so you want a HARD integral from the Berkeley Math Tournament
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- Опубликовано: 28 май 2024
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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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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 RUclips 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 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.
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?
@@user-ox2qt2cm4c 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
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: ruclips.net/video/Vuu4y8UZXJ4/видео.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
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.
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:
Excelent! You should post more challenging ones like this
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
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!!
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
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*
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.
Another beautiful use of Feynman's technique!!
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 love this channel. I hated calculus and did really poorly in it. Found this channel and it reignited my love for maths. Now I can follow along and even do the more difficult integrals thanks to these videos. Thank you a lot for igniting that flame again!!
Can you tell me how to start from scratch i want to learn calculus but i am bad at it
@@Legend-mr6ci my advice would be 1) try to make it fun for yourself by seeing where the topics are actually used in the real world, 2) go very slowly and take your time learning and understanding the concepts. Watch as many videos as possible and 3) practice a lot. So find videos you can practice along with. Calculus is a toolbox and if you don’t polish your tools in the toolbox then they won’t work that well
This was a very, very satisfying demonstration BPRP. I didn't know about the Feynman technique until recently. Thank you!😸
Op
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.
Gracias por existir este canal.....
Es de lo mejor que he encontrado. ..
wow wow , u explained it in a very simple way thank you ! please do more videos
Excellent problem and wonderful explanation!
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
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.
I really enjoyed that. Best part of my day so far :D
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..
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
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.
Watching this gives me so much relief🤩
Check out BMT official solution: www.ocf.berkeley.edu/~bmt/wp-content/uploads/2022/04/calculus-solutions-1.pdf
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
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?
@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)
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?!
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.
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).
You explain so well that i want all day watch you)
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)
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.
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.
U can also solve by parts. first xtanx = xcotx then by parts. integration from 0 to pi/2 ln(sinx) is -pi/2 ln2
thought of applying beta fn. but who knew.
hands down! i was literally smiling at the end
🙌
It's always good to see a nice calculus problem......
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))
It can be done by using definite integral property, a limit and the beautiful integral of 0 to π/2, ln(secx)
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
It is questions like these that just make me very happy
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.
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.
Really interesting. Well done!
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
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. :)
This was very cool to watch!
More like this please
Very nice solution!
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
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!
Nicely done 👍💯
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!
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!
Added it to my Math YT video collection!
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...
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
If we consider the integral as xcotx we can do it integration by parts and ans comes out to be π/2 ln2
This question was in my calc 1 final in 2021... I now know why this was so freaking hard.
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
Beautiful integral
Thank you sir.
Good luck on the BMT2022, Man, do your best
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 😂
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.
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!!
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)
You can also use By parts method
You must be a math GOD. Just your skills with a marker are proof enough. 😮
this was enlighting!
Love your videos :) Love your enthusiasm more!
Thank you!
Great
Thank you teacher
Please more like this
*bP🖋️rP🖍️* ❤️
Use a+b-x property of definite integrals it will be easier
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.
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.
This can be solved very easily using the properties of definite inegrals and complex analysis to get π/2ln2
Use integral by parts after converting 1/tanx into cotx.
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.
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
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
Thank you sir
Sin palabras... ¡genial!
Integration by parts gives us -Int(ln(sin(x)),x=0..π/2)
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
excellent work
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
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
Well played
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.
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.
We can directly solve this by integral x cotx and applying integration by parts( assuming first fnx as x and second fnx as cotx)..... We already know integral of ln(sinx) [ which is integral of cotx ] from 0 to π/2 is -π/2 ln2
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
Challenging! Good class and exercise!
im ngl but i thought this was one of the easier definite integral questions ive got lol, but that may be because i already knew what integral of 0 to pi/2 ln(sinx) is
okay i just saw your approach and i do see why this was thought of as difficult, however u can do this another way too write the expression as xcotx at the begginig and i change the bounds from h to pi/2, where lim h--> 0+, then use integration by parts with x as the one which gets differentiated and cot x getting integrated, the first part is easy to solve with limits, the second part u can write it as integration from 0 to pi/2 ln(sinx), which is -pi/2 .ln2, and you can prove this with kings rule and some manipulation, the first part once u solve becomes 0, the second part is -(-pi/2 ln2) which makes the final answer pi/2 ln2
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.
this is nuts
but tbf, it was quite enjoyable :D
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.
I have not learned any calculus yet but this is very interesting!! 10/10
Snap. Can't wait until I get to Calc III and can really understand what the multivariable stuff is all about.