"This is one of those integrals that could waste a lot of time if you don't know what to do or [...] you haven't seen it before." Oh, so you mean an integral.
Started watching you today... This gives me much more dopamine than anything. It is such a great feeling when everything connects. Your style is perfect! I want a chalkboard myself now 😭
The changing from lnsinx to lncosx, you intuitively used the King's Property, which is literally what you used! But the way of showing graphs to effectively show how the king's propery works is amazing! Integral of f(x) from a to b= integral of f((a+b-x) from a to b!!
Great video . Thanks . I got this problem at last in a test of university in 1980 . Thanks for remembering . Integrate 1( 1+ x ³ ) I think I can solve that even today .
The primitive is not an elementary function. It consists of polylogarithms and complex numbers. Which requires too much work and is not desirable. Most times when solving definite integral it is best to use definite integral techniques rather than relying too much on finding a primitive and plugging in the limits.
An interesting related result: I = int_0^π/2 ln(sin(x)) dx Substitute x -> π/2-x: I = int_0^π/2 ln(cos(x)) dx = int_0^π/2 (ln(cos(x))+ln(i) - ln(i)) dx = iπ²/4 + int_0^π/2 ln(cos(x)/i) dx Subtracting this from the initial expression for I gives: 0 = -iπ²/4 + int_0^π/2 ln(i tan(x)) dx Or in other words: int_0^π/2 ln(i tan x)/i dx = π²/4 I tried to verify using Wolframalpha, but it wasn't able to solve the integral, although the numerical output matched.
I remember when my Calc II professor did one of these. He gave it to us as homework and then showed the solution the next day in class. When he was done some guy called out "witchcraft!" 😄
@@JdeBP Yes, but if he created his own website or something like that, the demand for it would be better and easier and the support would be greater if he marketed it in his videos.
Usually calculating the derivative of this seems easier cuz it is just Chain Rule so you do Outside: Ln(x) = 1/x and Inside: Sinx = Cosx and for Chain Rule you evaluate the derivative by what is given on the inside and multiply it with the outside like 1/sinx*cosx would be the answer if you were to calculate the derivative of that Integrand, but calculating the Integral of that seems harder than you may think, haha :)
I'd just point out there is a lot of 'heavy lifting' via the symetry of sin/cos and the particular limits of this integral. If it was from 0 to pi, or something else... this would get ugly. Still, pretty neat trick.
I don't know if this is correct but by inspection, this seems to be equal to the integral of ln(x) from 0 to 1, which by parts is x*ln(x)-x from 0 to 1, which is then equal to -1 Edit: Nevermind, i am wrong. Apparently because sin(x) is not a linear function you can't just transform the integral into ln(x) from 0 to 1
In my opinion, to set the value of the integral to I requires the precondition that it converges. If the integral diverges, it is an error to set the value of the integral to I. For example, if we let 1-1+1-1+1-1... = I, then -1+1-1-1+1-1... = -1+I. The combination of the two equations is 0+0+0+0+....=0=-1+2*I, so I=1/2. This is an error, because we calculated that the divergence was I. In order for this solution to be a correct one, I think we must first prove that the integral interval contains ln(0) but does not diverge.
This one is frustratingly close to be doable by replacing sin(x) with its exponential version (e ²ᶦˣ - e ⁻²ᶦˣ)/2i . If you write sin that way and use logarithm rules you can get to ∫ix + ∫ln(2) + ∫ln(1 - e ⁻²ᶦˣ), but that last summand unfortunately seems like a roadblock. 🤷♂️
@@doctorb9264 Yeah I don’t see integration by parts working, at least as far as finding a closed form. 🤷♂️ But if he replies with a full solution then credit to him.
Ever since your last video I've grown to love "Never stop learning: the train is coming".
"This is one of those integrals that could waste a lot of time if you don't know what to do or [...] you haven't seen it before."
Oh, so you mean an integral.
The chalk break was ... HISTORIC.
Truly
Started watching you today...
This gives me much more dopamine than anything. It is such a great feeling when everything connects.
Your style is perfect! I want a chalkboard myself now 😭
I was wondering where you were headed with this! I thought you were going around in circles! But there was a surprise ending!
Hello Mr newton. I had my exam going on so I couldn't watch your videos. Now I'm gonna binge them all!
Never stop learning those who stopped learning have stopped living.
Yes
First time watching you and already witnessed a historic channel event...
Woww.... always wanted to see integrals of trigonometric function with log... thanks for the video 😊
Excellent video. One of your best that I have seen. "A for the day".
The changing from lnsinx to lncosx, you intuitively used the King's Property, which is literally what you used! But the way of showing graphs to effectively show how the king's propery works is amazing! Integral of f(x) from a to b= integral of f((a+b-x) from a to b!!
Great video . Thanks . I got this problem at last in a test of university in 1980 . Thanks for remembering . Integrate 1( 1+ x ³ ) I think I can solve that even today .
time=3:08 Congrats ! Finally
you broke a piece of chuck !!!!
👏👏👏
🤣
Very clever. There sure are a lot of calculus tricks. You explain it well.
Lovely example of alternate strategies
That's not a math trick you used. It's math magic! Very clever approach!
What if you just calculate the primitive using the method of integration by parts first and then plug in the values?
The primitive is not an elementary function. It consists of polylogarithms and complex numbers. Which requires too much work and is not desirable.
Most times when solving definite integral it is best to use definite integral techniques rather than relying too much on finding a primitive and plugging in the limits.
That was a very good video, thank you! Very insightful
An interesting related result:
I = int_0^π/2 ln(sin(x)) dx
Substitute x -> π/2-x:
I = int_0^π/2 ln(cos(x)) dx
= int_0^π/2 (ln(cos(x))+ln(i) - ln(i)) dx
= iπ²/4 + int_0^π/2 ln(cos(x)/i) dx
Subtracting this from the initial expression for I gives:
0 = -iπ²/4 + int_0^π/2 ln(i tan(x)) dx
Or in other words:
int_0^π/2 ln(i tan x)/i dx = π²/4
I tried to verify using Wolframalpha, but it wasn't able to solve the integral, although the numerical output matched.
I remember when my Calc II professor did one of these. He gave it to us as homework and then showed the solution the next day in class. When he was done some guy called out "witchcraft!" 😄
Marvelous 🎉
Doesn't the final result suggest a negative value for the area under the curve?
Yes it does. Which is ok
Did you make your own merch? If not, you should create a store for us to buy your hoodies and support you.
There was an offer on Patreon a year and a bit ago, I believe.
@@JdeBP Yes, but if he created his own website or something like that, the demand for it would be better and easier and the support would be greater if he marketed it in his videos.
Integrate[Ln[Sin[x]],{x,0,π/2}]=-Log[e^2,2^π] Input
integral_0^(π/2) log(sin(x)) dx = -log(e^2, 2^π)
Result
True It’s in my head.
Good job and I m gonna try this by myself to fixe this definitely
Usually calculating the derivative of this seems easier cuz it is just Chain Rule so you do Outside: Ln(x) = 1/x and Inside: Sinx = Cosx and for Chain Rule you evaluate the derivative by what is given on the inside and multiply it with the outside like 1/sinx*cosx would be the answer if you were to calculate the derivative of that Integrand, but calculating the Integral of that seems harder than you may think, haha :)
This is amazing
Gran video, gracias!
06:19 Who is going to send Newton a pack of coloured chalks? (-:
🤣🤣🤣🤣
Nahh.. the way this White chalk caress this Black board is my favorite sound !
I'd just point out there is a lot of 'heavy lifting' via the symetry of sin/cos and the particular limits of this integral. If it was from 0 to pi, or something else... this would get ugly. Still, pretty neat trick.
hi, is that university stuff? or highschool? or nobel price? i dont understand a thing , find it just fascinating...
university
(π/2)Ln(1/2)=-Log[e^2,2^π] Input
π/2 log(1/2) = -log(e^2, 2^π)
Result
True
True
I do not understand the concept of graph often. 😢
an interesting result indeed. however, for me, the real question is that how can we integrate ln(sinx) indefinitely.
I don't know if this is correct but by inspection, this seems to be equal to the integral of ln(x) from 0 to 1, which by parts is x*ln(x)-x from 0 to 1, which is then equal to -1
Edit: Nevermind, i am wrong. Apparently because sin(x) is not a linear function you can't just transform the integral into ln(x) from 0 to 1
Thanks Sir
🗣🔥🔥🔥
very enjoyable
Wah. That’s a trip.
Awesome
good solution
Ans is π/2 ln(0.5)
In my opinion, to set the value of the integral to I requires the precondition that it converges. If the integral diverges, it is an error to set the value of the integral to I.
For example, if we let 1-1+1-1+1-1... = I, then -1+1-1-1+1-1... = -1+I. The combination of the two equations is 0+0+0+0+....=0=-1+2*I, so I=1/2. This is an error, because we calculated that the divergence was I.
In order for this solution to be a correct one, I think we must first prove that the integral interval contains ln(0) but does not diverge.
Coloca a prova aqui para nós vermos.
Can you do it for us? We're interested, at least I am, to learn.
I love math
Nice
This one is frustratingly close to be doable by replacing sin(x) with its exponential version (e ²ᶦˣ - e ⁻²ᶦˣ)/2i . If you write sin that way and use logarithm rules you can get to ∫ix + ∫ln(2) + ∫ln(1 - e ⁻²ᶦˣ), but that last summand unfortunately seems like a roadblock. 🤷♂️
use the substitution u = sinx then integration by parts gives the same answer
Substituting u=sinx doesn’t work cleanly here because the expression is missing cosx outside the ln. (I’m assuming that’s why he didn’t try it.)
@@Bodyknock nonsense, putting u = sinx gives cosx = SQRT( 1 - u^2) etc...
@Ignoranceisbliss-i2e Keep going with the etc. … Also don’t you mean dx = 1/√(1-u²) du ?
u = sinx
x = arcsin u
dx = 1/√(1-u²) du
@@Bodyknock I don't think this is working ?
@@doctorb9264 Yeah I don’t see integration by parts working, at least as far as finding a closed form. 🤷♂️ But if he replies with a full solution then credit to him.
wow wow wow
I have a doubt. On differentiating the obtained answer we wont get back to ln(Sinx). So may I know the reason in this difference?
The answer is just a number, not a function. A definite integral was evaluated between 0 and pi/2.
@@NecrozeneYou are correct. This solution depends on it being a definite integral.
-πln√2
I remember solving this in class 12 in India for boards.
In india it is highschool stuff this question is in maths ncert😅