A cool log trig inverse trig integral

[@CM63_France]

Hi, "ok, cool" : 0:17 , 1:37 , 2:17 , 3:30 , 6:24 , "terribly sorry about that" : 1:32 , 3:08 .

[@circuitcraft2399]

I always like when everything cancels and the result is really simple

[@akirakato1293]

But that does imply, in the case of constants cancelling out, that there’s a more elegant easier solution vs brute forcing with special functions. I think special functions are cool when necessary, otherwise kinda ugly.

[@SamLindskougMBGY]

Interestingly, the integrand accually (for once) has an anti-derivative. Meaning you can evaluate the integral using completely normal calc two methods

[@Sugarman96]

You can't just leave us hanging, what's the anti-derivative?

[@unturnedd]

true i just checked with wolframalpha, too big to show here though

[@francis6888]

@@Sugarman96 Use integration by parts

[@BadlyOrganisedGenius]

At 1:35 we can integrate by parts: u = ln(sin(x)) dv/dx = sin(x) du/dx = cot(x) v = 1 - cos(x) (let the integration constant be 1 so that u*v stays bounded as x->0) Then we get [(1-cos(x))*ln(sin(x))] evaluated from 0 to pi/2, minus the integral of (cos(x) - cos^2(x))/sin(x) which is easy to solve using trig identities

[@raghavendraPi]

Is it a observation based way to use constant values? Or is it a class of approach in integration. Have not seen this kind of application. Thanks for sharing

[@BadlyOrganisedGenius]

@@raghavendraPi I've found it useful in the past during integration by parts, because sometimes naive IBP gives a value for u*v that doesn't make sense. I didn't learn it in a class or anything tho, just inspired by a video by Dr Peyam I think

[@ericthegreat7805]

So what youre saying is IBP is a variation of the Beta function

[@BadlyOrganisedGenius]

@@ericthegreat7805 ?

[@quentinrenon9876]

You can also, at the very beginning, notice that tan(arcsin(x))= x/√(1-x²) Therefore the integrand is : x lnx / √(1-x²) We can now integrate by parts, and we'll derive x lnx and integrate 1/√(1-x²) Notice that the term we'll need to evaluate goes to 0, and our integral takes a negative sign and the integrand is (1 + 1/x) arcsin(x) We can separate the integral into 2 and thus the integral of arcsin between 0 and 1 is 1 which becomes -1 because of our previous integration by parts, and then we are left with an integral which is the same as the one of x/tan(x) if you do a variable change. I think this goes to -ln2 and therefore we get the same result but idk I didn't actually try to solve it

[@neg2sode]

Gorgeous derivation leading to taking the derivative of the Beta function! Thank you for your great solution. Note that, in dealing with the integral of we can look for a Taylor series expansion for and using we can write the original integral as a series containing easy Beta terms which eventually gives us a result.

[@sandem4592]

Ooh what an interesting appraoch. Personally I used the definition of tan(x) = sin(x)/cos(x) = sin(x)/sqrt(1-sin^2(x)) and simplified to ln(x)*x/sqrt(1-x^2). From there using Feynman's was my first idea, but actually a approach I haven't seen too often worked as well. Using the limit definition of ln(x) = lim h->0 (x^h-1)/h. Lead to some neat calculations.

[@RobertChan2003]

At 1:40 apply integration by parts, resulted in (-cosθ) ⋅ ln(sinθ) + ∫ (cos²θ/sinθ) dθ and use ½-angle formulae (-cosθ) ⋅ ln[ 2⋅sin½θ⋅cos½θ ] + ∫ ( 1 - sin²θ )/sinθ) dθ , next we have (-cosθ) ⋅ [ ln(2) + ln(sin(½θ)) + ln(cos(½θ)) ] + ∫ (1/sinθ) dθ - ∫ sinθ dθ , perform the integration, (-cosθ) ⋅ ln(2) - cosθ ⋅ ln(sin½θ) - cosθ ⋅ln(cos½θ) + ln(tan½θ) - (-cosθ) + C. Evaluate at interval {0,½π}, grouping the 1st & last term (-cosθ) ⋅ ln(2) + cosθ = ln(2) - 1. Rewrite ln(tan½θ) as ln(sin½θ) - ln(cos½θ), in between terms cancelled out and are zero. PS: as limit θ→0, sin½θ→½θ, cosθ →1

[@manstuckinabox3679]

Darnnn it’s been so long since I heard your voice; MY YOUTH HAS BEEN RESTORED

[@mcalkis5771]

It's been a while and I had forgotten how brilliant your solutions are. Edit: Proof by hoodie.

[@maths_505]

Thanks homie

[@AlexDrug64]

in general, you can expand ln(sinx) into a series and then tinker a little with the series 1/((2n+1)n) ln(sinx)sinxdx = 1/2 int(0, п/2) ln(sin^2(x))sinx dx = 1/2 * sum (n=1 to inf) (-1) ^n * int (0/ п/2) (-cos^2(x))^n*dcosx = -1/2*sum (n=1 to inf) 1/((2n+1)n)=ln2-1

[@joelhernanvillavicenciocac7861]

Bro, I hope you can read this comment. I'm from Peru, I've been watching your videos for a while, and now, in your latest uploads, I see that you activated a mode for other languages, including Spanish, my language. I appreciate the initiative, but I recommend that someone translate you with their voice in the language you want to work with, and not use an automatic translator, which does it in a simplistic way and does not have the emphasis and enthusiasm that you give it in your original voice. Greetings, brother and good luck.

[@raghavendraPi]

Kind of Feynman technique on Beta integral 🔥

[@threepointone415]

"Therapy sorry about that"

[@emberdied]

Thx bro, through about 1 year you teach me a lot. Now, I can finally only look and I have same way to you, I still hope some cool integrals like using contour integral or some MIT problems (its like math competition methods)😃👍

[@Anmol_Sinha]

At 2:17,(white) we could substitute theta=arcsinx back to get an expression which is easy to integrate?

[@MrWael1970]

Very cool integral. thanks.

[@xanterrx9741]

Great result i hope for more videos from you .

[@Aplicapitagoras]

Very good bro👏👏

[@holyshit922]

First simplify expression tan(arcsin(x)) Calculate integral by parts but with clever choice of integration constant u = ln(x) , dv = x/sqrt(1-x^2)dx du = 1/x dx , v = -(sqrt(1-x^2)-1) We are left with integral \int_{0}^{1}\frac{\sqrt{1-x^2} - 1}{x}dx Integrand suggest to use Euler's substitution \frac{\sqrt{1-x^2} - 1}{x} = t \sqrt{1-x^2} - 1 = xt \sqrt{1-x^2} = xt + 1 After Euler's substitution another substitution u = t^2+1 will simplify integral and finally we get ln(2) - 1

[@holyshit922]

Here choosing +1 as integrating constant is important because allows to use integration by parts directly to definite integral, moreover simplifies a little bit remaining integral I switched the interval of integration so i had wrong sign , now it is correct In my opinion you overcomplicated this integral and my solution is the easiest understandable also for beginners in calculus No special functions , basic integration techiques

[@ashyes8626]

At 3:00, how does differentiating dB/dx add a ln(sinx) to the integrand?

[@abdulllllahhh]

Another integral where the γ gets cucked and cancels out by the end of the

[@giuseppemalaguti435]

I=INT(lnx*x/√(1-x^2))...x=sinθ..I=INT(lnsinθ*sinθ)dθ(θ=0,π/2)..per parti ...I=lim(θ>0)lnsinθ+ln(cscθ+ctgθ)-1=ln2-1

[@nicolasgobert8324]

Well played

[@yurfwendforju]

I feel like this would've been a lot easier if we just simplified tan(arcsin(x)) = (1-x^2)^(-1/2)

[@bandishrupnath3721]

wonderful video sir ,btw why didnt u converted the sin inverse to tan inverse? and that theta symbol in the first of the video lmao that theta got a cancer.

[@alexchan4226]

infinity

[@ericthegreat7805]

Nooo :( Oily Macaroni :(

[@yplayergames7934]

Audio track: *portuguese Brasil* Me: No

[@ardayazar1705]

first xD