your thumbnail doesn't have ln x

@2kchallengewith4video

19 күн бұрын

Welp, someone got fired

@jordonm5675

19 күн бұрын

ngl i was disappointed when i saw the bait and switch math problem when i clicked on the video. we all got played.

@unflexian

18 күн бұрын

@@jordonm5675 nah he isn't a click bait channel like that, it's probably just an honest mistake imo

@KilamSabba

18 күн бұрын

I saw the thumbnail, didn’t play the video, solved it to an answer of zero, then watched the switched integral. Well, at least the final answer is the same. (Oops, it might be pi/2 - (-pi/2) = pi , not 0 )

@andrewhone3346

13 күн бұрын

​@@KilamSabbayes without the ln x it's pi: integrand is positive over the whole interval so can't be zero!

You never mentioned that technically this is an improper integral, as the denominator vanishes when x=4, 0. And the numerator is undefined when x=0.

Man that's a use of a really satisfying method to get a really unsatisfying result.

Euler's substitution sqrt(4x-x^2) = ux Split into two integrals 2ln(2)\int\limits_{0}^{\infty}\frac{1}{1+u^2}du - 2\int\limits_{0}^{\infty}\frac{\ln(1+u^2)}{1+u^2}du - 2\int\limits_{0}^{\infty}\frac{\ln(1+u^2)}{1+u^2}du This integral can be calculated by u = tan(theta) substitution

MEGA! Thank you for presenting this integral! 😊

Just let x = 4sin^2(u) and dx = 8(sinu)(cosu)du. This will simplify much more rapidly. ☺

@thejelambar82

19 күн бұрын

But such substitution is not "regular". Personally, I prefer to work out from known substitution like rewrite it as 4-(x-2)² then use x-2=2sin(t)

@fartsniffa8043

19 күн бұрын

so you did both of his substitutions in one step?

@anshumanmondal8317

18 күн бұрын

Please explain

@kjl3080

18 күн бұрын

I don’t think anyone could possibly see that substitution- this may be shorter but it’s probably harder

@suwapete9761

15 күн бұрын

​​@@kjl3080its fairly easy to see if the denominator is split into sqrt x and sqrt(1-x)

Can’t imagine a tough time consuming exam question like this unless it’s multiple guess😂 In real life it would take hours to hammer it out and it proves little as an exam question ( other than probing your character).

Good use of the 'magic box that erases part of the blackboard' at 5:51 and 15:04. It's what got me interested in his videos in the first place!

Wonderful editing :)

Back in 1980 I got a 5 on AP Calc AB, and a 4 on AP Calc BC, but I would not have gotten into Cambridge back in 1886. Dang! All of these steps are things I would have known how to do (though in my aged rustiness I need to be reminded), but to put this together? Really cool stuff.

You missed the fraction line at the cover picture

Nice problem professor! I just stumbled over the integrand ln(sin x) on smaller math channel, and I remembered how you solved that problem using a phase shift. I feel like this should be a field of study: the class of non trivial integrands whose integral vanish over some interval with nonzero measure.

Since when does sin(x-pi/2) ever equal sin(x)… no matter the interval

@cheesetheorange4503

6 күн бұрын

yeah have no idea where that came from

@FishSticker

6 күн бұрын

@@cheesetheorange4503 this makes me feel so weird about the whole proof

@LaurentPaulMontaron

2 күн бұрын

Came here to say this. What is he smoking?

@hydropage2855

Күн бұрын

Yeah that definitely caught me off guard. I feel like every video he makes a really noticeable mistake, but then he’s going off his notes which are correct, so he doesn’t realize he makes these huge mistakes

@FishSticker

Күн бұрын

@@hydropage2855 I don't think this happens nearly every video, but it happens like a quarter of the time

Easy freshman calc. Complete the square, substitute. Arc sine. Answer pi

Very relaxing! Thanks Michael!

10:30 sin(u - pi/2) = sin(u) Is this right? 🤔 Not -cos(x)?

@buldysk1537

19 күн бұрын

It is correct. The equation says that sine is symmetric around x = pi/2.

@nerdatmath

19 күн бұрын

Right; take u = 2pi/3, which is in the interval of u. sin(u-pi/2) = sin(pi/6) = 1/2. But sin(u) = sin(2pi/3) = sqrt(3)/2. So definitely not a valid substitution.

@yulflip

19 күн бұрын

it is wrong: at u=pi/2, sin(0) is NOT sin(pi/2)!! But, the integrals should be the same: instead of sin(u-pi/2) going from 0 to 1, it's sin(u) going from 1 to 0

@thomashoffmann8857

19 күн бұрын

​@@buldysk1537this would mean sin(pi/2 + x) = sin(pi/2 - x). Just insert x=pi/2 in his formula.

@michaelguenther7105

19 күн бұрын

You are correct and Michael is wrong. Since the integral of ln(sin x) from 0 to pi/2 is the same as the integral over ln(cos x), the third term should be a copy of the second term (with the cosine), and then the substitution u = x + pi/2 works to give the result Michael wants.

Another cool way to compute ∫ln(sinx)dx is to use Riemann sums, as pi/2n∑ln(sin(k*pi/2n) tends to the value of the integral as n tends to infinity. you then have to compute ∏sin(kpi/2n) which is classic and is equal to n/2^n. You then conclude with the uniqueness of the limit.

If you do the substitution x=2(1-cos t) then this becomes integral of the even function ln(2-2 cos t) from 0 to pi, so 1/2 of the same integral from -pi to pi. Now this is a special case of Jensen's Lemma, which says that integral of ln | a - exp(it) | is equal to ln max (1, |a| ), for an complex number a. Just set a=1 to recover the case at hand. (This result is used to define what is called the Mahler measure of a polynomial)

20:09 This is not the thumbnail 🤨

@Ricardo_S

18 күн бұрын

From what it said in the thumbnail the answer is π First √(4x-x²)=√(4-x)*√(x) u=√(4-x) du=-[1]/[2√(4-x)]dx -2du=[1/[√(4-x)]dx u²=4-x u²-4=-x 4-u²=x √(4-u²)=√(x) Change of limits u(0)=2 u(4)=0 So {0}S{4}([dx]/[√(4x-x²)])= {2}S{0}([-2du]/[√(4-u²)]) We can take the 2 out from the integral, and change the limits of integrations with the negative sign 2*{0}S{2}([du]/[(4-u²)]) Now trig substitution u=2sin(θ) du=2cos(θ)dθ θ=arcsin([u]/[2]) θ(0)=arcsin(0)=0 θ(2)=arcsin(1)=[π]/[2] u²=4sin²(θ) 4-u²=4-4sin²(θ)=4(1-sin²(θ))=4cos²(θ) So 2*{0}S{[π]/[2]}([2cos(θ)dθ]/[√(4cos²(θ))])= 2*{0}S{[π]/[2]}([2cos(θ)dθ]/[|2cos(θ)|]) Because 2cos(θ) its positive on the interval [0,π÷2] |2cos(θ)|=2cos(θ) 2*{0}S{[π]/[2]}([2cos(θ)dθ]/[2cos(θ)])= 2*{0}S{[π]/[2]}(dθ)= 2*θ{[π]/[2]\0}= 2*([π]/[2]-0)=π

@forcelifeforce

14 күн бұрын

Obi Wan Kenobi: "These are not the droids you are seeking."

Wow, I haven't seen this coming, that the area on the interval from 1 to 4 (no pi around) exactly compensates the negative area from 0 to 1!

10:55 I got lost at this step. How can sin(x) subbed to make cos(t) result in cos(x) with the same bounds?

There is a way simpler way to do the first integral: Just substitute t^2 for x. You get 2 times integral of 1/sqrt(1-t^2) from zero to one 2 times a half circle area of radius one. pi.

Magnificent!

Nice to see my old room in the thumbnail :)

This got me thinking, I always pause when I do a trig substitution, because the range of (cos and sin) trig functions is -1 to 1, so you can't always use it, but with the explicit example of scaling the domain by a change of variables to x = 4y and thus enabling the use of trig substitutions, but this only works for definite integrals, I wonder if there's ever a case where you can do something similar with indefinite integrals and you take a limit or something. To give a crude example x = y * 1/h where h -> infinity. This won't work in most cases, but you might be able to get a cancelation in the integral, and therefore you technically convert an indefinite integral to a definite one where the bounds are between -1 and 1 so you can use cos or sin substitution. I've not fully thought this through, I've only had the idea seeded by the video.

He always know the good place to stop 😮

this is simpler to solve by substituting x=t**2 and the t=cos(y) and the integral becomes trivial

does the integral from 0 to 1 of the same function have a nice value?

What are the limitations as to which types of integrals that are computable in practice? - when do we just have to give up straight away?

integral of 1/sqrt(x-x^2)dx,x=0,1 is a lame integral. You perform the change of variable u=sqrt(x) then you perform the change of variable u=sin(t).

I did it in my head. The answer’s 7.

I don't understand why you could say u and x are both just dummy variables when u is in terms of x.

I solved the thumbnail integral before watching the video 😭

What level of the Tripos did this question appear in?

So the first step is to change the problem? is that how they do things at Cambridge?

Help me out, you call it a 'box' but isn't it a square?

THANKS PROFESOR. !!!!, VERY INTERESTING !!!!

@forcelifeforce

14 күн бұрын

Stop yelling your post in all caps. Also, watch your spelling and grammar.

Seems like a ton of these types of integrals end up with a ln2 in the solution. Just an observation

20 minutes of viewing to end with a zero . 😀

W Respect for Cambridge math!

Awesome job wit facebook

Integ 0 to 2 is (-2 Catalan). The question is why? Something is going on. Maybe the world is flat we are just looking at it wrong. I doubt anyone that has not seen the solution could solve this on a test. So what were they testing?

The integral in the thumbnail is equal to pi. I have not tried to solve the one in the video yet.

so, it seems i would fail it

I was doing this in my head and read it wrong, so I completed the square with the substitution u = (x-2), and ended up with arcsin of -2 and 2, which are imaginary! But the imaginary parts cancel and you get a real number. So that was a fun mistake.

I might have known ;-)

Rewrite the part under the root as 4(1-(x-2)^2/4) and substitute (x-2)/2=tanh t

I just did a u sub to get it in a different form (for the thumbnail one)

@Ricardo_S

18 күн бұрын

From what it said in the thumbnail the answer is π First √(4x-x²)=√(4-x)*√(x) u=√(4-x) du=-[1]/[2√(4-x)]dx -2du=[1/[√(4-x)]dx u²=4-x u²-4=-x 4-u²=x √(4-u²)=√(x) Change of limits u(0)=2 u(4)=0 So {0}S{4}([dx]/[√(4x-x²)])= {2}S{0}([-2du]/[√(4-u²)]) We can take the 2 out from the integral, and change the limits of integrations with the negative sign 2*{0}S{2}([du]/[(4-u²)]) Now trig substitution u=2sin(θ) du=2cos(θ)dθ θ=arcsin([u]/[2]) θ(0)=arcsin(0)=0 θ(2)=arcsin(1)=[π]/[2] u²=4sin²(θ) 4-u²=4-4sin²(θ)=4(1-sin²(θ))=4cos²(θ) So 2*{0}S{[π]/[2]}([2cos(θ)dθ]/[√(4cos²(θ))])= 2*{0}S{[π]/[2]}([2cos(θ)dθ]/[|2cos(θ)|]) Because 2cos(θ) its positive on the interval [0,π÷2] |2cos(θ)|=2cos(θ) 2*{0}S{[π]/[2]}([2cos(θ)dθ]/[2cos(θ)])= 2*{0}S{[π]/[2]}(dθ)= 2*θ{[π]/[2]\0}= 2*([π]/[2]-0)=π

@AnakinSkywalker-zq6lm

18 күн бұрын

@@Ricardo_S my initial u was different I think. If I remember correctly I got it into difference of squares by saying: u=2+x Then du=dx And sqrt(x(4-x))= sqrt((2+u)(2-u)) =sqrt(4-u^2) And the rest was trig subs I think. You can do both subs at once but I didn’t want to type the whole trig function every time.

thanks a lot from UZBEKISTAN

In my opinion, you have to pay attention to the fact that 0 and 4 are not in the domain of the integrate.

@Hiltok

19 күн бұрын

The bounds of integration are indeed 'improper' for this case, but are simply handled by taking them to be limits approached by x.

@lrlp2007

18 күн бұрын

@@Hiltok Ok, that is true, but I am afraid that the chain of reasoning, as presented by Michel Penn, is no longer valid.

If you graph 1/sqrt(4x-x^2), it is entirely above the x axis, meaning that it cannot have an integral of 0

@5alpha23

19 күн бұрын

Well, if you consider that the thumbnail is wrong and look at the actual function ... Did you even watch the video? 😅

Mistake in the thumbnail.

@Ricardo_S

18 күн бұрын

From what it said in the thumbnail the answer is π First √(4x-x²)=√(4-x)*√(x) u=√(4-x) du=-[1]/[2√(4-x)]dx -2du=[1/[√(4-x)]dx u²=4-x u²-4=-x 4-u²=x √(4-u²)=√(x) Change of limits u(0)=2 u(4)=0 So {0}S{4}([dx]/[√(4x-x²)])= {2}S{0}([-2du]/[√(4-u²)]) We can take the 2 out from the integral, and change the limits of integrations with the negative sign 2*{0}S{2}([du]/[(4-u²)]) Now trig substitution u=2sin(θ) du=2cos(θ)dθ θ=arcsin([u]/[2]) θ(0)=arcsin(0)=0 θ(2)=arcsin(1)=[π]/[2] u²=4sin²(θ) 4-u²=4-4sin²(θ)=4(1-sin²(θ))=4cos²(θ) So 2*{0}S{[π]/[2]}([2cos(θ)dθ]/[√(4cos²(θ))])= 2*{0}S{[π]/[2]}([2cos(θ)dθ]/[|2cos(θ)|]) Because 2cos(θ) its positive on the interval [0,π÷2] |2cos(θ)|=2cos(θ) 2*{0}S{[π]/[2]}([2cos(θ)dθ]/[2cos(θ)])= 2*{0}S{[π]/[2]}(dθ)= 2*θ{[π]/[2]\0}= 2*([π]/[2]-0)=π

English is impressed with tamil ! Grandfathers and babies learn maths ?

20 sec

2th problems is the key that is KING's Property

More than half the video was a waste! Just set x-2=2cos(theta) and go from there.

Aren't these integrals improper? Shouldn't that be part of the conversation? It drives me up a wall that being improper is just ignored. I have students that gloss over this fact so many times. In Michael's integrals here, it ultimately doesn't matter. But in general it does.

@markprange2430

14 күн бұрын

Rationalise denominators if necessary or convenient.