Is this hard integral from the 2020 Berkeley Math Tournament really that hard?

[@vadymshtabovenko3532]

blackpenredpen as well as you is guy whom i can be grateful for me being able to solve math-olimpiad integrals, both of you are the best

[@maths_505]

Ah man you almost made me cry tears of joy there🥺 Thank you so much

[@GiornoYoshikage]

Simple and insane integral! Sure the experience helps finding clear and beautiful approaches!

[@ernestschoenmakers8181]

One can also apply Feynman's trick where you transform x into arctan(tanx) and where you put the variable a inside this arctan function. This becomes arctan(a*tanx) and differentiate under the integral sign with respect to a.

[@genosingh]

ngl the writing tanx as cotx part is prolly the hardest part, as I never would've even thought about using ibp in this.

[@TanmaY_TalK]

Whenever I see 0 to pi/2 my brain always tells to apply king's property

[@francis6888]

09:42 an even easier thing to do is substitute t for pi/2 - t since the ln of an even function is still an even function.

[@peternyekete6802]

Very interesting observation. Kindly rework the problem and post it for me. Am trying to learn the techniques

[@bjornfeuerbacher5514]

Geometrically, this is the area under the "quadratrix" (of Hippias) - or is proportional to that area, depending on the exact values you choose for the parameters of the quadratrix.

[@TanmaY_TalK]

Ans is pi/2 ln2 not negetive pi/2 ln2 ! [xlnsinx]0 to pi/2 then -I1 that's why your answer has an extra spicy '-' (-ve)

[@そるそる-e5r]

My friends solved lim[x→0]xln(sinx) in this elegant way: lim[x→0]xln(sinx) =lim[x→0] x/sinx * sin(x)ln(sinx) =1 * 0=0 (∵lim[x→0]sinx/x=0, lim[t→0]tln(t) =-lim[s→∞]ln(s)/s=0)

[@genosingh]

For proving that int from 0 to pi is same as 2 times(int from 0 to pi/2), of ln(sinx), you actually don't even have to look at the graph, you can straight up divi the integral into 2 parts one from 0 to pi/2 and the other from pi/2 to pi, and apply king's rule on the second one and sub x - pi/2 = t, and you'll prove that statement.

[@renesperb]

Another way of calculating the integral of ln(sinx) is to write sinx as 1/2i *Exp[i x]*(1-Exp[-2 i x]. Then one can simplify the ln of this expression as -ln 2 - i π /2 + i x+ ln(1- Exp[- 2 i x). The integral from 0 to π of the last term is zero and the integral of the first three terms gives -π ln2. But I find your solution more elegant.

[@maths_505]

Thanks That's why I wanted to share it I often look for solutions using real techniques only

[@vinculum_mhm]

great approach

[@師太滅絕]

What about X/tan x = Xcos x / sin x, then we do substitution?

[@ronnykazadi352]

5:27 where does the phase shift come from? Why replace x by pi/2 - x? We know cosx=sin(pi/2 - x) but i dont understand your step

[@maths_505]

It's like a substitution t=pi/2-x but we rename the dummy variable "t" back to "x" all in one step

[@ronnykazadi352]

@@maths_505 Ohhhhh I understand. But if that's the case then, you missed minus sign because by substitution, the integral would be from "pi/2 to 0" but to switch it back to "0 to pi/2" there is going to be a minus sign in front of the integral. That explains why your final answer is " minus pi/2 ln2" instead of "pi/2 ln2" check the minus sign that you forgot. Great video by the way.

[@maths_505]

There is no missing negative sign🤦‍♂️🤦‍♂️🤦‍♂️....t=pi/2 - x means dt=-dx.... The extra negatives cancel out!

[@takemyhand1988]

​@@ronnykazadi352it's called King's property

[@emanuellandeholm5657]

Taking notes... Edit: huge fan of bprp too :)

[@utuberaj60]

Appreciate your elementary method to solve this integral in contrast to the Feynman technique used by Blackpenredpen- who's videos I watch and subscribe too. But I see a discrepancy in the end result using the above two techniues- please pause at 4:30 of your video. Your answer is (-) Pi/2 * (ln2), whereas Blackpenredpen result is the same BUT is positive. I tried boththe methods, and confirmed this. Can you explain why this discrepancy? The methods seem Ok.

[@ガアラ-h3h]

They both use a lengthy tiring method use kings rule and multiply to instantly kill the tan x then just integrate a polynomial and donedio

[@MarceloKatayama]

His end result is, in fact, positive. This is because his integral has a negative sign, which means that (-pi/2 ln2) is multiplied by -1, which nets pi ln2 /2

[@leesweets4110]

I took complex analysis. Very interesting subject. But it was only introductory level. Im not quite sure how complex analysis can be used to solve an integral. Can we have a lesson on that?

[@maths_505]

Yeah I plan on uploading an entire course on that Probably once I'm done with the course on DEs Oh snap I forgot to upload the homework solution 🤦‍♂️ Ah well...I'll do it tomorrow with another video too

[@maths_505]

In the meantime I plan on uploading some content on complex integration too

[@moeberry8226]

Be ready for Euler’s formula Boys.

[@kamalsaleh6497]

So how do we use complex analysis for the ln(sin x)?

[@ガアラ-h3h]

Add I and subtract I then use log rules and Euler identity and do integration by parts cancel split integral into two piece then you basically have very easy integral

[@par22]

You lost me at 5:30 when you did the phase shift. How can you equate sin x = sin (pi/2 - x) = cos x ? sin x is not cos x, so it wouldn't be 2*I1... very confused.

[@maths_505]

Oh there's a video proving this transformation kzbin.info/www/bejne/a2rQeYp9ntV2fLM It's in the first 3 or 4 minutes (1st property)

[@par22]

@@maths_505 Awesome, thank you! I get it now. I completely forgot about that transformation

[@maths_505]

@@par22 no problem mate It's a pretty useful one too especially for trig integrals

[@Shadow-Presentations]

So what is the actual solution of this? Like when do we take what value??

[@Krishnajha20101]

How did you get the negative sign in the end?

[@thealternativefactor6694]

My head hurts..

[@farhadazadi]

Hey may I ask what software you are using to write?

[@alexliu801]

awesome!!!

[@holyshit922]

Do you want indefinite integral which Wolfram alpa is unable to calculate ? Int(x/sqrt((x+2)^2+exp(x)),x) Hint 1 Rewrite integral as Int(x/sqrt((x+2)^2+exp(x)),x)=Int(1+(x/sqrt((x+2)^2+exp(x))-1),x) then you have to integrate two integrals but first one is easy Int(1,x)+Int((x-sqrt((x+2)^2+exp(x)))/sqrt((x+2)^2+exp(x)),x) Hint 2 To calculate integral Int((x-sqrt((x+2)^2+exp(x)))/sqrt((x+2)^2+exp(x)),x) multiply numerator and denominator by x+2+sqrt((x+2)^2+exp(x)) to get Int(1/(x+2+sqrt((x+2)^2+exp(x)))((x-sqrt((x+2)^2+exp(x)))(x+2+sqrt((x+2)^2+exp(x))))/sqrt((x+2)^2+exp(x)),x) then multiply numerator and you should see suitable substitution

[@maths_505]

Damn! That is a good idea I'll keep it in mind and I'll mention you in the video too

[@holyshit922]

Int(ln(sin(x)),x=0..Pi) = Int(ln(sin(x)),x=0..Pi/2)+Int(ln(sin(x)),x=Pi/2..Pi) u = Pi-x du=-dx Int(ln(sin(x)),x=0..Pi) = Int(ln(sin(x)),x=0..Pi/2)+Int(ln(sin(Pi-u))(-1),u=Pi/2..0) Int(ln(sin(x)),x=0..Pi) = Int(ln(sin(x)),x=0..Pi/2)+Int(ln(sin(u)),u=0..Pi/2) Int(ln(sin(x)),x=0..Pi) = 2Int(ln(sin(x)),x=0..Pi/2)

[@moeberry8226]

You can say what you want about blackpenredpen but he actually speaks good English and has over a million subscribers which means there’s over a million people and me who don’t agree with you.

[@maths_505]

Definitely one of the GOATs of KZbin maths

[@spore124]

My friend you have formed a parasocial relationship with a person who does math problems on KZbin.

[@moeberry8226]

@@spore124 so when I state a fact about another person that means I’m in a para social relationship with that person? You sound like a 5 year old kid.

[@spore124]

@@moeberry8226 Well, ya. You didn't post the comment as a reply to a comment and it doesn't seem relevant to anything in the video, and then you conflated 1 million subscribers as being a hivemind that thinks the same. A textbook case, right? We're not members of devout Pythagorean societies, this is just a nice place to work out math problems and learn new methods. Cool off with a nice walk or calculate a couple stress free derivatives.

[@moeberry8226]

@@spore124 your 100 percent wrong, I posted in response to another person saying he speaks bad English and saying he doesn’t like Blackpenredpen and I stated that a lot of people who view math channels do not agree with him including myself. Nothing is conflated other than you and your sarcasm. Your opinions don’t matter to facts.

[@zahari20]

There is no such thing as Feinman integration. This is the Leibniz rule for integrals with parameter.

[@user-he4em5zx8s]

Explain x->ㅠ/2-x Plz

[@bjornfeuerbacher5514]

Essentially, that only uses that if you integrate over the interval 0 to pi/2, sin(x) and cos(x) will take on exactly the same values, only in a different order.

[@takemyhand1988]

It's called King's property of integrals For a definite integration, if the bounds are a to b. Then the f(x) can be substituted as f(a+b-x).

[@bjornfeuerbacher5514]

@@takemyhand1988 That's just restating the method in general, not explaining why it actually works.

[@takemyhand1988]

@@bjornfeuerbacher5514 maybe he was confused where pi/2- x came from.

[@brunojani7968]

Google variable substitution. Image you let t = pi/2-x, and a variable is only a label, and so you change it back to x

[@prollysine]

Hi Math 505, do you think ? int xctgx dx, /D, I/, --> =xlnsinx |0,pi/2) |0,pi/2|=0, stays: -int(ln(sinx)dx, -->/f(x)=f(a+b-x)/, int (ln(sinx)+ln(cosx))dx=--2I, -2I =int ln(cosxsinx)dx, --> Int ln((1/2)sin(2x))dx, int ln(1/2)dx + int ln(sin2x) dx = -I + -I, int -ln(2)dx = -xln2, -2I = -xln2 -I /-I=ln(sin2x)dx/ -I = -xln2 |0,pi/2) --> I = (pi/2)*ln2 = 1,08879... 2023.02.05. papa

[@BikoleLege]

exellent !