Man I missed that channel! 12:09 - I just solved this integral yesterday without the general trig sub!, I just used cos(x)= cos(2x/2) and used some double angle formula to turn it into something useful. However now it's way cooler with a method that I haven't used in a while! Glad to see the quality of your vids are still on the rise, keep going brother!
14:41 our base case for I(a) was for a = 0 but notice that if we plug a = 0 into this derivative function we end up with the square root of a negative which is fine if we’re accounting for complex numbers but then we have the inverse tangent of i*infinity rather than positive infinity
It's better to write arc-cosh(7) as 2*ln(2+sqrt(3)), because the final integral substitution doesn't have to be sech. It could be sec instead, and you will arrive at the natural log form, which is a more fundamental expression. In fact, this problem has a general form in that Integral_0^(pi/2){dx ln(a^2*tan^2+b^2)} = pi*ln(a+b).
Why in the world most of ur integral looks easy and then when u start solving the integral things starts to aproach in a totally different way, But must admit ur integral are so interesting and relaxing to watch.❤
I more like doing electrical engineering nowadays. Really miss those old days. Even my profile picture is still that monsteruous integral from your TH-cam channel Edit: and also my name 😄
@14:40 To evaluate integral, devided by (Alpha - 1), Alpha shall not equal to 1. Therefore integral result may not valied for case Alpha =1. So, is that evaluation constant C is valied ?? Can you explain this ?? 😢
I will check this. But also another issue raises with negative Alpha values. 'Intergration 'log mod (Alpha+ cos x)', which I(Alpha) well defined at Alpha =-1. But after Applying Fymens trick, I (-1) is not Exits/ Not real since the term involved with Arccosh (-1). So that part of the solutions vanishes. What is the reason for this ??'
@@sumith_086 I think Math505 just forgot about it since it doesn't matter in this case, but the int(dt/(a+cost)) from 0 to pi = pi/(a-1)*sqrt(a-1/a+1) = pi/sqrt(a^2-1)*sign(a-1) Integrating this is well defined for "a " less than or equal to -1, giving **cosh^-1(|a|)+C**. cosh^-1(|-1|)=0 This is because int(dt/sqrt(t^2-1)) bounds 1 to x = cosh^-1(x) and int(dt/sqrt(t^2-1)) bounds -1 to -x = -cosh^-1(x) by the substitution x = -cosh(u) This means **int(dx/sqrt(x^2-1)) = cosh^-1(|x|)*sign(x)** Compare this to the rule you are taught: d/dx(cosh^-1(x))=1/sqrt(x^2-1) where cosh^-1 is just undefined for x>>cosh^-1(|x|)*sign(x)
If coshx different of zero with I(0) give -πlog2 this can't by normally I mean so we can't try I(1) as long as I(0) exist I mean it useless to do I(1) cuz I(0) exist as -πlog2 Wich mean something went wrong all the way in the beginning or middle cuz the first integral it's the same that the result so if the original integral can't be equal to the results there's something wrong
@ maths 505 Intergration 'log mod (Alpha+ cos x)', which I(Alpha) well defined at Alpha =-1. But after Applying Fymens trick, I (-1) is not Exits/ Not real since the term involved with Arccosh (-1). So that part of the solutions vanishes. What is the reason for this ??
@@maths_505 Yes. Just I needed to Discuss intergral result when case of α ≤ -1, since above result is not valied. I gone through evaluation and get the Idea. Thanks
What's about the transformation cos²x+sin⁴x=cos²x+sin²x•(1-cos²x)= cos²x+sin²x-sin²x•cos²x= 1-1/4•sin²2x=1-1/8•(1-cos4x)= (7+cos4x)/8 It's much simpler, isn't it?
Intanto , c'è integrale di ln(secx)^4=4ln2π/2=πln4...+ ln(1-(sin2x)^2/4) se non ho sbagliato i calcoli...ho provato con feyman,ma risultano dei /0..boh
Man I missed that channel! 12:09 - I just solved this integral yesterday without the general trig sub!, I just used cos(x)= cos(2x/2) and used some double angle formula to turn it into something useful. However now it's way cooler with a method that I haven't used in a while!
Glad to see the quality of your vids are still on the rise, keep going brother!
Thanks homie.
Where ya been? How's everything?
@@maths_505On that math major Grind mah boii!! I'm starting to share your channel like cigarretes around campus
Thanks mate
Good Idea, thank you for your featured effort.
It’s only a matter of time before we see “GONE SEXUAL” in the title
14:41 our base case for I(a) was for a = 0 but notice that if we plug a = 0 into this derivative function we end up with the square root of a negative which is fine if we’re accounting for complex numbers but then we have the inverse tangent of i*infinity rather than positive infinity
It's better to write arc-cosh(7) as 2*ln(2+sqrt(3)), because the final integral substitution doesn't have to be sech. It could be sec instead, and you will arrive at the natural log form, which is a more fundamental expression. In fact, this problem has a general form in that Integral_0^(pi/2){dx ln(a^2*tan^2+b^2)} = pi*ln(a+b).
Better 2*cosh⁻¹(2)
amazing video ❤
Why in the world most of ur integral looks easy and then when u start solving the integral things starts to aproach in a totally different way, But must admit ur integral are so interesting and relaxing to watch.❤
Man where do you find such intimidating integrals
or πcosh⁻¹(2) = (π/2)cosh⁻¹(7)
cosh⁻¹(7) = 2cosh⁻¹(2)
As soon as I saw the thumbnail, I knew that arccosh(7) would feature in the answer. Obvious.
I more like doing electrical engineering nowadays. Really miss those old days. Even my profile picture is still that monsteruous integral from your TH-cam channel
Edit: and also my name 😄
@14:40 To evaluate integral, devided by (Alpha - 1), Alpha shall not equal to 1. Therefore integral result may not valied for case Alpha =1. So, is that evaluation constant C is valied ?? Can you explain this ?? 😢
While 1/sqrt(a^2-1) goes to inf. at a=1, its integral, cosh^-1(a)+C is well defined there. Cosh^-1(a) simply has an infinite slope at a=1.
I will check this. But also another issue raises with negative Alpha values.
'Intergration 'log mod (Alpha+ cos x)', which I(Alpha) well defined at Alpha =-1. But after Applying Fymens trick, I (-1) is not Exits/ Not real since the term involved with Arccosh (-1).
So that part of the solutions vanishes. What is the reason for this ??'
@@sumith_086 I think Math505 just forgot about it since it doesn't matter in this case, but the int(dt/(a+cost)) from 0 to pi = pi/(a-1)*sqrt(a-1/a+1) = pi/sqrt(a^2-1)*sign(a-1)
Integrating this is well defined for "a " less than or equal to -1, giving **cosh^-1(|a|)+C**. cosh^-1(|-1|)=0
This is because int(dt/sqrt(t^2-1)) bounds 1 to x = cosh^-1(x) and int(dt/sqrt(t^2-1)) bounds -1 to -x = -cosh^-1(x) by the substitution x = -cosh(u)
This means **int(dx/sqrt(x^2-1)) = cosh^-1(|x|)*sign(x)**
Compare this to the rule you are taught: d/dx(cosh^-1(x))=1/sqrt(x^2-1) where cosh^-1 is just undefined for x>>cosh^-1(|x|)*sign(x)
@@lih3391 Yeah. Thanks. I will gone through this.
If coshx different of zero with I(0) give -πlog2 this can't by normally I mean so we can't try I(1) as long as I(0) exist I mean it useless to do I(1) cuz I(0) exist as -πlog2 Wich mean something went wrong all the way in the beginning or middle cuz the first integral it's the same that the result so if the original integral can't be equal to the results there's something wrong
15:06 you can intergrate from[0,7]
for the value of 0 the functions dne
@ maths 505 Intergration 'log mod (Alpha+ cos x)', which I(Alpha) well defined at Alpha =-1. But after Applying Fymens trick, I (-1) is not Exits/ Not real since the term involved with Arccosh (-1).
So that part of the solutions vanishes. What is the reason for this ??
???
The solution doesn't need alpha = -1....we needed α >= 1.
@@maths_505 Yes. Just I needed to Discuss intergral result when case of α ≤ -1, since above result is not valied.
I gone through evaluation and get the Idea.
Thanks
Cool video! But, nothing went wrong here and you should change the title of the video 🙂
It was faster to substitute t=tan(x) and then consider Feynman technique on log(t^4+a(t^2+1))/(t^2+1)
cos u is symmetric about u = pi, but how do u know ln of (7 + cos 4u) is symmetric about u = pi?
What's about the transformation
cos²x+sin⁴x=cos²x+sin²x•(1-cos²x)=
cos²x+sin²x-sin²x•cos²x=
1-1/4•sin²2x=1-1/8•(1-cos4x)=
(7+cos4x)/8
It's much simpler, isn't it?
Well hindsight is always 20/20 😂
Natural logarithm (ln)
If you had worked with theorems and proof I don't think you would have get confused trust me it will easen up the work
Why does your cos look like ω3
Bro feynman is so OP they should nerf it :D
Intanto , c'è integrale di ln(secx)^4=4ln2π/2=πln4...+ ln(1-(sin2x)^2/4) se non ho sbagliato i calcoli...ho provato con feyman,ma risultano dei /0..boh
These questions are from advanced calculus
In higher degree in math or PhD in math we can learn and understand this
If you had worked with theorems and proof I don't think you would have get confused trust me it will easen up the work