which method do you prefer??
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- เผยแพร่เมื่อ 28 ก.ย. 2024
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The second way looked weird at first but in fact it's the faster route to the result.
Very nice example. I prefer the Weierstrass substitution. Thanks a lot for your clear presentation - great👌
Definitely like Weierstrass in this particular case, and save the series for stuff that really needs it.
15:19 y need to be less then 1 to the sum will final and not infinite
One note: Weierstrass had absolutely nothing to do with this substitution. His name got attached to it based on one paper in the 60s saying so, and then Stewart repeated the claim in his popular calculus textbook. The substitution was used by Euler a half century before Weierstrass's birth.
We keep getting y^2-1 in the denominators. What happens if y^2-1=0. Do those cases converge?
Let's try it out. y²-1 = 0 implies y = +-1. For y = 1, we have the integral over 1/(2 - 2 cos(x)) from 0 to pi/2. Which is the same as 1/4 times the integral over 1/sin²(x/2) from 0 to pi/2. Which gives -1/2 times cot(x/2), evaluated from 0 to pi/2. Since cot diverges for x = 0, this integral does not exist. In the other case, y = -1, we get 1/2 times tan(x/2), evaluated from 0 to pi/2, which yields a finite result.
Alternatively, use the result at 15:50 directly. For y going to +1, this diverges; for y going to -1, it yields a finite value (you have to use l'Hopital's rule for that).
Don’t u need |y| |-t^2|
Using de Moivre theorem for substitution looks like cheating. t-substitution like using forgotten magic. I just discovered t-subtitution when my private student asked it last year. Sadly my uni didnt teach us about it, thats very useful
I’m a fan of the weierstrauss sub, it’s easy to see what the trig functions turn into and I’d rather work with the polynomials
The second method is how I would usually approach this integral, while the first one is more of a lecture to illustrate those integration techniques needed for more difficult integrals.
This integrand is used as a "base" for many integrals E.G Malmsten's and similar ones. The general procedure to solve them is series expansions of the denominator. So, I prefer the first which mimics it closely.
If you want a suggestion for a video, something I was playing around with was the exponential of the derivative. Something interesting I noticed was that any function I applied it to followed the equation e^(d/dx)f(x)=f(x+1). I made a loose proof of it using power series (it involves manipulation of the power series and expansions of binomials, switching order of summation, and other fun algebraic manipulations) but are there any circumstances where it doesn't work? One thing I noticed is that it doesn't play nice for functions with asymptotes, lnx, 1/x, they do converge for some values but don't for others, although when it does converge it does converge to the equation mentioned above. If this already exists, I just haven't seen it yet so sorry if that's the case.
It has been known for at least 100 years, I'd estimate. It's heavily used in quantum mechanics, where the momentum operator "creates" translations of the wave function.
And if I remember correctly, Michael already made at least one video about that.
i think there's a video of this on the channel
20:31
I have almost a gut instinct now to use and prefer Euler whenever I think it can help.
I never even remember trig identities. That's how useful it is. So it's in my toolbelt as my multitool for pretty much any trig problem that is sufficiently complicated.
wolfram|alpha gives same ans. but omits 🥧/2 .... very strange
Your students are very lucky people.
A beautiful journey thru integration techniques Rock Star ⭐
It seems to me that mathematically it would be stricter to first indicate that for y = 1 the integral diverges, for y = -1 it is 1/2. And then perform the calculations assuming that y≠ ± 1.
Aha 2nd comment:
Let denominator in integral be considered as
A²+2AX+1 then by looking at squares we have (A+X)² + (1-X²)
and as A≡y and X≡cos(x) we have
(y+cos(x))² + (1-cos²(x)) or identically (y+cos(x))² + sin²(x)
ps: I don't know if this helps or not
signed ∞8∞ 🙂
This is assuming y is bounded enough that we can do the geometric series trick?
Cuz doesn’t mention that heds
This question was in jee main 2024 January attempt
no doubt, I take the second method :-) Thank you, Prof 😅
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i like it 🤩🤩
不過呢
Yeah, I did second method....way quicker but the first method is richer in the sense that the material is more applicable in other examples as well. For here though, a little overkill so I place my bets on method 2
2nd method for sure
Thanks!
I think the result for this integral should be:( 2/(y^2-1) )arctan((y+1)/(y-1))
Why this and not the other one?