Thanks for the very nice integral!

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Michael Penn

Michael Penn

Күн бұрын

Пікірлер: 57
@aashishkapoor1468
@aashishkapoor1468 2 жыл бұрын
since michael asked, here's the vid where he calculates the integral of ln(sinx) from 0 to pi/2: kzbin.info/www/bejne/bajMqpWDoKqZmMU :)
@lawliet2263
@lawliet2263 2 жыл бұрын
Thanks
@cicik57
@cicik57 2 жыл бұрын
great, only this stopped me from solving it )
@CM63_France
@CM63_France 2 жыл бұрын
And that's a good place to pin. Sound recording is better now, I guess, because Michael spoke louder at that time.
@bottlecapbrony366
@bottlecapbrony366 2 жыл бұрын
Finding the old video is left as an exercise for the viewer
@goodplacetostop2973
@goodplacetostop2973 2 жыл бұрын
13:03
@barsercan7203
@barsercan7203 2 жыл бұрын
I solved it with by making it a double integral first and then changing the order of integration. I chose writing x^2 as integral of 2y from 0 to x and the region in xy plane is a simple triangle and it worked out pretty nicely. I think it's really helpful to consider this strategy if the bounds are finite and nice, because it unlocks the tools available for double and triple integrals. Thanks Michael great video
@mathflipped
@mathflipped 2 жыл бұрын
Interesting problem! Great job, Michael.
@jesusalej1
@jesusalej1 2 жыл бұрын
If you put down one of the x's, you get sinx/x in the denominator whos limit is 1. Above you get x cosx whos limit is 0
@yowaimo890
@yowaimo890 2 жыл бұрын
Ew that's a very generic method let's go with applying first principle ;-;
@giuseppemalaguti435
@giuseppemalaguti435 2 жыл бұрын
piln2+(1/12)pi^3
@levanjikiya4454
@levanjikiya4454 2 жыл бұрын
I just multiplied by 1, which is (sin^2(x)+cos^2(x)) and the whole thing was just simple integration from what we learn in Calc 2.
@aadhaarmurty1180
@aadhaarmurty1180 2 жыл бұрын
You can also proceed by integration by parts if you know your antiderivatives really well. The integral is equivalent to the integral of x^2 csc^2(x) minus pi^3/24. Just differentiate x^2 (x^2 ---> 2x --->2) and integrate csc^2(x) (csc^2(x) ---> - cotx ---> - ln(sinx) ) apply the formula, evaluate the limits and you're done.
@sagarmajumder7806
@sagarmajumder7806 2 жыл бұрын
Sir, in second limit just use 1)fundamental theory of engineering is sinx=x, then L(xlnx)=0.(2) when x tends to 0 then sinx= x. For both reasons calculation will be shorter.🙏🙏
@thsand5032
@thsand5032 2 жыл бұрын
A friend gave it to me and I just nuked it with the residue theorem (separate x²/tan²x = x²/sin² x- x², integrate x²/sin²x by parts to get 2x/tan(x), which turns into arctan(x)/x(1+x²) from 0 to infinity. Write arctan(x) = 1/2i [log(1+ix)-log(1-ix)], separate in two, sub in the 2nd to get an integral over R, residue nails it). I just love the residue thm so much
@mryip06
@mryip06 2 жыл бұрын
4:26 Should we use lim(x->0) x/sin x = 0 ?
@gregsarnecki7581
@gregsarnecki7581 2 жыл бұрын
Please tap the blackboard when you make the jump to a cleaned-up half of a blackboard - it enhances your already magician-like status as the maths YT wizard!
@panadrame3928
@panadrame3928 2 жыл бұрын
When calculating limits, why don't you use equivalents (dunno if this is the english term) ? Like sin x ~ x and then since it doesn't tends toward 1 when x -> 0, ln(sin(x)) ~ ln (x) then x ln(sin x) ~ x ln (x) -> 0 Quite faster... And smoother It kinda works for any "hard" limit so you should consider using this method
@autumn_auburn
@autumn_auburn 2 жыл бұрын
This approach isn't rigorous. You'd probably only use it in exams like SAT where no reasoning has to be provided for the answer and there is a time constraint.
@SuperSilver316
@SuperSilver316 2 жыл бұрын
Do u = tanx, then the integral becomes Int(arctan^2(u)/((u^2)*(u^2+1))) from 0 to inf You can use partial fractions to break up the integrals into int(arctan^2(u)/u^2)-int(arctan^2(u)/(u^2+1)) Both of these from 0 to inf Then you can compute these integral individually easily, one by Feynman integration, and the other by u substitution. Pretty cool integral!
@victormd1100
@victormd1100 2 жыл бұрын
How do you propose to evaluate the one by feynmann integration?
@SuperSilver316
@SuperSilver316 2 жыл бұрын
Sorry just saw this! You can integrate it parts first and that turns that first integral into int(arctan(x)/((x)(x^2+1))) from 0 to inf the uv parts will vanish due to the limits. Then define I(t) as I(t) = int(arctan(xt)/((x)(x^2+1))) From there take a derivative with respect to t, the x on the bottom gets canceled dI/dt = int(1/((1+x^2)(1+x^2t^2)) From here it’s a partial fractions for the integral, then integrate one more time in the t variable, and you should be able to arrive at the answer.
@victormd1100
@victormd1100 2 жыл бұрын
@@SuperSilver316 thank you, it was really helpful
@holyshit922
@holyshit922 2 жыл бұрын
I calculated it the same way as Michael but before watching video
@txikitofandango
@txikitofandango 2 жыл бұрын
I can't easily find the video, but with simple u substitutions (u = π/2 - x or u = π - x) you can show that, if I = integral from 0 to π/2 of ln(sin x)dx J = integral from 0 to π/2 of ln(cos x)dx K = integral from π/2 to π of ln(sin x)dx, then I = J = K. Then 2I = I + J = integral from 0 to π/2 of ln(sin x * cos x)dx = integral from 0 to π/2 of (ln(sin(2x)) - ln(2))dx With a simple u-sub u=2x this becomes 2I = (1/2)(I + K) - integral from 0 to π/2 of ln(2)dx I = - integral from 0 to π/2 of ln(2)dx = -πln(2) / 2
@8jhjhjh
@8jhjhjh 2 жыл бұрын
Aha that’s pretty neat
@dariusbic
@dariusbic 2 жыл бұрын
Nice exercise por a Calculus exam!!
@richardheiville937
@richardheiville937 2 жыл бұрын
d/dx tan(x)=1+tan(x)^2 and tan(0)=0 therefore lim x=0, tan(x)/x=1
@tgx3529
@tgx3529 2 жыл бұрын
And what about use the idea(similar) d/dx(g(x)/tgx)=(g'(x) tgx-g(x)(1+(tgx)^2))/(tgx)^2 Then g(x)/tgx=integral g'(x)/tgx dx- integral g(x)/( tgx)^2 dx- integral g(x) dx For g(x)=x^2 we can see there the original integral......
@cernejr
@cernejr 2 жыл бұрын
Approx. 0.885658
@GirishManjunathMusic
@GirishManjunathMusic 2 жыл бұрын
This video was very fun to watch, and the detective-work challenge was enjoyable to complete, thank you!
@random19911004
@random19911004 2 жыл бұрын
Just about every time there is an indeterminate form in the 'UV' part of an integration by parts, the answer turns out to be 0. There must be some reason why...
@ConManAU
@ConManAU 2 жыл бұрын
Mostly because they’re usually a polynomial going to 0 times a more complicated function going to infinity, and those usually come down to the limit being either 0 (if the polynomial wins) or ♾ (if the other function wins), and answers of infinity are often unsatisfying for these sorts of problems.
@belalahmed2826
@belalahmed2826 2 жыл бұрын
Peace upon you Integral of (tan x)/((cos x sqrt(xsin^5sqrtx)) dx Substitute t= sin sqrtx
@winky32174
@winky32174 2 жыл бұрын
Nice!
@leif1075
@leif1075 2 жыл бұрын
Whybuse natural log of sine x at all when there is no log and nonsine by itself here..I don't see why anyone would.think of it..why not natural log of tangent..
@manucitomx
@manucitomx 2 жыл бұрын
An easy to follow complicated-ish integral. Thank you, professor.
@anunrecognizedgenius3329
@anunrecognizedgenius3329 2 жыл бұрын
The hypothesis Andrew Beal, was wrong, 128^5+32^7=8^12, do you want to know more?
@qm_turtle
@qm_turtle 2 жыл бұрын
Your example doesn't violate the Beal conjecture, since 128, 32, and 8 all have a the common prime factor 2.
@anunrecognizedgenius3329
@anunrecognizedgenius3329 2 жыл бұрын
@@qm_turtle in general, it violates, another common divisor is the number 4.
@qm_turtle
@qm_turtle 2 жыл бұрын
@@anunrecognizedgenius3329 I don't know enough about the Beal conjecture, so I needed to look it up on wikipedia in the first place. There it is stated for the Beal conjecture that if A^x + B^y = C^z, with x,y,z,A,B,C non-zero integers and x,y,z >= 3, then A,B,C have a common prime factor. There is nothing stated about other common divisors. So how does your statement about the common divisor 4 come into play here?
@anunrecognizedgenius3329
@anunrecognizedgenius3329 2 жыл бұрын
@@qm_turtle Wikipedia says that in the hypothesis there is only a simple divisor, as I understand it.
@anunrecognizedgenius3329
@anunrecognizedgenius3329 2 жыл бұрын
@@qm_turtle no i sostavnym. I can prove that in the equation, the divisor can be not only prime, but also composite
@沈栋-w9h
@沈栋-w9h 2 жыл бұрын
Nice!
@MrFtriana
@MrFtriana 2 жыл бұрын
This is the link: kzbin.info/www/bejne/bajMqpWDoKqZmMU
@michelbernard9092
@michelbernard9092 2 жыл бұрын
TY !
@chmjnationalsuperarmygener8564
@chmjnationalsuperarmygener8564 2 жыл бұрын
Hi
@richardheiville937
@richardheiville937 2 жыл бұрын
antiderivative of 1/tan(x)^2 is -1/tan(x)-x and antiderivative of 1/tan(x) is log(sin(x))
@NovoBoss
@NovoBoss 2 жыл бұрын
I am definitely not first
@juliang8676
@juliang8676 2 жыл бұрын
Yeet
@replicaacliper
@replicaacliper 2 жыл бұрын
Yeat
You won't believe all the tricks for this integral!!
18:55
Michael Penn
Рет қаралды 29 М.
Thanks viewer, for this nice integral!!
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We Attempted The Impossible 😱
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Арыстанның айқасы, Тәуіржанның шайқасы!
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One of the most beautiful and powerful tools in mathematics!
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Can you solve this integral?
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Michael Penn
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A nice integral.
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A very unfriendly integral problem!
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so you want a VERY HARD math question?!
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We Attempted The Impossible 😱
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