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.
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.
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.
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 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.
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
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?
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
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.
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
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.
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
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.
@@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.
@@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.
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.
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
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
Ah man you almost made me cry tears of joy there🥺
Thank you so much
Simple and insane integral! Sure the experience helps finding clear and beautiful approaches!
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.
ngl the writing tanx as cotx part is prolly the hardest part, as I never would've even thought about using ibp in this.
Whenever I see 0 to pi/2 my brain always tells to apply king's property
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)
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.
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.
Very interesting observation. Kindly rework the problem and post it for me. Am trying to learn the techniques
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)
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.
What about X/tan x = Xcos x / sin x, then we do substitution?
great approach
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.
Thanks
That's why I wanted to share it
I often look for solutions using real techniques only
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
It's like a substitution t=pi/2-x but we rename the dummy variable "t" back to "x" all in one step
@@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.
There is no missing negative sign🤦♂️🤦♂️🤦♂️....t=pi/2 - x means dt=-dx....
The extra negatives cancel out!
@@ronnykazadi352it's called King's property
So how do we use complex analysis for the ln(sin x)?
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
Taking notes...
Edit: huge fan of bprp too :)
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?
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
In the meantime I plan on uploading some content on complex integration too
Be ready for Euler’s formula Boys.
How did you get the negative sign in the end?
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.
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
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
My head hurts..
So what is the actual solution of this? Like when do we take what value??
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.
Oh there's a video proving this transformation
ua-cam.com/video/59mHWGipCFQ/v-deo.html
It's in the first 3 or 4 minutes (1st property)
@@maths_505 Awesome, thank you! I get it now. I completely forgot about that transformation
@@par22 no problem mate
It's a pretty useful one too especially for trig integrals
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
Damn!
That is a good idea
I'll keep it in mind and I'll mention you in the video too
Hey may I ask what software you are using to write?
awesome!!!
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)
There is no such thing as Feinman integration. This is the Leibniz rule for integrals with parameter.
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.
Definitely one of the GOATs of UA-cam maths
My friend you have formed a parasocial relationship with a person who does math problems on UA-cam.
@@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.
@@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.
@@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.
Explain x->ㅠ/2-x
Plz
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.
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).
@@takemyhand1988 That's just restating the method in general, not explaining why it actually works.
@@bjornfeuerbacher5514 maybe he was confused where pi/2- x came from.
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
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
exellent !