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
2 жыл бұрын
And then multiply everything by BMT2020
@Jha-s-kitchen
2 жыл бұрын
Equate to BMT2021 and solve for x 🤣
@harith_khaleel
2 жыл бұрын
😂😂😂😂
@gauravsonawane
2 жыл бұрын
🤣🤣🤣
@admink8662
2 жыл бұрын
😂
@troym856
2 жыл бұрын
You clown 🤣🤣🤣🤣
@sabrinagiang
2 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 KZitem channel. It’s crazy how much this channel has grown. Congrats!! 🎉
@blackpenredpen
2 жыл бұрын
Thank you!! I do recognize this name. And wow, it’s been over 6 years!! Hope you are doing well 😃
@sphrcl.
2 жыл бұрын
@@blackpenredpen where do you teach?
@wartex3561
2 жыл бұрын
@@sphrcl. harvard
@satyagaming996
2 жыл бұрын
Op
@ChannelTerminatedbyYouTube
Жыл бұрын
@@sphrcl. but how is your math now 💀💀
@aidankwok1475
2 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.
@limpinggabriel6658
2 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
2 жыл бұрын
Op
@ibrahimel-g7x
Жыл бұрын
by the same method?
@limpinggabriel6658
Жыл бұрын
@@ibrahimel-g7x yeah, same method. I knew where to apply it though!
@bruhe8895
10 ай бұрын
Same, Ibdidnt use at the exact same place but i knew i would need this technique
@AriosJentu
2 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
2 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
2 жыл бұрын
@@yoto60 I can't understand🤔
@shiviarora4173
2 жыл бұрын
how did you apply IBP on this x/tanx dx mate? Did you convert it to ∫x tan-1x dx
@ayoubmff7834
2 жыл бұрын
@@wondersoul9170 ooh thanks broo😁🙏❤️
@AriosJentu
2 жыл бұрын
@@shiviarora4173 of course not, its just x * cotx, x to be differentiate, cotx to be integrate
@gustavozola7167
2 жыл бұрын
Excelent! You should post more challenging ones like this
@silversleezy4953
2 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: kzitem.info/news/bejne/t6urmq1ujoyJfWk
@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
@everytime865
2 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
@ガアラ-h3h
Жыл бұрын
No it’s Leibniz rule but this one east with king property
@Happy_Abe
2 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
2 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
2 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
2 жыл бұрын
@@Happy_Abe yeah and we don't We take the value and put it in the integral definition
@Happy_Abe
2 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
2 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
@SpringySpring04
9 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!!
@EmpyreanLightASMR
Жыл 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
11 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
11 ай бұрын
@@robertkb64 i need to rewatch this or another video explaining this concept again. it's been awhile haha
@aeroeng22
8 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).
@kaanetsu1623
2 жыл бұрын
Another beautiful use of Feynman's technique!!
@blackpenredpen
2 жыл бұрын
Check out BMT official solution: www.ocf.berkeley.edu/~bmt/wp-content/uploads/2022/04/calculus-solutions-1.pdf
@sairithvickgummadala2688
2 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
2 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
2 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
2 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?!
@xd_metrix
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
11 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
8 ай бұрын
Differentiations*
@xd_metrix
8 ай бұрын
@@yoelit3931 you are right mb, funny thing is I still don't understand, isn't feynman technique?
@chennebicken372
2 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.
@tarouk8446
19 күн бұрын
Dear Professor BPRP, Thanks to your video, I was finally able to solve a problem that had stumped me for over three years. The problem was to integrate x/tanx log(1/tan x) from 0 to π/2. By using the techniques you introduced and combining them with a few other methods, I was able to show that the answer is π^3 / 48. If you’re interested, I’d love to see you take on this challenge and make a video about it. thank you very much, greetings from Japan☺
@rishabsaini8347
2 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
2 жыл бұрын
pi/2 log(2) is about half the amount BPRP got.
@kozokosa9289
2 жыл бұрын
@@HappyGardenOfLife I think they meant (pi/2)ln2
@seroujghazarian6343
2 жыл бұрын
Good luck integrating ln(sin(x))
@sam-gooner
2 жыл бұрын
@@seroujghazarian6343 🤣
@dsfdsgsd644
2 жыл бұрын
@@seroujghazarian6343 :tf:
@12wholepizzas13
2 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
2 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
2 жыл бұрын
@@BrollyyLSSJ thanks so much, was wondering about this caveat too
@UglyRooks
2 жыл бұрын
@@asparkdeity8717 me too...
@alexandervanhaastrecht7957
2 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
2 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*
@bjornfeuerbacher5514
2 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
2 жыл бұрын
You don't really need to multiply top and bottom by f sec²(u). Just substitute u=tan x straight off.
@Assterix
2 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
@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. :)
@Jack_Callcott_AU
2 жыл бұрын
This was a very, very satisfying demonstration BPRP. I didn't know about the Feynman technique until recently. Thank you!😸
@satyagaming996
2 жыл бұрын
Op
@dankmortal4911
6 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
@tgx3529
2 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.
@BuggyDClown-yc8ws
7 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
@Nick-wh4jt
2 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
@itz_adi.g
9 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..
@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.
@AhirZamanSairi
2 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.
@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.
@redroach401
3 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).
@musicngamesyo6628
2 жыл бұрын
Watching this gives me so much relief🤩
@neypaz8054
2 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
2 жыл бұрын
Thank you!
@shubhmittal77
2 жыл бұрын
I can't imagine I really used to solve all this a few years back 😂
@ShanBojack
2 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)
@abhishekchoudhary4689
2 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
2 жыл бұрын
How do you integrate ln(sinx)?
@rad858
2 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
@bakeyourownshit4137
2 жыл бұрын
thought of applying beta fn. but who knew. hands down! i was literally smiling at the end 🙌
@bheriraju722
2 жыл бұрын
Take x/tanx is x cotx Take x as first function cotx as second function and integrate by parts We get xlog |sinx|- integral of log| sinx| dx between limits 0 and pi/2 xlog|sinx| when limits applied becomes zero And we are left over with - integral of log|sinx| dx between limits 0 and pi/2 Let I =-int log|sinx| dx between 0 and pi/2 I=-int log|sin((pi/2)-x)dx between 0 and pi/2 Adding both 2I=-int log|(sin2x)/2| dx between limits 0 and pi/2 2I=-int log|sin2x|dx-int log2 dx between limits 0 and pi/2 2I=- int log |sin2x| dx- xlog2 between limits 0 and pi/2 -2I+(pi/2) log 2= int log| sin 2x| dx between limits 0 and pi/2 Put 2x = t 2dx = dt dx = dt/2 Also limits change from 0 to pi -2I+(pi/2)log 2=int (log|Sint|)dt/2 between limits 0 and pi Since sin (pi-t)=sin t -2I+(pi/2)log 2=(2int log|Sint| dt )/2 between limits 0 and Pi/2 -2I+(pi/2)log 2=-I I=(pi log 2)/2
@sathwikpatibandla1159
2 жыл бұрын
Super
@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.
@garyhuntress6871
2 жыл бұрын
I really enjoyed that. Best part of my day so far :D
@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
@8DVIBEZ
2 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
2 жыл бұрын
Bc it’s the same principle and it’s faster.
@fritzartfan
6 ай бұрын
That's a piece of cake if you know integral of ln(secx from 0 to π/2 is π/2 (ln2). Use product rule
@moeberry8226
2 жыл бұрын
There is no need to use Feynmans trick for this integral it can be solved using integration by parts by rewriting 1/tan(x) as cot(x) and then differentiating x and integrating cot(x) you will reach the integral of ln(sin(x)) which easily can be handled with doing a little phase shift.
@GrouchierThanThou
2 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
2 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.
@chiragsitlani5333
Жыл 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
@stevemonkey6666
2 жыл бұрын
It's always good to see a nice calculus problem......
@ANASzGAMEOVER
2 жыл бұрын
Good luck on the BMT2022, Man, do your best
@ANASzGAMEOVER
2 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 😂
@AlexeyGodin
Жыл бұрын
Taking t=pi/2-x substituting, summing two integrals and noticing we get a function symmetric vs x=pi/4 we get I=\int_0^{pi/4} 2x/tan(2x) dx+ pi/2 \int_0^{pi/4} tan(x)dx =I/2+pi/4 ln(2). And that's all. Two minutes of work instead of all the hassle and Feinamn's tricks.
@Lyrical-lounge-music
13 күн бұрын
I think you can just use by parts here, of the two terms first is 0 and second is int lnsinx 0 to pi/2 which can be calculated by applying di properties to get pi/2ln2
@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
@OussamaYaqdane2
2 жыл бұрын
I have a question sir, First you said that a>0, so when x--> ∞, arctan(ax)= π/2 and later you added that a≥0, but if a=0, when x--> infinity, ax=0, cuz a is exactly 0 isn't this a mistake?
@iamtraditi4075
2 жыл бұрын
If you evaluate arctan(au) as u -> infinity when a = 0, you get lim_{u -> infinity} arctan(0) = 0. Following this through in the rest of the work, I’(0) evaluates to pi/2, which is exactly what you get if you take the form of I’(a) listed and evaluate at 0 (i.e. we can manually check that the formula we got to works for a = 0). From here, it’s fine to do the rest of the working
@ganeshreddykomitireddy5128
2 жыл бұрын
Use integral by parts after converting 1/tanx into cotx.
@richardheiville937
2 жыл бұрын
No need double integral (or derivation under integral sign) integration by part is ok here since 1/tan(x)=cos x/sin x=f'/f has an obvious antiderivative.
@bradleybeauclair8282
Жыл бұрын
Listen lady, all I heard you say was, "when you're integrating a derivative of a trigonometric function, therefore you're going to put the natural log in the answer in front of a coefficient (a), and that comes after the suffix when the suffix equals 1 and add C (in this case, c = 0). You squared the denominator and used that number and multiplied it by 1/pi to get the number you would multiply across to make 1=1 (.....4/pi * pi/2). That was the number you put for a. This works for some reason because quadratic equation and that suffix, (pi/2) has to go in front of ln and 2 in Chain Rule such that you get pi/2 * ln * 2 + 0."
@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))
@abhiramkadaba9564
Жыл бұрын
We can convert the definite integral to integral tanX(pi/2-X) dX, Then use integration by Parts, we get 0 -integral log(cosX) dX. Which can be solved to get Pi/2 log2
@procerpat9223
Жыл бұрын
Geometrically it’s equivalent to the integral of the area as function of a of 1/(1+(a tan (x))^2) from o to pi/w evaluated at a=1!
@Noam_.Menashe
2 жыл бұрын
An harder integral is from 0 to pi/4. It took me three days to solve it.
@ericmccormick1639
2 жыл бұрын
Never heard of Catalan’s constant until I researched this.
@MadScientist1988
Жыл бұрын
how did you solve it? Please show us.
@Noam_.Menashe
Жыл бұрын
@@MadScientist1988 integrate by parts and then watch video "how to integrate ln(cosx) and ln(sinx) from 0 to pi/4" for answer.
@chemistrytable7347
2 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...
@renesperb
Жыл бұрын
There is a much simpler way to calculate this integral. First you integrate by parts to get x*ln(sinx) in the limits 0 and π which gives 0. It remains the integral from 0 to π/2 of ln(sinx) .This can be done as follows : write sinx = 1/2i *(exp[i x]- exp[- ix]). Then , ln(1/2i *(exp[i x]- exp[- ix]) = -ln2 - i*π/2 +ln(exp[i x]- exp[- ix]). But ,clearly , the original integral must have a real value , and hence it has the value - π/2 *ln2 , and the other integrals vanish.
@renesperb
Жыл бұрын
It is not hard to show that the other integrals in fact vanish.
@trelosyiaellinika
4 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!
@holyshit922
2 жыл бұрын
Integration by parts gives us -Int(ln(sin(x)),x=0..π/2)
@neilgerace355
2 жыл бұрын
2:05 Is it also called Feynman's Technique?
@blackpenredpen
2 жыл бұрын
Yes. I am not why for I was forgetting that lol
@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.
@cesarluishernandezpertuz8794
Жыл бұрын
Gracias por existir este canal..... Es de lo mejor que he encontrado. ..
@BigDBrian
8 ай бұрын
while doing the cover up method there was just casually u²=-1 meaning we visited the complex world along the way
@jonatansvensson8346
Жыл бұрын
When a=0, can You be sure that a*u=inf when u goes to inf? I feel like you included a=0 after without proving that it was ok. Any thoughts?
@siyuanhuo7301
Жыл бұрын
I was thinking the same thing. I scrolled down here just to find this comment.
@vijaichikatimalla3211
Жыл бұрын
wow wow , u explained it in a very simple way thank you ! please do more videos
@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!!
@antormosabbir4750
2 жыл бұрын
It can be done by using definite integral property, a limit and the beautiful integral of 0 to π/2, ln(secx)
@karabodibakoane3202
2 жыл бұрын
I'm still at the beginning of the video but I think IBP will yield. It results in the integral -\int_0^{\pi/2}\ln\sin x\,dx which is a standard integral that at one point was thought to be computable only with the use of complex analysis techniques and has a value of \frac{\pi\ln 2}{2}. No need for Feynman's trick.
@advaykumar9726
2 жыл бұрын
I did it the same way
@vasubhalani3938
2 жыл бұрын
- π÷2ln(1÷2) it is easy
@karabodibakoane3202
2 жыл бұрын
@@advaykumar9726 Nice 👍.
@vasubhalani3938
2 жыл бұрын
Sol the int 0toπ÷2 log(sinx) slove without complex number it is slove by KING RULE
@karabodibakoane3202
2 жыл бұрын
@@vasubhalani3938 I see what you did there 😉.
@armaanhussain7105
7 ай бұрын
im not getting why we cant just write x/tanx as xcotx and then do DI or by parts? can anyone explain that to me?
@oloyt6844
3 ай бұрын
Idk
@not-a-school-account
3 ай бұрын
you can try to do DI method yourself (spoiler: it doesn't work). infact i advise you to try a method before you comment
@amineaoussar5015
Ай бұрын
The integral of cotan(x) is ln(sinx)
@agrushnev
4 ай бұрын
It's easy to see that I = - \int_0^{Pi/2} ln(sin(x)) = - \int_0^{Pi/2} ln(cos(x)) which implies 2I = (Pi/2)ln(2) + I
@CDChester
2 жыл бұрын
Added it to my Math YT video collection!
@saumyatheallrounder6193
10 ай бұрын
Is this Feynman technique ???
@Kanekikun007
Ай бұрын
Yup
@harshchoudhary2691
Жыл бұрын
So basically using king's property we get 2I= int 0 to pi/2 pi/2/cot x + int 0 to pi/2 x(1/tan x - 1/cot x) which can be simplified as tan (pi/4-x) then we can use integration by parts and apply limits
@famillemagnan1313
Жыл бұрын
What is hard here, is the circonvoluted method to solve it! Kudos by the way. But you do a simple integration by part and you end up with the ultra classical integral from 0 to pi/2 of ln( sinx) which is well known, I’m pretty sure, by all the nerds taking the test….
@maalikserebryakov
Жыл бұрын
Wrong
@superlinux
Жыл бұрын
Why didn't you convert the tanget to co-tangent in order to get : x/tan(x) = x cot(x) ? And then do something like - not sure though - by parts method .. that's faster i think. You took a very long solution path.
@maalikserebryakov
Жыл бұрын
You can’t do that. There is NO elementary antiderivative for that function. Why can’t you understand this
@Zia568
Жыл бұрын
This technique is called feynman integration technique and its very useful in solving the integrals.
@yahirdominguez8222
2 жыл бұрын
It is questions like these that just make me very happy
@aashcharyagorakh2459
Жыл бұрын
If we consider the integral as xcotx we can do it integration by parts and ans comes out to be π/2 ln2
@armanavagyan1876
Жыл бұрын
You explain so well that i want all day watch you)
@Sproggo
Жыл бұрын
18:33 I lost it 😂
@akashsunil7464
2 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)
@reo1a
8 ай бұрын
Just note that generally one should discuss the justification for exchanging the integration and differentiation (in accordance with the Leibniz integral rule) before applying it.
@TheHellBoy05
Жыл бұрын
This can be solved very easily using the properties of definite inegrals and complex analysis to get π/2ln2
@meet_2480
Жыл бұрын
M2: substitute x= tan(inv)theta the using by parts
@DrLiangMath
Жыл бұрын
Excellent problem and wonderful explanation!
@saharhaimyaccov4977
2 жыл бұрын
More like this please
@totiodfanzsai6079
Жыл бұрын
I=Integral(x/tanx)dx [0,pi/2] _____(1) => I=Integral((pi/2 - x)/tan(pi/2 - x))dx [0,pi/2] As integral(f(x))dx [a,b] = integral(f(a+b-x)dx [a,b] we know tan(pi/2 -x)=cotx So, I = Integral((pi/2 -x)cotx)dx [0,pi/2]_____(2) multiply 1 and 2:- I^2 = Integral(x(pi/2 - x))dx [0,pi/2] Solving it, we get:- I = 0.806(approx) CORRECT ME KINDLY, IF ANY MISTAKES FOUND
@Zavstar
2 жыл бұрын
Use a+b-x property of definite integrals it will be easier
@yoyoezzijr
2 жыл бұрын
Beautiful integral
@looney1023
Ай бұрын
It would also be quite complicated but I wonder if there's an approach that uses either the series expansion of x*cotx or the identity x/sinx = gamma(1+x/pi)*gamma(1-x/pi)
@saxbend
Жыл бұрын
If you're allowing a to be equal to zero, then don't you have to consider that zero times infinity might in fact not be equal to positive infinity in the earlier step?
@ebenezerasiama5868
10 ай бұрын
Can’t we use the King property?
@marionfelty7247
2 жыл бұрын
You must be a math GOD. Just your skills with a marker are proof enough. 😮
@siddheshphadke3719
7 ай бұрын
There’s another method you can do, integratjon xcotxdx, then use integration by parts, considering x and Ist fnc, cotx as IInd fnc, and then it reduces to integration ln(sinx)dx which is really easy to find out
@h.m8811
5 ай бұрын
How do you find ln(sinx) dx
@grant1390
Жыл бұрын
Seems this integral is all over KZitem lately. Instead of this relatively simple one try integrating over the same limits x / (tanx)^1/3. Numerically it is approx. 1.013596558 but it can be written in terms of elementary constants and functions.
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