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Seegee Aye
Добавлен 21 сен 2021
the correction of one method used in the last video
the correction for a mistake in the last video ruclips.net/video/nGHgEBMeid8/видео.html
Просмотров: 40
Видео
interesting integral done by interesting method
Просмотров 2905 дней назад
an interesting integral done by an interesting method. there is a mistake. Pls go see the correction ruclips.net/video/JXDjhzq9vAI/видео.html. Learning the mistake together with correction just makes the example better!
an interesring integral with an interesting method
Просмотров 3478 дней назад
an integral done by 3 methods one of which is very interesting
interestig integral equation
Просмотров 53725 дней назад
an equation containing the function and its definite integral
to find the sum of an infinite series by using differential equation
Просмотров 40026 дней назад
a wonderful application of differential equation
a limit involving a definite integral
Просмотров 277Месяц назад
a nice way to find the limit involving a definite integral
an interesring integral
Просмотров 99Месяц назад
a good example to improving basic techniques of integration
an interesring integral
Просмотров 295Месяц назад
a nice integral to improve basic technieque of integration
Nice. You can use de l'Hospital's rule a second time for the first method: lim x->0 (tanx/2x) = lim x->0 (sec²x/2) = lim x->0 (1/2.cos²x) = 1/2.1 = 1/2
Thanks to this correction video I was able to learn a new integration technique using the Tangent substitution method. Phenomenal work! may god bless you sir.
Can you pls explain why the derivative on the second integral (on the 1st method), is secant squared of theta minus two? How is it equal to secant theta?
The derivative of (sqr of secθ - 2) = 2 secϑ dθ which is the original second part of the integral after multiply 1/2
@@seegeeayeI appreciate the clarification, but I am still a little bit confused, because when I take the derivative of (Sqrt of sec(θ)-2), and multiply it by 1/2, I get sec^2(θ)*tan(θ)dθ, not sec(θ) dθ.
@@rogeliovargas6359 You are right, I’ll post the correction a little later. Thank you so much❤
ruclips.net/video/JXDjhzq9vAI/видео.html
Very cool way to solve a tricky ahh integral
Favorite method : the last one
from Morocco....thank you very much ...for this clear and nice proof
Fantastic
Και όμως αυτά τα ολοκληρώματα λύονται έξυπνα με μια όμορφη αντικατάσταση!!! Θέσε x= 1+2sin^2(θ) dx= 4sinθ.cosθ.dθ .......................
Μπράβο, πολύ έξυπνες λύσεις!! Η 3η μέθοδος η πιο ωραία!!!
i love u for this video as a math lover this really interested me
path ahead : constant-coefficient second-order equation (in general, if the homogeneous equation does not have constant coefficients ,the solutions y(1), y(2) cannot be written as elementary functions)
nice
It was fantastic
try to put the limit you want calcule in the tilte like : lim x->0 (sec(x))^1/x^2
Nice
Very neat, thank you!
nice
How about taylor expansion method ?
the Taylor expansions of exponential and sine are different series, to put them together matching the same “n’ will be very complicated
Hii from Perú 🇵🇪
Dutch is a such But
Keep going sir , you are the greatest in the whole youtube
Very good method 🌹
Really helpfull
Nice solution
thanks sir please don't stop posting those integrals
Wonderful explination , thanks sir keep going
Well explained
thanks sir
good explanation sir
thank you for your efforts sir , never stop posting those integrals please
thanks sir , that was a really wondrful video , we need more integrals like this
Is there any way to solve this without the Taylor series ?
it might be impossible
It's a nice integral. Sound is awful. To continue, you can compute integral of log(tan(x))/(1+tan(x)^3),x=0,Pi/2 (there is a recent vidéo on YT about this integral)
you hack the game
Really amazing
Thank you Sir !!!🙏
I don't think your answer is correct. You made at least one sign error.
I checked by taking the answer's derivative, it's correct. But I do make errors sometimes, thank you for pointing out if any.
Nice
Thank you so much! This helped a lot
Thank you sir! It was very useful, very clear explanation!
Thank you !
very nice question
It's wrong : You can add COS(x/(2pi)) Or all 1-périodic function and it would satisfy the problem.
P is a polynomial, no trig function in it
@@seegeeaye Ok, but it's not said in the title and not in the video... The other problem is that it's not a total proof because you never prouved that there are no othe polynomial that satified the condition. The error is that your deduction is that P(x) = a*X*(X-1) +C And no... it would be better to say P(x) = X*(X+1)*Q(X) +C, where Q(X) is a polynomial. Then prouve that Q(X) is a constant (same as your alpha).
@@user-mj2ng3ry3e P stands for polynomial。 it’s a solution,not saying unique
@@seegeeaye Solving the problem is presenting all the solutions with valid proof. You presented all solutions but not in a valid proof.
I solved similar question like this their is an infinte series in that question
Let J = ∫ⁿ₀arctan(cosx)dx.= ∫ⁿ₀arctan(cos(π - x)dx. = -∫ⁿ₀arctan(cosx)dx = -J ∴ J = 0 and I = π²/4.
Well done! Nice demo, you might explain why you decided to take the complex way. Is it because you found out from the beginning that we are looking for 5 solutions and the real one only is obvious? Some others take the algebric path which is more painfull and more subject to luck in finding the way out. Thanks from France.
We know from beginning the equation has 5 roots.one real and four complex, using complex should be better. Merci
not bad!
Sir what is your profession ?
math, music
Sir are you alive ?
Sir are you fine ?