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.

We can also calculate indefinite integral by substitution u = x/sqrt(1+sqrt(1+x^4)) This substitution will rationalize integrand

tedious integral, but stunning solution. Thank you.

you're one of the amazing mathematics channel which i'm studying hard with them , thank prof

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

this is ridiculously cool.

Amazing Video Sir

Amazing solution

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.

Awesome!!!

this integral looks like when the professor is on vacation and the TA gives the homework instead

Integrals are the best puzzles

Nice bro

W for using Wolfram alpha for partial fraction

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

just a normal problem they throw at you in your first year 😢

Bro, Make videos on I.M.O problems

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

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]])

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.

(Almost) impossible integral 😃💯

2.5

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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.

Do you have yor tablet already?