integral = 0"?

[@andrewkarsten5268]

Depends on the test, but all the tests I’ve taken want you to show your work. Only well known results can be taken without your own proof of them

[@mathisnotforthefaintofheart]

Well, no. Technically you should show that the area under the curve (from 0 to 1) and above the curve (from 1 to infinity) are both finite and THEN since they are equal, it cancels. Not that this is difficult to show BUT if both areas are infinite, one cannot just "cancel" and call declare the integral zero. But no worries here. Comparison theorem lnx against sqrt(x) will do the job

[@shivansh668]

Tooo much substitution 😕

[@viktor-kolyadenko]

int{1/(t^2+a^2)} = ...*atan(t/a) + C

[@srikanthtupurani6316]

Nice.

[@ahmednasr6171]

3:37 it dose not make sense to me we already assume that x=1/t how we make it x=t

[@dicksonchang6647]

At this point even t is not x But the intergal are the same Because they have the same form and same interval

[@yueno0727]

It's just interchanging the variable of integral, and completely irrelevant to the prior variable transformation.

[@pjmmccann]

It's just a dummy variable, so he's really just *renaming* the variable of integration rather than substituting.

[@TheVillan1980]

@@pjmmccann For anyone still struggling with this, perhaps substitute t = u in the second substitution, then t = v and so on. You get the same form, just with a different variable, so x is just as valid as any other identifier even though we did already use it.

[@tylerteh952]

Substituting x = 1/t and then substituting t = x at the end: doesn't that mean that the first x and the final x isn't the same x? This makes the whole tool wrong and hence the question wrong

[@2false637]

This was just a trivial substitution, purely notational. He just replaced it with x because it’s standard to do so. He could just as well have replaced it with “u” or “g” or “🎈”

[@Helalll294]

@@2false637 more details plz

[@JorgeGomez-li9td]

Very nice

[@serbanudrea9429]

While I definitely learned something new from this interesting video I would say that there are some issues with it. First nothing is said about the case in which the polynomial has at least a positive real root. I don't think that the trick ca be applied just like that in such a case. Second, we deal here with an improper integral, thus a trick while formally correct may not lead to the right result. How about proving that this improper integral is expected to behave, if I can say so. Finally, the whole calculation could have been a little shorter. For instance, no need to divide by 9 when 9x^2+6x+1 is simply (3x+1)^2, thus one immediately gets 9x^2+6x+9=(3x+1)^2+8. At this point then it is quickest to divide by 8 and directly make the substitution u=(3x+1)/√8.

[@hadireg]

👍👍awesome!!

[@Craznar]

Substitute teacher.

[@wesleydeng71]

Why can't we just go directly from 6:35 to 9:11? Would save 2.5 minutes.

[@gigino05]

Wolfram Mathematica gives a different answer. The procedure looks right, but there might be a computation glitch somewhere...

[@gigino05]

Sorry, actually it DOES give the same answer.

[@theguythatmakesyoumad3834]

*In proving the trick, you resub x = t, but wasnt x = 1/t?*

[@Eidolon2003]

The variable name that you use is irrelevant really. Call it t, x, u, v whatever, the integral will still evaluate to the same answer. He only changed it back to x for convenience.

[@Helalll294]

@@Eidolon2003 In details plz!

[@Eidolon2003]

@@Helalll294 There are some other people talking about the same thing in here, but here goes. You have a function, f The integral of f(x)dx = f(t)dt = f(u)du = ... etc. As long as you don't change the bounds or anything, what you call the variable has no effect on the answer that you get. When making a substitution it's a good idea to change variable names to avoid confusion in the conversion process obviously. Say you're doing the integral of f(x)dx from a to b. You make the substitution x=2u, dx=2du, and you have to change the bounds You end up with twice the integral of f(2u)du from a/2 to b/2 Then, like I said before, the letter you use doesn't matter, so you can just change the u's back to x's. Pick a function and some numbers to check with a calculator that the integral of f(x)dx from a to b is equal to twice the integral of f(2x)dx from a/2 to b/2. For example the integral of x dx from 2 to 4 = 6 and twice the integral of 2x dx from 1 to 2 = 6 Same values after the substitution, the variable makes no difference. Hopefully that makes sense :)

[@theguythatmakesyoumad3834]

@@Eidolon2003 Ah! Thank you :) Forgot we were talking definites :)

[@mcwulf25]

Always seems a slog to work out those 1/quadratic integrations. There is a formula for it. I mean, first principles are good but so many videos do it while none prove the arctan bit.

[@Helalll294]

How could you assume that x=1/t at the beginning and then finaly assumed that t=x !! Can any one explain this to me plz

[@drchaffee]

Imagine the original integral had been written with a y instead of an x. And then the substitution would be y = 1/t. Then, when the integral is written in terms of t he could have substituted t = x. What would the final result be? It would be identical to what he shows. It's a progression of substitutions, and there's no built-in memory of what the prior variables were called or their relations. In the example problem he works, he could have started with x, made a substitution with t, then a wholly different substitution with x, the again with t, and so on. They're just dummy variables. Let me know if that helps.

[@nellvincervantes6233]

"Dummy variables" are weird for me yet still works.

[@kawrno5396]

You transformed x=t^-1, then you inverse transformed t=x? How's that logical?

[@Catilu]

I'm a time traveler!!

[@Mystic_hermit]

Prove it.

[@pardeepgarg2640]

How even you do this???? Susssssss

[@Oleg-mv6cx]

How

[@emanuelvaldes7166]

Muy enredada y poco clara la resolución de ésta integral !!! .

[@bopaliyaharshal2399]

11:11 mistake limit wrong 1 to 0 you did 0 to 1 😅

[@davidgould9431]

11:21 "multiplied the minus one through to switch the bounds of integration"