so you want a HARD integral from the Berkeley Math Tournament

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Күн бұрын

You can now register for BMT 2022 here: www.ocf.berkeley.edu/~bmt/bmt...
The first-place team wins $1000 bprp scholarship and I will be there!
We will integrate x/tan(x) from 0 to pi/2. We will use Feynman's technique (aka Feynman's trick of integration or differentiation under the integral sign) to compute this integral. This problem is from the Berkeley Math Tournament in 2020. Here's the link to the exam and the solution: www.ocf.berkeley.edu/~bmt/arc...
If you are also interested in sponsoring BMT and making an impact, please contact communications@bmt.berkeley.edu
Thank you!
0:00 we haven't done a hard integral for a while
2:33 the steps of Feynman's trick of integral
5:30 differentiate I(a) first, then do the integral
18:59 integrate I'(a) with respect to a to get I(a)
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Пікірлер: 713

@blackpenredpen Жыл бұрын

You can now register for BMT 2022 here: www.ocf.berkeley.edu/~bmt/bmt-2022/ The first-place team wins a $1000 bprp scholarship and I will be there!

@orangesite7625 Жыл бұрын

Please solve -{integral (0-->π/2) {log[sinx]}dx}

@vijaichikatimalla3211 Жыл бұрын

@@orangesite7625 cos c + tan 90

@anko6999 Жыл бұрын

what language is this?

@solo_driven Жыл бұрын

Thank you very much for such a valuable content. I have a question why did you so casually change the conditions of a from greater than 0 to greater or equal to 0. Shouldn't you have returned to the integration part to see if it doesn't change anything? Indeed it doesn't but still ...

@holyshit922 Жыл бұрын

@@orangesite7625 answer is in this video but it can be done easier

@chinesecabbagefarmer Жыл бұрын

And then multiply everything by BMT2020

@Jha-s-kitchen Жыл бұрын

Equate to BMT2021 and solve for x 🤣

@harith_khaleel Жыл бұрын

😂😂😂😂

@gauravsonawane Жыл бұрын

🤣🤣🤣

@admink8662 Жыл бұрын

😂

@troym856 Жыл бұрын

You clown 🤣🤣🤣🤣

@sabrinagiang Жыл бұрын

Hello! You probably don’t remember me but I took one of your classes back in 2016. I was struggling with math and you advised me to refer to your KZfaq channel. It’s crazy how much this channel has grown. Congrats!! 🎉

@blackpenredpen Жыл бұрын

Thank you!! I do recognize this name. And wow, it’s been over 6 years!! Hope you are doing well 😃

@sphrcl. Жыл бұрын

@@blackpenredpen where do you teach?

@wartex3561 Жыл бұрын

@@sphrcl. harvard

@satyagaming996 Жыл бұрын

@ChannelTerminatedbyYouTube Жыл бұрын

@@sphrcl. but how is your math now 💀💀

@limpinggabriel6658 Жыл бұрын

I actually solved this one on my own! I would never have been able to do that without guys like you teaching.

@satyagaming996 Жыл бұрын

@user-ox2qt2cm4c Жыл бұрын

by the same method?

@limpinggabriel6658 Жыл бұрын

@@user-ox2qt2cm4c yeah, same method. I knew where to apply it though!

@bruhe8895 8 ай бұрын

Same, Ibdidnt use at the exact same place but i knew i would need this technique

@aidankwok1475 Жыл бұрын

I have another way of doing it I = Integ (x cos x / sin x) dx = Integ x/sin x d (sin x) = Integ x d ln (sin x) Applying integration by parts, I = [pi/2 ln sin (pi/2) - 0 ln sin (0)] - Integ (ln sin x) dx The integral from 0 to pi/2 of ln sin x dx is a classic integral (can be calculated by king’s property)

@renesperb Жыл бұрын

I also used this integration by parts.For the integral from 0 to pi/2 of ln sin x dx I write sinx as 1/(2 i )*exp[i x]*(1-exp[-2 i x]) If you take the ln you get - ln 2- π i/2+i x+ ln[1-exp[-2 i x]. Next you use that ln[1- z] = sum (z^k/k ),k =1 to inf..Here z=exp[-2 i x]. If you now integrate term by term it is easy to see that you get 0 . The rest of the integration for the other terms is then trivial and you get the answer - π/2 ln2 .

@renesperb Жыл бұрын

I forgot a detail : because of the symmetry of sin x with respect to π/2 you can take 1/2 of the integral from 0 to π. This makes the last integral 0.

@LouisEmery Жыл бұрын

That's the way I was going to go, but didn't see the ln.

@honeythapa9489 Жыл бұрын

I did the same way

@ReaperUnreal Жыл бұрын

Yeah, that was my first reaction as well. Glad to see someone else had the same thought process.

@gustavozola7167 Жыл бұрын

Excelent! You should post more challenging ones like this

@SpringySpring04 8 ай бұрын

Just finished my first semester in Calculus 1, and now I finally understand a lot of the concepts presented in this video, as if it summarized my entire semester into one big challenge (the only thing I haven't really learned yet is the partial derivatives, but they seem to make sense from the video) This was an AMAZING experience!!

@silversleezy4953 Жыл бұрын

If you treat the integral as xcotx Then apply the power series of expansion for cotx we have the integral as: x ( 1/x−x/3−x^3/45−2x^5/945 - ⋯) After multiplying the x into the expansion, we can then easily integrate and evaluate the function from 0 to pi/2 and see it converges to pi/2 ln2 (By factoring Pi/2 out and then consider power series of ln functions)

@vivekraghuram2459 Жыл бұрын

This was my first thought! But I enjoyed the solution that he presented quite a lot!

@RathanAadhi Жыл бұрын

this is a much better solution and answer can be obtained easily even by using byparts

@tangsolaris9533 Жыл бұрын

Wow, Maclaurin series are good

@irshad334 Жыл бұрын

There is one serious downside that no one here is quite pointing out, but assuming that you are not provided this in the competition, then you are asking someone to be able to either recite the Maclaurin series for cot x or to be able to reproduce it on their own. Which is an incredibly impractical thing to do. You appear to need to know Bernoulli numbers etc. to reproduce this expansion, or just memorise the not so friendly fractions here. You can't empirically produce cot x as easily as you would for say sin x - unless of course you know of a more succinct and beautiful way of doing it then what I seen on in the Internet.

@seegeeaye Жыл бұрын

A different way: integral = int of x cotx dx, then using By Parts, to get - int of ln sin x dx, by using the property of int of f(x) = int of f（pi/2 - x）, we have 2I = int of (ln sinx + ln cosx) dx. then follow some basic algebra and trig, we get I =( pi/2)(ln2) detail: kzfaq.info/get/bejne/jNulZ9xpusC8emg.html

@ssj_brownie6447 Жыл бұрын

When he said “you give it a go,” I did this exact technique!

@TomJones-tx7pb Жыл бұрын

Exactly so.

@ankursingh.1.2m Жыл бұрын

Exactly I solved the same

@Sparky1_1 Жыл бұрын

I was looking for this comment Thanks

@AriosJentu Жыл бұрын

I've done this integral with a little bit different technique. First of all I've done IBP in first place to get "- int from 0 to pi/2 of ln(sinx) dx". Of course this integral also not so easy, and I've done separation of integration ranges from 0 to pi/4 and from pi/4 to pi/2. On second integral it was a u-sub like "x = pi/2 - u", to get from sine function - cosine. There is integral of sum of logarithms of sine and cosine functions from 0 to pi/4. Using log identities, we can get "int from 0 to pi/4 of ln(sin(2x)/2) dx", and with a u-sub "u = 2x". Assuming algebraic-integral equation "J = int from 0 to pi/2 ln(sin(x)) dx" and "J = -pi/2 ln2 - 1/2 J", we can get the same result as in this video. I've tried this first :D

@yoto60 Жыл бұрын

This was also my method, but since int ln(sin(x)) dx = int ln(cos(x)) dx (for this region, using u=pi/2-x), I let 2A = ln(2*sin(x)*cos(x)) - ln(2) dx :)

@ayoubmff7834 Жыл бұрын

@@yoto60 I can't understand🤔

@shiviarora4173 Жыл бұрын

how did you apply IBP on this x/tanx dx mate? Did you convert it to ∫x tan-1x dx

@ayoubmff7834 Жыл бұрын

@@wondersoul9170 ooh thanks broo😁🙏❤️

@AriosJentu Жыл бұрын

@@shiviarora4173 of course not, its just x * cotx, x to be differentiate, cotx to be integrate

@chennebicken372 Жыл бұрын

I just love it, when pi and e show up together. Well, actually it makes sense… The pi comes from the trig-function and the log_e(2) from the integration.

@kaanetsu1623 Жыл бұрын

Another beautiful use of Feynman's technique!!

@garyhuntress6871 Жыл бұрын

I really enjoyed that. Best part of my day so far :D

@everytime865 Жыл бұрын

This type of integration is also known as integration by reduction formula here in India. Nice video btw!

@biswakalyanrath966 Жыл бұрын

I think reduction formula is something like I n = a I (n-1)+ b I (n-2) which gives a recursive relation among the integrals but this question has nothing to do with it

@user-uh9bo2im1h Жыл бұрын

No it’s Leibniz rule but this one east with king property

@vijaichikatimalla3211 Жыл бұрын

wow wow , u explained it in a very simple way thank you ! please do more videos

@Jack_Callcott_AU Жыл бұрын

This was a very, very satisfying demonstration BPRP. I didn't know about the Feynman technique until recently. Thank you!😸

@satyagaming996 Жыл бұрын

@Gabi-we1ff Жыл бұрын

22:20 we can see the passion and appreciation this gentleman has for maths with just a look, he found his happiness and that's bittersweet af, nice video as always fella.

@cesarluishernandezpertuz8794 Жыл бұрын

Gracias por existir este canal..... Es de lo mejor que he encontrado. ..

@rishabsaini8347 Жыл бұрын

Why so complicated? Here are the steps I took:- 1) Do integration by parts taking x as the first function and 1/tan x as the second function as 1/tan x is easy to integrate over 0 to pi/2. (~20 seconds) 2) Find integral of 1/tan x over 0 to pi/2 by doing u-substitution, u = sin x , you'll get integral of 1/tan x as log(sin x) in like 2 sub-steps. (~1 minute) 3) When we put value of integral 1/tan x in our Integration by parts equation of step 1, we get [x log(sin x) - integral (log(sin x) dx ] 0 to pi/2 Second term looks wild but It's easy to solve using properties of definite integrals (It was in my 12th grade NCERT book, I even remember the result). Int 0 to a F(A-x) = F(x) (~10 minutes) We get -pi/2 log(2) for the second term. Bam! We get pi/2 log(2) as the answer

@HappyGardenOfLife Жыл бұрын

pi/2 log(2) is about half the amount BPRP got.

@kozokosa9289 Жыл бұрын

@@HappyGardenOfLife I think they meant (pi/2)ln2

@seroujghazarian6343 Жыл бұрын

Good luck integrating ln(sin(x))

@sam-gooner Жыл бұрын

@@seroujghazarian6343 🤣

@dsfdsgsd644 Жыл бұрын

@@seroujghazarian6343 :tf:

@DrLiangMath Жыл бұрын

Excellent problem and wonderful explanation!

@musicngamesyo6628 Жыл бұрын

Watching this gives me so much relief🤩

@5Stars49 Жыл бұрын

Nicely done 👍💯

@davidkemball-cook559 Жыл бұрын

Really interesting. Well done!

@Happy_Abe Жыл бұрын

Technically we can’t let a=0 at the end because then the argument from earlier doesn’t work so maybe we can take a limit as a approaches 0 to get the value for c

@paokaraforlife Жыл бұрын

we can since if we take a to be greater than or equal to 0 from the start(as we should) when we get to the point in the limit, we don't calculate it we just say ''well if a=0 then we go back to the integral and see that I(0)=0 so for a>0.......''

@Happy_Abe Жыл бұрын

@@paokaraforlife problem is if we just say a is greater than OR equal to 0 then when evaluating inverse tan of au as u approaches infinity we can’t do what we do in the case where a=0

@paokaraforlife Жыл бұрын

@@Happy_Abe yeah and we don't We take the value and put it in the integral definition

@Happy_Abe Жыл бұрын

@@paokaraforlife sorry I’m confused From what I understand the integral can’t be evaluated at a=0, that is not within the domain of the I(x) function as previously mentioned. We can’t just evaluate functions at points outside the permitted domain

@paokaraforlife Жыл бұрын

@@Happy_Abe ok I'll take it from the beginning We define the function using he integral for a greater than or equal to 0 For a=0 we plug the value into the integral and we get the integral of 0 which is 0 For a>0 we do the same work as in the video and get a value through the limit

@metrix7513 Жыл бұрын

I'am in high school I understand only the derivations and integrations but this is just.. wow. Thanks for such great content and the reassurance that I want to do math

@ashirdagoat 10 ай бұрын

Lol me watching this after not even being one semester through calc in high school and wanted to do a premed biochem major anyways: 😂

@yoelit3931 7 ай бұрын

Differentiations*

@metrix7513 7 ай бұрын

@@yoelit3931 you are right mb, funny thing is I still don't understand, isn't feynman technique?

@armanavagyan1876 Жыл бұрын

You explain so well that i want all day watch you)

@brandd7395 Жыл бұрын

This was very cool to watch!

@bakeyourownshit4137 Жыл бұрын

thought of applying beta fn. but who knew. hands down! i was literally smiling at the end 🙌

@rage_alpha Жыл бұрын

Hi, I have another way of doing it. we can change the 1/Tan(x) part to cot (x) and then Integrate X.Cot(x) using integration by parts. It would've left us with integral of ln|Sin(x)| from 0 to pi/2 which can be easily integrated using king's Rule further. :)

@bjornfeuerbacher5514 Жыл бұрын

I would have used the substitution u = tan(x) first, leading to \int_0^{\infty} arctan(u)/(u (1+u²)) du. Then in that integral, replace the arctan(u) with arctan(a*u), so you have an integral depending on a. Differentiating with respect to a then gives the same integral as in the video at about 12:10. I think that's more intuitive than coming up with the trick x = arctan(tan(x)) first, and additionally, coming up with the idea that one should insert a factor of sec²(u).

@blitzer2062 Жыл бұрын

You don't really need to multiply top and bottom by f sec²(u). Just substitute u=tan x straight off.

@EmpyreanLightASMR 11 ай бұрын

I was just telling someone yesterday about Feynman "re-discovering" differentiating under the integral, but I'd never seen it in action. This was awesome!

@robertkb64 9 ай бұрын

I came to add that this is just the “Feynman Rule” but you beat me to it :) When you’re stuck just try integrating under the integral. Worst case you get a different problem you still can’t solve :p

@EmpyreanLightASMR 9 ай бұрын

@@robertkb64 i need to rewatch this or another video explaining this concept again. it's been awhile haha

@aeroeng22 6 ай бұрын

it appears to me that Feynmann found an undiscovered or rarely used way to use the Leibniz rule for differentiating under the integral. More generally, the Leibniz rule incorporates differentiation of the two limits (if they aren't constants).

@utuberaj60 Жыл бұрын

Very intersting solution -using a nice trig manipulation and the usual Differentiation Under Integral Sign (aka the "Feynman Trick"). But what I liked best here was the way you decomposed the irreducible quadratic factors (1+a^2u^)*(1+u^2) into linear fractions- by cleverly substituting the (a*u) term as a linear term -enabling to use the "cover-up" to get the constants "A" and "B" very easily. I watched your detailed videos on "Partial Fractions"- and the "cover up" short-cut method- never heard of it during my student days (I am a 60+ engineer). Enjoy your videos and it makes me feel young again. Thanks a ton BPRP, and also your partner Dr Peyam's videos-espeacially the integration ones.

@12wholepizzas13 Жыл бұрын

20:30 by putting a=0 here and having a>=0 won't we have an issue while inputting the limits in tan(au) when u approaches infinity. If a can be equal to zero won't we have a possible 0×infinity situation

@BrollyyLSSJ Жыл бұрын

You're right. Instead, we should consider that lim(a->0+) I(a) = lim(a->0+) (pi/2 * ln(a+1) + C) = C. We can see from definition of I that it is continuous and I(0) = 0, so C = 0 as well.

@asparkdeity8717 Жыл бұрын

@@BrollyyLSSJ thanks so much, was wondering about this caveat too

@UglyRooks Жыл бұрын

@@asparkdeity8717 me too...

@alexandervanhaastrecht7957 Жыл бұрын

@Brollyy there you assume that a*u (with a ->0 and u->infinity) equals infinity, else I(a) isn’t continuous for a = 0. The problem with this limit is that a*u can be any (positive) real number or infinity. The definition of I(a) you used assumed that a was not equal to zero, so if you change that, you have to calculate everything again

@carstenmeyer7786 Жыл бұрын

@@alexandervanhaastrecht7957 You are right - if you really want to be rigorous, you need to show continuity from above at *a = 0* using the integral definition of *I(a).* Luckily, the integrand (call it *f_a(x)* ) can be continuously extended to *x = 0* via *f_a(0) = a.* The extended integrand converges uniformly to zero on *[0; 𝞹/2]* as *a -> 0^+,* proving continuity from above of *I(a)* at *a = 0*

@babo105 Жыл бұрын

Very nice solution!

@ChoCho-yi2sw Жыл бұрын

Thank you sir.

@neypaz8054 Жыл бұрын

Just wanted to say that I have subscribed to your channel. I've seen your videos since 2020, and thanks to you I learned a lot of cool math things I didn't know beforehand. =)

@blackpenredpen Жыл бұрын

Thank you!

@Assterix Жыл бұрын

Idk if you read the comments on your old videos but you're basically my math professor for calculus :)

@zinswastaken Жыл бұрын

@@English_shahriar1 stop advertising your yt channel in comments

@ddognine Жыл бұрын

An alternative method is to integrate by parts with u = x and dv = cotxdx which will result in the integral of ln|sinx|dx over the same interval which can be easily evaluated using power series and setting y = 1 - sinx. The challenge is showing that the resultant integral converges to pi/2ln2, but honestly, I couldn't really care as long as you're able to show a solution.

@saharhaimyaccov4977 Жыл бұрын

More like this please

@stevemonkey6666 Жыл бұрын

It's always good to see a nice calculus problem......

@SuperYoonHo Жыл бұрын

Thank you sir

@mr.grantelkade4073 Жыл бұрын

this was enlighting!

@AhirZamanSairi Жыл бұрын

This reminded me of another hard one I can't forget, integral of (sqrt 1 + x^2)/x (non-trig solution takes up whole board) U already did that, but only trigonometrically, I'd like to see u do it with the "u world" as well, it's a bit harder that way, and I'd love it.

@tgx3529 Жыл бұрын

You can use integral x* cotg x, then by parts, you have then - integral log( sinx), It's also on internet And some tric to use.

@UmairAli-dw2ny Жыл бұрын

Well played

@ryanxing2526 Жыл бұрын

excellent work

@Nick-wh4jt Жыл бұрын

Kings property of integration and Integration by parts result: x*ln(sinx)|(0->π/2)-(0->π/2)definite integral ln(cosx)dx Where x*ln(sinx)|(0->π/2)=0 And -(0->π/2)definite integral ln(cosx)dx = -1/4*dβ/dn(1/2,1/2) = -1/4*β(1/2,1/2)*[ψ(1/2)-ψ(1)] = -1/4* Γ(1/2)^2/Γ(1)*[-2ln2-γ-(-γ)] = -1/4*π*[-2ln2] = πln2/2

@yoyoezzijr Жыл бұрын

Beautiful integral

@CDChester Жыл бұрын

Added it to my Math YT video collection!

@dankmortal4911 5 ай бұрын

Bro we can directly apply by parts and we would end up with the common integral int(0toπ/2)ln(sinx) which is directly -π/2ln2

@yahirdominguez8222 Жыл бұрын

It is questions like these that just make me very happy

@siddhantdas6401 Жыл бұрын

Fantastic!

@alvaroarizacaro3451 Жыл бұрын

Sin palabras... ¡genial!

@itz_adi.g 7 ай бұрын

Its basically an easy question use by parts Derivative of x and integral of cotx and then int(ln(sinx)) from 0 to π/2 will be -(π/2)ln2 using king's property and some basic evaluation of integrals and there is -ve sign in by part term so it will be (π/2)ln2 and initial part xlog(sinx) from 0 to π/2 would be zero so final answer would be (π/2)ln2..

@abhishekchoudhary4689 Жыл бұрын

Use integration by parts U will get[ xln(sinx)]0 to π/2 - inte(lnsinx) 0 to π/2 now those who know limits will know xlnx x tends to zero is zero So the final result will be -I I = inte ln sinx from zero to π/2 which is equal to -π/2ln2

@leonardolazzareschi9347 Жыл бұрын

How do you integrate ln(sinx)?

@rad858 Жыл бұрын

@@leonardolazzareschi9347 split into 0 to π/4 and π/4 to π/2 integrals, then sub u=π/2-x in the second one, giving a cos(u). Use sinxcosx = (1/2)sin2x, and you'll find you have the same integral on both sides and you can get the result. Much quicker than the I(a) method

@aneeshgupta2002 Жыл бұрын

there was a pretty easy solution . we could use the product rule for integration by treating x as first function and cotx as second it would come out to be a-b where a = xlnsin(x ) evaluated from 0 to pi/2 ( it comes out to be zero) b= integral of lnsin(x) from 0 to pi/2 which is -pi/2*ln(2) so answer becomes -(- pi/2*ln(2))

@BuggyDClown-yc8ws 5 ай бұрын

I actually learned about the value of integral of lnsinx in class beforehand so it was just a matter of applying ILATE rule You first prove that integral of ln(sinx)=ln(sin2x)=ln(cosx).The value will come out to be π/2ln2. Then just convert the given integral into xcotx and apply ILATE rule

@owenmath1146 Жыл бұрын

I think I should add a few comments on the last parts, where you cancel a-1 directly and sub a=1 to the solution. There should be a limit instead of just sub a in directly, because what you did in the integral part w.r.t. x is an extension of the original integral, which has singularity when a=1 during partial fraction. At last when we sub a=1 to the solution, I think the continuity should be specified using convergence of original integral.

@shubhmittal77 Жыл бұрын

I can't imagine I really used to solve all this a few years back 😂

@ShanBojack Жыл бұрын

And now?

@navjotsingh2251 Жыл бұрын

@@ShanBojack usually, you study all this rigorously in school just to realise it’s not used as much in the working world. He probably hasn’t touched this since graduating.

@jose4877 Жыл бұрын

@@ShanBojack Keep using it or lose it. 😅😂

@BigDaddyGee85 Жыл бұрын

@@ShanBojack dude its actually worthless and not efficient at all to solve this by yourself...nobody will pay you for this nowadays. Of course its cool and stuff but in reality its just wasting time at the end of the day.

@AnkitThakur-rp6gp Жыл бұрын

@@navjotsingh2251 cmon this stuff is beautiful , and if you dont do anything you dont need anything to do what you are doing and that would be nothing hence you need nothing to do nothing (in any field)

@blackpenredpen Жыл бұрын

Check out BMT official solution: www.ocf.berkeley.edu/~bmt/wp-content/uploads/2022/04/calculus-solutions-1.pdf

@sairithvickgummadala2688 Жыл бұрын

We can use integration by parts, x is the one to be differentiated and cotx must be integrated, after doing that we get I = -int(lnsinx) from 0 to pi/2 which is a standard integral and we get I = (pi/2)ln2

@bogdanmarandiuc2895 Жыл бұрын

Hey in the future videos could you explain why cos(π/2^2)•cos(π/2^3)•...•cos(π/2^(n+1))=1/2^n•sin(π/2^(n+1))??? Because i saw it in the solutions of a problem and i can't explain myself why, and also why does its limit tends to 2/π?

@vasilis500 Жыл бұрын

@blackpernredpen Can you actually solve this ? Pr(m>=N/2) = sum from m=N/2 to N of (n/m)* 0.25^m * 0.75^(n-m)

@amirmahdypayrovi9316 Жыл бұрын

if n>0 , a>0 و a=0 ---> integral from 0 to (pi/2)^1/n of arctan(a.tan(x^n))/tan(x^n) dx =pi/2.ln(a+1)........... Is this true?!

@user-yu9mc6pu3q Жыл бұрын

Beautiful

@Sproggo Жыл бұрын

18:33 I lost it 😂

@trelosyiaellinika 2 ай бұрын

Nice method, one of many! I enjoyed it. I would have solved it a bit differently though, by integrating by parts x cotx to get the integral of ln(sin x) then through a phase shift demonstrating that the integral of ln(sin x) is the same as the integral of ln(cos x) and then adding them together to show that 2I=I-xln2 from 0 to π/2, i.e. Integral of ln(sin x) = - (π/2)ln2 and since the integral of x cot x = the integral of - ln(sin x) then the answer is (π/2)ln2... But all roads lead to Rome and it is interesting to explore all possible roads! Thanks for this!

@marionfelty7247 Жыл бұрын

You must be a math GOD. Just your skills with a marker are proof enough. 😮

@hachemimokrane8013 5 ай бұрын

Bravo!

@chemistrytable7347 Жыл бұрын

I solved this question with the help of complex numbers. x over tan(x) equals to x times cot(x). And from here we find x times ln(sinx) minus ln(sinx) in the integral. sin(x) equal to (e^(2ix)-1)/2ie^(ix). İt's easy after that😉. İf it wasn't for BPRP, I wouldn't have such a great mindset. Thanks...

@magicponyrides 4 ай бұрын

This is so cool.

@numericalcode Жыл бұрын

That’s a really cool problem

@MathwithMarker Жыл бұрын

Good job

@GrouchierThanThou Жыл бұрын

If you would substitute a = 1 into the intermediate steps of the calculation you would get zero denominators. Isn't that a problem?

@zunaidparker Жыл бұрын

No because the (a-1) is only a factor that comes out of the manipulation in the intermediate steps. It is canceled by an (a-1) factor in the numerator at the end, so it was never a "real" divide by zero problem to begin with.

@BloodHawk31 Жыл бұрын

Anothe method to bring linearity is substituting 1/tanx with its trigonometric/parabolic equal which is ix(e^ix + e^-ix)/e^ix - e^-ix. It is easier to work from here. Another easier way is simplifying the equation to xcotx, it will bring a ln to integrate, but it is still easier to find methods past that than the method explained. But that is the beauty if differentiation and integration, there are many ways of getting the same answer, trig/hyperbolic functions are my choice, but whatever is easiest for each person.

@tb-cg6vd Жыл бұрын

I too can fill a whiteboard with lots of integral formulae in red and black. Usually takes a lot of beer first. See you at the math comp!!

@vinayt7605 Жыл бұрын

Love your videos :) Love your enthusiasm more!

@blackpenredpen Жыл бұрын

Thank you!

@studentofcounterpoint Жыл бұрын

Powerful!

@dfh291 Жыл бұрын

This question was in my calc 1 final in 2021... I now know why this was so freaking hard.

@fable4315 Жыл бұрын

Are we allowed to use a=0 to find c, because as we looked at the limit of tan^-1(u*a) as u -> infinity, ist it still infinity if a=0? Or how can we allow this to be consistent?

@TrueRyoB 2 ай бұрын

this is nuts but tbf, it was quite enjoyable :D

@saumyatheallrounder6193 8 ай бұрын

Is this Feynman technique ???

@Kanekikun007 8 күн бұрын

Yup

@ANASzGAMEOVER Жыл бұрын

Good luck on the BMT2022, Man, do your best

@ANASzGAMEOVER Жыл бұрын

Holy shritting crap, a *heart* from the legend him self, you don't know how much I learned from you, I didn't even take Calculus in school yet, yet here I am, solving with you, thank you so very much, and hopefully, *hopefully* , you get first place 💜💜💜💜💜💜 Edit: I now realize how funny this word "shritting" is 😂

@user-wu8yq1rb9t Жыл бұрын

Great Thank you teacher Please more like this *bP🖋️rP🖍️* ❤️

@chiragsitlani5333 11 ай бұрын

U can also solve by parts. first xtanx = xcotx then by parts. integration from 0 to pi/2 ln(sinx) is -pi/2 ln2

@Shubhanshu05 Жыл бұрын

It's simple take it as integral of xcotx and then apply integration by parts And after that put the limiting values from 0---> π/2

@hdufort Жыл бұрын

The result is absolutely beautiful. The path to get there is horrific. I think I'll have mathmares (math nightmares) now.

@joshuawalsh6968 11 ай бұрын

So good

@antormosabbir4750 Жыл бұрын

It can be done by using definite integral property, a limit and the beautiful integral of 0 to π/2, ln(secx)

@akashsunil7464 Жыл бұрын

I applied the limit rule then I got tan(x)(pi/2-x) then I integrated tan(x)x using the power series expansion of tan(x)

@redroach401 Ай бұрын

I solved the integral by doing integration by parts with x = u and dv = cotx. uv from pi/2 to 0 is 0 - integral from 0 to pi/2 of ln(sinx)dx. To solve, set integral equal to I, make u-sub pi/2-x = u. You will find I also equals integral from 0 to pi/2 of ln(cosx). Therefore, 2I = integral from pi/2 to 0 of ln(sinxcox)dx. The inside is just 1/2sin(2x) and then separate the ln to get integral - pi/2ln(2). That integral is just I so bring to other side and you'll see that your answer when multiplied by -1 yields pi/2ln(2).

@tomiokashw Жыл бұрын

That's crazy!

@padraehsani3908 8 ай бұрын

Fantastic

@thanoscube8573 Жыл бұрын

I have not learned any calculus yet but this is very interesting!! 10/10

@frederic0chopin Жыл бұрын

i dont understand anything from these videos but i watch bcs its fun :D

@ajithv Жыл бұрын

Is it really OK to allow a = o at the end? Wouldn't that cause an issue a few steps earlier when evaluating tan (au) as u goes to infinity?

@ClaudioCP Жыл бұрын

Challenging! Good class and exercise!

@uzdefrederic1055 Жыл бұрын

Hello what a nice integral. It seems Feyman's technique does work everytime. I'd be curious to know if there are exceptions or not

@ironman85000 Жыл бұрын

Is this result valid for a=1? In the intermediate steps we had a 1/(a^2 - 1) term, which has a 0 in the denominator when a=1

@8DVIBEZ Жыл бұрын

Hey I have a doubt, why can't you use partial fraction method instead of cover up method? any advantages in cover up method? I'm eager to know this from you :)

@blackpenredpen Жыл бұрын

Bc it’s the same principle and it’s faster.

@booguy2636 Жыл бұрын

Amazing

@ronin4923 Жыл бұрын

easy integral. Use kings property to convert it into integral (pi/2-x) tanx dx, then you split it into two parts. Tan x integration is direct, for x *tan x we use by parts, upon which we have to solve the integral sec^2xdx. Now for this, assume this integral to be I, now we use kings property and get integral cosec^2x. Adding both, we get 2I = integral (1/(sin^2x*cos^2x)), and now we multiply the numerator and denominator by 4 to convert the integral into 2/(sin^2(2x)) = I. Now divide both the numerator and denominator by cos^2(2x), which will give us (2sec^2(2x))/(tan^2(2x)), which is easily solvable by the substitution tan(2x) = t. The method you used feels like you wanted to make the question way harder and impressive than it really is

@EminTuralic Жыл бұрын

Shouldn't a>=0 create a problem in the multiplication with tan(x) when x=pi/2? We'll get 0×infty which is an indeterminate form... which also beats the purpose of setting a=0 in order to calculate C. Or am I missing something?