The return of big integrals.

[@jimschneider799]

I tackled this by substituting sin(t) into the Taylor series for atan(x) about x = 0, then dividing the result through by sin(t), and doing term-by-term integration on the series. This gave me the same series as you derived at the end. Of course, then I got stuck because I didn't recognize the power series for asinh(x) ...

[@Swybryd-Nation]

Yes this technique should be called Euler’s Technique. he used it to evaluate integral of sin(ln(t))/ln(t)*dt from 0 to 1. This gives pi/4 see Dunham’s Calculus Gallery book for details. You can use this for Achmed’s integral as well.

[@Reboxy1]

Oh man the same thing happend to me I was about to tipe that

[@0xTJ]

Thanks for the re-upload!

[@englematics]

I propose a solution using Feynman's technique which I think is simpler. Let f(x) = int_0^pi/2 (arctan(x*sint))/sint dt. Then h'(x) = int_0^pi/2 1/(1+x^2*sin^2 t) dt. Multiply top and bottom by sec^2 t. Then substitute u = sqrt(1+x^2)*tan t. As a result h'(x) = pi/2 * 1/sqrt(1+x^2). Then h(x) = pi/2 * ln|x+sqrt(1+x^2)|+C. Since h(0) = 0, C = 0. Then h(1) = pi/2 * ln(1+sqrt2) which is the same answer!

[@heungseoblee5980]

I have doubts about mathematical validity of Feynman's technique - Maybe I have no talent about mathematics :(. NEVERTHELESS, this way is brilliant!

[@xizar0rg]

Given the quantity and variety of videos done, I feel like enumerating them in the title (similar to how lockpickinglawyer does) would be valuable for reference. This is especially true given that many of the videos have frivolous titles, such as this one ("the return of big intervals").

[@gastonsolaril.237]

Nice one, M! Though aren't you missing the "2n + 1" denominator in the leftmost integral term that you are writing at 5:38?

[@SuperSilver316]

I mistakenly read this video as arcsin(sin(t)) and was wondering why you didn’t just do t/sin(t), and now I see the error of my ways. That’s still a good problem to have tackled before I watched this though!

[@noahtaul]

22:57

[@agamanbanerjee9048]

16:21 Shouldn't that be dx then dy?

[@holyshit922]

(-1)^n*(2n-1)!!/(2n)!! simplifies to -1/2 choose n and from binomial expansion we have \sum_{n=0}^{\infty}{{-1/2 \choose n }\cdot x^{2n}} = \frac{1}{\sqrt{1+x^2}} If we integrate both sides we will get arsinh(1)

[@FreshBeatles]

and thats a good place to start

[@SuperSilver316]

One other idea I had you can integrate by parts and make some substitutions and the pictured integral reduces into int(arccosh(x)/(x^2+1)) from [1,inf) Which at least from a cursory check through wolfram gives the same answer. Curious to see if anyone has any ideas on how to tackle it? I think contour integral should work theoretically, Collects residues at i and -i, and you are integrating along the principle branch of the hyperbolic inverse cosine, so I think it shouldn’t be too hard?

[@jorex6816]

If I'm not mistaken the derivative of ln(1+√(1−y²)) is actually y/(y²−√(1−y²)−1). At least WolframAlpha says so and I got the same result. Or is there a possible factorization I'm missing?

[@jorex6816]

Yeah okay, after quiet a lot of manipulations I got the same result as the one stated in the video

[@bjornfeuerbacher5514]

@@jorex6816 You don't need much manipulation. The derivative is 1/(1+√(1−y²)) times (-y/√(1−y²)). Multiplying the numerator and denominator of the first fraction with (1-√(1−y²)) gives (1-√(1−y²))/y² times (-y/√(1−y²)). And then you are almost done.

[@jorex6816]

@@bjornfeuerbacher5514 Stimmt! I multiplied all the inner derivatives together and only then rationalized the denominator which lead to more steps than necessary.

[@a_llama]

Great problem! Also, the audio's somewhat tinny

[@thomashoffmann8857]

The fraction in front of the integral temporarily got invisible. But luckily it reappeared around 7:20 🤗

[@tgx3529]

Peace for all people!! arctg(sinx)=sum[(-1)^n *(sinx)^(2n+1)/(2n+1)] n=0;....., We have then integral suma(-1)^n (sinx)^(2n)/(2n+1) =suma (-1)^n/(2n+1) integral(sinx)^(2n)dx , we can use finally Wallis formula for the integral, there Is also uniform convergention of series by Dirichlet test.

[@carstenmeyer7786]

As an alternative to _Feynman's Trick,_ you can also rewrite the given integral as a double integral: *I := ∫_0^{ 𝜋/2 } arctan( sin(t) ) / sin(t) dt* *=: ∫_0^{ 𝜋/2 } ∫_0^1 f(r, t) dr dt | f(r, t) := 1 / ( 1 + r^2 * sin^2(t) )* The integrand *f* is absolutely integrable and the integration bounds are constant, so we may change the order of integration via _Fubini's Theorem._ The substitution *s := tan(t)* yields a much nicer double integral: *I = ∫_0^1 ∫_0^∞ 1 / (1 + s^2 + r^2s^2) ds dr* *= ∫_0^1 [ arctan( s√(1 + r^2) ) / √(1 + r^2) ]_0^∞ dr* The upper bound yields *𝜋/2 / √(1+r^2)* while the lower bound vanishes. We are left with *I = 𝜋/2 * ∫_0^1 1 / √(1 + r^2) dr = 𝜋/2 * [ arsinh(r) ]_0^1 = 𝜋/2 * arsinh(1)*

[@dattierarbre9294]

Fantastic

[@richardheiville937]

Anyway, if you want to use series expansion, what about Wallis' integrals?

[@jorgepresto8823]

My solution to the problem is the following: In te target integral, change variable x=sin(t); in the next integral develop in power series Arctan(x) ( it can be done because x is in [0,1]). In the resulting integral change variable y=(1-x^2)^0.5 and finally change variable y=sin(w). The final integral is integral(cos^(2n)(w) dw) which drive you to the solution.

[@DWestheim]

This escalated quickly.

[@MultiWiskid]

Impressive!

[@faresberarma3349]

can be done easily within one line

[@ChefSalad]

I think your mic's gain was set a bit too hot for this recording.

[@threstytorres4306]

I hope you edit it, if not i'm bad for you

[@SuperSilver316]

I figured it out!!! You can integrate arccosh(x)/(x^2+1) through contour integration!! Look at the integral of (arccosh(x))^2/(x^2+1). Take the principle branch where the function is analytic everywhere except for (-inf,1], and use a keyhole contour with an indented circle around z = 1. The integral around the large semi circle vanishes and so does the indented circle around z =1, so all you have are the contributions from the residues, and you can collect the integral you want from the negative real axis, but you have to be very careful about how you define the arccosh(x) about the negative real axis, specifically in the regions (-inf,-1] and (-1,1). But it can be done!!

[@holyshit922]

I calculated it with double integral arctan(sin(x))=Int(sin(x)/(1+y^2sin(x)),y=0..1)

[@StarsManny]

Audio works now 👍

[@mihaipuiu6231]

Beautiful,....but ..hard demonstration. The first part was easier finally, I understood all. The formula with arcsinh(1) I have to look at it separately.For me math. is a hobby.

[@richardheiville937]

Consider F(a)=\int_0^{\pi/2} \frac{\arctan(a\sin x}{sin x}dx ,compute F'(a) and it's easy to compute F(1).

[@bjornfeuerbacher5514]

I wouldn't say it's "easy" to compute F(1) - but at least it's _way_ easier than what Michael did do here. :D

[@richardheiville937]

@@bjornfeuerbacher5514 1/sqrt(1+x^2) has an antiderivative that is well-known. The harder part is to compute \int_0^{\frac{\pi}{2}} \frac{1}{1+a^2\sin^2 x}dx The change of variable u=\cot x is probably a good one here.

[@bjornfeuerbacher5514]

@@richardheiville937 I did it this way: \frac{1}{1 + a² sin²x} = \frac{1}{cos²x + sin²x + a² sin²x} = \frac{1}{1+ (a²+1) tan²x} \frac{1}{cos²x}, and then use u = \sqrt{a²+1} tan x.

[@richardheiville937]

@@bjornfeuerbacher5514 Your change of variable is weird. It will introduce the variable a in the bounds. IMHO, u=cot x is better.

[@bjornfeuerbacher5514]

@@richardheiville937 No, the bounds will simply become 0 and infinity.

[@StarsManny]

Audio is broken ☹️

[@0xTJ]

It works fine for me.

[@federicopagano6590]

Its just insane lol

[@franksaved3893]

When the professor says there will be only one question in the calculus 2 exam. The question:

[@ChefSalad]

The inverse function of sinh(x) is not called the hyperbolic arcsine of x nor is it written arcsinh(x). The actual inverse function is called the "hyperbolic area sine" of x and is written arsinh(x). Thought I'd let you know.

[@nasim09021975]

I waited 23 minutes to hear the coveted “…and that’s a good place to stop” ☺️😃

[@reggaetyro9659]

Sin sinner peculiar morals. Etymology. Science. Scientific law. Law.

[@hassanalihusseini1717]

OK, nice video, but that was a mouthful really! 🙂

[@theoliveinyourmartini283]

There's no audio

[@0xTJ]

Try closing and re-opening, there is audio.

[@johns.8246]

Horrible. Can't you just use a Taylor series expansion and then integrate?