1886 Cambridge University Exam Integral 

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.

your thumbnail doesn't have ln x

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

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

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

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!

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.

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

You missed the fraction line at the cover picture

MEGA! Thank you for presenting this integral! 😊

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.

Making the substitution x=4sin^2(p), after making the preliminary factorization on the quantity under the radical, √X(4-X), reduces everything very quickly!

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

Nice to see my old room in the thumbnail :)

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.

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!

20:09 This is not the thumbnail 🤨

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?

Wonderful editing :)

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

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.

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 😮

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

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?

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

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 solved the thumbnail integral before watching the video 😭

W Respect for Cambridge math!

Very relaxing! Thanks Michael!

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?

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

so, it seems i would fail it

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

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

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).

Awesome job wit facebook

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

Magnificent!

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)

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

I did it in my head. The answer’s 7.

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

God help me if after doing a trig substitution like in that first integral my bounds of integration go outside the bounds or what sine can provide. There has to be a better way.

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

I feel like I might be tripping but he says sin(u-pi/2)=sin(u) which is just incorrect? How does this work? It's around 10:15 for reference.

20 minutes of viewing to end with a zero . 😀

thanks a lot from UZBEKISTAN

I might have known ;-)

What level of the Tripos did this question appear in?

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

Mistake in the thumbnail.

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

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

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

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.

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

20 sec

sin⁡(u+π/2)= cos ⁡u