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Great integral, bro. One ick I had about it was that it didn't evaluate to a nice value at the end and instead ended up with a bunch of radicals; however, apart from that, the integral was 🔥🔥.
@@daddy_myers this integral really did take by surprise. The journey with all those transformations was awesome but the simplification towards the end really was annoying.
We can also simplify integrand by integration by parts twice Int(sqrt(1+sqrt(1+x^4)),x) = x*sqrt(1+sqrt(1+x^4))-1/2Int(x/sqrt(1+sqrt(1+x^4))*(2x^3/sqrt(1+x^4))) Int(sqrt(1+sqrt(1+x^4)),x) = x*sqrt(1+sqrt(1+x^4))-Int(x^4/(sqrt(1+sqrt(1+x^4))*sqrt(1+x^4)),x) Int(sqrt(1+sqrt(1+x^4)),x) = x*sqrt(1+sqrt(1+x^4))-1/2*Int(2x^3/sqrt(1+x^4)*x/sqrt(1+sqrt(1+x^4)),x) d/dx x/sqrt(1+sqrt(1+x^4)) = (sqrt(1+sqrt(1+x^4)) - x*1/2*1/sqrt(1+sqrt(1+x^4))*2x^3/sqrt(1+x^4))/(1+sqrt(1+x^4)) d/dx x/sqrt(1+sqrt(1+x^4)) = ((1+sqrt(1+x^4)) - x^4/sqrt(1+x^4))/((1+sqrt(1+x^4))*sqrt(1+sqrt(1+x^4))) d/dx x/sqrt(1+sqrt(1+x^4)) = (sqrt(1+x^4)+1+x^4 - x^4)/((1+sqrt(1+x^4))*sqrt(1+sqrt(1+x^4))*sqrt(1+x^4)) d/dx x/sqrt(1+sqrt(1+x^4)) = (1+sqrt(1+x^4))/((1+sqrt(1+x^4))*sqrt(1+sqrt(1+x^4))*sqrt(1+x^4)) d/dx x/sqrt(1+sqrt(1+x^4)) = 1/(sqrt(1+sqrt(1+x^4))*sqrt(1+x^4)) Int(sqrt(1+sqrt(1+x^4)),x) = x*sqrt(1+sqrt(1+x^4))-1/2*(x*sqrt(1+x^4)/sqrt(1+sqrt(1+x^4)) - Int(sqrt(1+x^4)*1/(sqrt(1+sqrt(1+x^4))*sqrt(1+x^4)),x) Int(sqrt(1+sqrt(1+x^4)),x) = x*sqrt(1+sqrt(1+x^4))-1/2*x*sqrt(1+x^4)/sqrt(1+sqrt(1+x^4))+1/2*Int(1/sqrt(1+sqrt(1+x^4)),x) Now we can substitute u=x/sqrt(1+sqrt(1+x^4)) and it will be a liitle less calculations
If we like Euler substitutions second one will be good option sqrt(1+x^4)=x^2u^2-1 In fact this is reciprocal of substitution which i previously proposed
Get out of here... get out. that video should go down on why this channel is for the absolute mad men, and I can't be more happier to be part of it LOL! great video as always my dude.
Hi, the new layout you use for the problems makes it so I can’t view the whole thing in full screen on mobile. Idk if you know what I’m talking about, but the issue only appears on the newer videos.
I=-arctgu-arcthu+4(u+(2/5)u^5+(3/9)u^9+(4/13)u^13+(5/17)u^17...u=sqrt (sqrt2-1)...=1,4449..l'ultima parte è una serie binomiale,non ho trovato di meglio..Thanks for the integrals
This can be considerably simplified through the application of trig identities. Letting w = 2^(-1/4), we have I = (2sqrt(sqrt(2) + 1) + arccosh(sqrt(2) + 1) + arccos(sqrt(2) - 1))/4 = (sqrt(w^2 + 1)/w + arcsinh(w) + arccos(w))/2 As for the derivation, for brevity, I'll write T = tan(pi/8). The logarithm term is equivalently 2 arctanh(sqrt(tan(pi/8)). Then we can use cosh(2 arctanh(x)) = (1+x^2)/(1-x^2) and cos(2 arctan(x)) = (1-x^2)/(1+x^2), meaning those terms are equivalently arccosh((1 + T)/(1-T)) and arccos((1-T)/(1+T)). Then we can use (1 + tan(x))/(1-tan(x)) = tan(x + pi/4) to simplify to arccosh(1/T) and arccos(T). Now we can use T = sqrt(2) - 1 to simplify the fraction term to get the first form above. Applying the half-angle identities to the inverse trig terms then gets the second. Using the above method, we can also get an explicit antiderivative of the original function. It comes out to (2x sqrt(1 + sqrt(1 + x^4)) + arccosh(sqrt(1 + x^4) + x^2) + arccos(sqrt(1 + x^4) - x^2)) / 4 which gives the value above when x = 1. I leave verification that the above differentiates to the integrand as an exercise to the reader. It should also be noted that the expressions inside the inverse trig functions are tangent half-angle formulae, but I can't seem to figure out a way to further simplify.
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tan(pi/8)=sin(pi/4)/(1+cos(pi/4))=sqtr(2)-1 you could have used that
My dad always told me not to stick it in crazy, now I know exactly why he told me that.
😂😂
Great integral, bro.
One ick I had about it was that it didn't evaluate to a nice value at the end and instead ended up with a bunch of radicals; however, apart from that, the integral was 🔥🔥.
@@daddy_myers this integral really did take by surprise. The journey with all those transformations was awesome but the simplification towards the end really was annoying.
We can also calculate indefinite integral by substitution
u = x/sqrt(1+sqrt(1+x^4))
This substitution will rationalize integrand
you're one of the amazing mathematics channel which i'm studying hard with them , thank prof
tedious integral, but stunning solution. Thank you.
This channel is really phenomenal.
We can also simplify integrand by integration by parts twice
Int(sqrt(1+sqrt(1+x^4)),x) = x*sqrt(1+sqrt(1+x^4))-1/2Int(x/sqrt(1+sqrt(1+x^4))*(2x^3/sqrt(1+x^4)))
Int(sqrt(1+sqrt(1+x^4)),x) = x*sqrt(1+sqrt(1+x^4))-Int(x^4/(sqrt(1+sqrt(1+x^4))*sqrt(1+x^4)),x)
Int(sqrt(1+sqrt(1+x^4)),x) = x*sqrt(1+sqrt(1+x^4))-1/2*Int(2x^3/sqrt(1+x^4)*x/sqrt(1+sqrt(1+x^4)),x)
d/dx x/sqrt(1+sqrt(1+x^4)) = (sqrt(1+sqrt(1+x^4)) - x*1/2*1/sqrt(1+sqrt(1+x^4))*2x^3/sqrt(1+x^4))/(1+sqrt(1+x^4))
d/dx x/sqrt(1+sqrt(1+x^4)) = ((1+sqrt(1+x^4)) - x^4/sqrt(1+x^4))/((1+sqrt(1+x^4))*sqrt(1+sqrt(1+x^4)))
d/dx x/sqrt(1+sqrt(1+x^4)) = (sqrt(1+x^4)+1+x^4 - x^4)/((1+sqrt(1+x^4))*sqrt(1+sqrt(1+x^4))*sqrt(1+x^4))
d/dx x/sqrt(1+sqrt(1+x^4)) = (1+sqrt(1+x^4))/((1+sqrt(1+x^4))*sqrt(1+sqrt(1+x^4))*sqrt(1+x^4))
d/dx x/sqrt(1+sqrt(1+x^4)) = 1/(sqrt(1+sqrt(1+x^4))*sqrt(1+x^4))
Int(sqrt(1+sqrt(1+x^4)),x) = x*sqrt(1+sqrt(1+x^4))-1/2*(x*sqrt(1+x^4)/sqrt(1+sqrt(1+x^4)) - Int(sqrt(1+x^4)*1/(sqrt(1+sqrt(1+x^4))*sqrt(1+x^4)),x)
Int(sqrt(1+sqrt(1+x^4)),x) = x*sqrt(1+sqrt(1+x^4))-1/2*x*sqrt(1+x^4)/sqrt(1+sqrt(1+x^4))+1/2*Int(1/sqrt(1+sqrt(1+x^4)),x)
Now we can substitute u=x/sqrt(1+sqrt(1+x^4)) and it will be a liitle less calculations
omg u typed that all out?
If we like Euler substitutions second one will be good option
sqrt(1+x^4)=x^2u^2-1
In fact this is reciprocal of substitution which i previously proposed
Reading all that is gonna be hell painful
Amazing Video Sir
Amazing solution
this is ridiculously cool.
Awesome!!!
this integral looks like when the professor is on vacation and the TA gives the homework instead
Nice bro
hi couldnt we have used integration by parts? like we have that sec^2x term inside the integral which is nicely the derivative of tanx wrtx
W for using Wolfram alpha for partial fraction
Integrals are the best puzzles
Get out of here... get out. that video should go down on why this channel is for the absolute mad men, and I can't be more happier to be part of it LOL! great video as always my dude.
just a normal problem they throw at you in your first year 😢
Hi, the new layout you use for the problems makes it so I can’t view the whole thing in full screen on mobile. Idk if you know what I’m talking about, but the issue only appears on the newer videos.
I forgot to use the full screen mode on the notes app for this video. Sorry about that I'll fix it for tomorrow's video
Bro, Make videos on I.M.O problems
If u=Sqrt[1+x^4] then Mathematica gives:
Integrate[Sqrt[1+u],x] = (1/2)*(x*Sqrt[1+u] + ArcTan[x/Sqrt[1+u]] + ArcTanh[x/Sqrt[1+u]])
I=-arctgu-arcthu+4(u+(2/5)u^5+(3/9)u^9+(4/13)u^13+(5/17)u^17...u=sqrt (sqrt2-1)...=1,4449..l'ultima parte è una serie binomiale,non ho trovato di meglio..Thanks for the integrals
2.5
(Almost) impossible integral 😃💯
Like 👍
Do you have yor tablet already?
Yup
This can be considerably simplified through the application of trig identities. Letting w = 2^(-1/4), we have
I = (2sqrt(sqrt(2) + 1) + arccosh(sqrt(2) + 1) + arccos(sqrt(2) - 1))/4
= (sqrt(w^2 + 1)/w + arcsinh(w) + arccos(w))/2
As for the derivation, for brevity, I'll write T = tan(pi/8). The logarithm term is equivalently 2 arctanh(sqrt(tan(pi/8)). Then we can use cosh(2 arctanh(x)) = (1+x^2)/(1-x^2) and cos(2 arctan(x)) = (1-x^2)/(1+x^2), meaning those terms are equivalently arccosh((1 + T)/(1-T)) and arccos((1-T)/(1+T)). Then we can use (1 + tan(x))/(1-tan(x)) = tan(x + pi/4) to simplify to arccosh(1/T) and arccos(T). Now we can use T = sqrt(2) - 1 to simplify the fraction term to get the first form above. Applying the half-angle identities to the inverse trig terms then gets the second.
Using the above method, we can also get an explicit antiderivative of the original function. It comes out to
(2x sqrt(1 + sqrt(1 + x^4)) + arccosh(sqrt(1 + x^4) + x^2) + arccos(sqrt(1 + x^4) - x^2)) / 4
which gives the value above when x = 1. I leave verification that the above differentiates to the integrand as an exercise to the reader.
It should also be noted that the expressions inside the inverse trig functions are tangent half-angle formulae, but I can't seem to figure out a way to further simplify.