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 ...
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 TH-cam channel. It’s crazy how much this channel has grown. Congrats!! 🎉
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)
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 .
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☺
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
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 :)
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.
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)
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.
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!!
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: th-cam.com/video/Vuu4y8UZXJ4/w-d-xo.html
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!
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
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).
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
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).
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
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.......''
@@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 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
@@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
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
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
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. :)
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.
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
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.
@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
@@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*
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.
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
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
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 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
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.
@@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.
@@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.
@@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)
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
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.
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.
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….
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...
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..
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))
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. =)
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.
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.
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
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!
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.
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
15:00 my teacher and I both of us just hate partial fraction You may multiply numerator and denominator by a²-1 And then in numerator a²+a²u²-(1+a²u²) Although thanks for this thought
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).
This was excellent! This one really pushed into some nice little tricks/nuances. Is that whole approach (i.e. inserting a parameter, differentiating w.r.t. the parameter etc etc) known as Feynman's trick?
Yes! This is the famous Feynman's trick used here but the more fascinated approach here applied is converting x into arctan(tanx). That is really thinking out of the box, if not thing out of the Universe 😅 The one who discover this approach is admirable! I couldn't be able to model such an awesome approach even I have been hinted with Feynman trick.
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
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.
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.
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."
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.
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
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.
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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!
Please solve -{integral (0-->π/2) {log[sinx]}dx}
@@orangesite7625 cos c + tan 90
what language is this?
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 ...
@@orangesite7625 answer is in this video but it can be done easier
And then multiply everything by BMT2020
Equate to BMT2021 and solve for x 🤣
😂😂😂😂
🤣🤣🤣
😂
You clown 🤣🤣🤣🤣
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 TH-cam channel. It’s crazy how much this channel has grown. Congrats!! 🎉
Thank you!! I do recognize this name. And wow, it’s been over 6 years!! Hope you are doing well 😃
@@blackpenredpen where do you teach?
@@sphrcl. harvard
Op
@@sphrcl. but how is your math now 💀💀
I actually solved this one on my own! I would never have been able to do that without guys like you teaching.
Op
by the same method?
@@ibrahimel-g7x yeah, same method. I knew where to apply it though!
Same, Ibdidnt use at the exact same place but i knew i would need this technique
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)
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 .
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.
That's the way I was going to go, but didn't see the ln.
I did the same way
Yeah, that was my first reaction as well. Glad to see someone else had the same thought process.
Excelent! You should post more challenging ones like this
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☺
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
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 :)
@@yoto60 I can't understand🤔
how did you apply IBP on this x/tanx dx mate? Did you convert it to ∫x tan-1x dx
@@wondersoul9170 ooh thanks broo😁🙏❤️
@@shiviarora4173 of course not, its just x * cotx, x to be differentiate, cotx to be integrate
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.
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)
This was my first thought! But I enjoyed the solution that he presented quite a lot!
this is a much better solution and answer can be obtained easily even by using byparts
Wow, Maclaurin series are good
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.
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!!
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: th-cam.com/video/Vuu4y8UZXJ4/w-d-xo.html
When he said “you give it a go,” I did this exact technique!
Exactly so.
Exactly I solved the same
I was looking for this comment Thanks
Another beautiful use of Feynman's technique!!
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!
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
@@robertkb64 i need to rewatch this or another video explaining this concept again. it's been awhile haha
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).
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
Lol me watching this after not even being one semester through calc in high school and wanted to do a premed biochem major anyways: 😂
Differentiations*
@@yoelit3931 you are right mb, funny thing is I still don't understand, isn't feynman technique?
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).
You don't really need to multiply top and bottom by f sec²(u). Just substitute u=tan x straight off.
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
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.......''
@@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
@@Happy_Abe yeah and we don't
We take the value and put it in the integral definition
@@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
@@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
This type of integration is also known as integration by reduction formula here in India. Nice video btw!
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
No it’s Leibniz rule but this one east with king property
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
pi/2 log(2) is about half the amount BPRP got.
@@HappyGardenOfLife I think they meant (pi/2)ln2
Good luck integrating ln(sin(x))
@@seroujghazarian6343 🤣
@@seroujghazarian6343 :tf:
This was a very, very satisfying demonstration BPRP. I didn't know about the Feynman technique until recently. Thank you!😸
Op
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. :)
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.
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
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.
@@BrollyyLSSJ thanks so much, was wondering about this caveat too
@@asparkdeity8717 me too...
@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
@@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*
this was the moment I realized I was never going to be a great mathematician
Idk if you read the comments on your old videos but you're basically my math professor for calculus :)
@@English_shahriar1 stop advertising your yt channel in comments
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.
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
Super
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
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.
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
How do you integrate ln(sinx)?
@@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
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.
I can't imagine I really used to solve all this a few years back 😂
And now?
@@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.
@@ShanBojack Keep using it or lose it. 😅😂
@@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.
@@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)
It's always good to see a nice calculus problem......
Watching this gives me so much relief🤩
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
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.
I did it the same way
- π÷2ln(1÷2) it is easy
@@advaykumar9726 Nice 👍.
Sol the int 0toπ÷2 log(sinx) slove without complex number it is slove by KING RULE
@@vasubhalani3938 I see what you did there 😉.
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
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.
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….
Wrong
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...
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!!
wow wow , u explained it in a very simple way thank you ! please do more videos
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..
An harder integral is from 0 to pi/4. It took me three days to solve it.
Never heard of Catalan’s constant until I researched this.
how did you solve it? Please show us.
@@MadScientist1988 integrate by parts and then watch video "how to integrate ln(cosx) and ln(sinx) from 0 to pi/4" for answer.
You must be a math GOD. Just your skills with a marker are proof enough. 😮
More like this please
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))
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. =)
Thank you!
I really enjoyed that. Best part of my day so far :D
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
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.
I'm 17 and i got no idea what's going on but for some reason I enjoy this lol.
This question was in my calc 1 final in 2021... I now know why this was so freaking hard.
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.
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
Beautiful integral
Integration by parts gives us -Int(ln(sin(x)),x=0..π/2)
Added it to my Math YT video collection!
Gracias por existir este canal.....
Es de lo mejor que he encontrado. ..
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
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!
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
You explain so well that i want all day watch you)
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)
No one taught me that method. Thanks. After some investigations, I know now that the method is called for some people Feynman.
If you would substitute a = 1 into the intermediate steps of the calculation you would get zero denominators. Isn't that a problem?
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.
This technique is called feynman integration technique and its very useful in solving the integrals.
Excellent problem and wonderful explanation!
Use a+b-x property of definite integrals it will be easier
It is questions like these that just make me very happy
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
while doing the cover up method there was just casually u²=-1 meaning we visited the complex world along the way
15:00 my teacher and I both of us just hate partial fraction You may multiply numerator and denominator by a²-1
And then in numerator a²+a²u²-(1+a²u²)
Although thanks for this thought
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).
This was excellent! This one really pushed into some nice little tricks/nuances. Is that whole approach (i.e. inserting a parameter, differentiating w.r.t. the parameter etc etc) known as Feynman's trick?
Yes! This is the famous Feynman's trick used here but the more fascinated approach here applied is converting x into arctan(tanx). That is really thinking out of the box, if not thing out of the Universe 😅 The one who discover this approach is admirable! I couldn't be able to model such an awesome approach even I have been hinted with Feynman trick.
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
I think that the parameter "a" should be not equal to "1", because you're divided by (a-1), so we must suppose that 0
U can also solve by parts. first xtanx = xcotx then by parts. integration from 0 to pi/2 ln(sinx) is -pi/2 ln2
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.
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.
It is not hard to show that the other integrals in fact vanish.
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."
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.
Very nice solution!
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
Many integrals like this one can be handled more effectively and efficiently in the complex world by using contour integration
M2: substitute x= tan(inv)theta the using by parts
This was very cool to watch!
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.
Nicely done 👍💯
You are very smart
Use integral by parts after converting 1/tanx into cotx.
Apply queen's rule...
2 ₀∫⁰ ̇²⁵ᵖⁱ (π/2-x)/tan(π/2-x)
=>2 ₀∫⁰ ̇²⁵ᵖⁱ (π/2-x)/cot x
=>2 ₀∫⁰ ̇²⁵ᵖⁱ (π/2-x)tan x
Solve xtanx using integration by parts
The result is absolutely beautiful. The path to get there is horrific. I think I'll have mathmares (math nightmares) now.
Thank you sir.
Really interesting. Well done!
Well, There's a More Easy method....
WE can apply KING'S Rule by and then integrate by parts and we get the ans...
This seems like magic to me.
It can be done by using definite integral property, a limit and the beautiful integral of 0 to π/2, ln(secx)
18:33 I lost it 😂
Hi BPRP!
Differentiation under the integral sign might be my favorite math technique ever.
I love it, too!