Reddit Integral 2

[@hOREP245]

yay! This is actually the same method that I used in the original file, but I left out the steps to solving the integral at that point since I felt it would go on for too long and is a semi-trivial integral. Thanks for being the reader in the imaginary "this is left as an exercise to the reader".

[@Rundas69420]

If you made the same noises on the plane as you did at 4:57, the person sitting next to you had every reason to think that you are somewhat crazy :D.

[@drpeyam]

Hahahahaha

[@azmath2059]

Amazing integral. It was a pleasure to watch as you navigated through all the steps to a such an elegant solution.

[@6612770]

I got really all choked-up watching you do this one. So had to go to the medicine cabinet and have some COTH syrup...

[@dgrandlapinblanc]

Alleluia ! Thanks. Good week-end Dr Peyam's.

[@MurshidIslam]

4:55 Howard Anton's calculus book says that "csch" is pronounced "coseech".

[@GreenMeansGOF]

Does partial fractions work for 1/(u^2-t^2)? Also, the +C from the arctanh derivation, is it 0?

[@weerman44]

Yes! I managed to solve it using PFD

[@Whateverbro24]

14:29 why didnt you integrate the C? please reply

[@faresberarma3349]

Hello Dr Peyam, I found an easy way to do this integral using the same method but by introducing the parameter t only in the numerator I(t)=integral from 0 to inf ln((exp(x)+t)/(exp(x)-1))dx, I=I(1) and I(-1)=0 dI/dt=intgral from 0 to inf dx/(exp(x)+t) after substitution and simplification we obtain dI/dt=ln(1+t)/t dI=(ln(1+t))/t dt we integrate both sides from -1 to 1 Finally we obtain the same result pi^2/4 but more quickly Best Regards

[@drpeyam]

That’s the way I did it in another video

[@faresberarma3349]

@@drpeyam thank you for reply, can you send me the link of this vidéo ? Thanks a lot

[@Uni-Coder]

Please do something about probability theory. Mathematical statistics. Data science and machine learning. Artificial Intelligence.

[@drpeyam]

I literally know nothing about those subjects

[@Uni-Coder]

@@drpeyam Even the simple addition of two continuous random variables is convolution, what comes down to integrating. I know you love integrals :) It would be interesting how we can add, multiply and so on random variables. Sometimes we get gamma functions and very hard integrals :)

[@Gold161803]

You should make a full series of "Integrals, the Swag Way"

[@drpeyam]

There’s an Integral playlist with lots of swag integrals, haha

[@brianlamptey4823]

2:17 Shouldn't you multiply the derivatives by e^x?

[@TimesOfSilence]

No, because e^x is just a constant. Be careful, you derive with respect to t here, and not to x ;) And the derivative of t is simply 1, so you can leave it out :)

[@brianlamptey4823]

@@TimesOfSilence Oh I thought t was the constant. Just realized the function was with respect to t.

[@weerman44]

Yeaahhh managed to solve this one on my own! :D

[@terminate5888]

At 2:07 you forgot to use the chain rule: ln(e^x+t)- ln(e^x-t) dy/dx = e^x/(e^x+t) -e^x/(e^x-t) so your integral should be: integral (e^x(1/(e^x+t) -1/(e^x-t))) however I do not understand why you did this as this will just give you back the function that you wanted to integrate. couldn't of you instead set up this integral: integral (ln(e^x+1) -ln(e^x-1))dx and knowing that the integral of lnx=xlnx -x +c then the function can be obtained: (e^x+t)ln(e^x+t) -(e^x+t)- (e^x-t)ln(e^x-t) -(e^x-t) +C (e^x+t)ln(e^x+t)-(e^x-t)ln(e^x-t) -2e^x +C I think I may of misunderstood your method, can you elaborate? thanks.

[@2070user]

You don't have to multiply by the derivative of (eˣ+t) and etc because it is deriving with respect to t, not with respect with x.

[@dp121273]

It's funny why the 'h' for 'hyperbolic' has found its way to the end of the (abbreviated) function. It would be more normal to write hsin, hcos, htan, hcsc, hsec, hcot and from that maybe to pronounce it hysine, hycos(ine), hytan(gens), hycosec(ans), hysec(ans), hycot(an(gens)).

[@drpeyam]

Cosh sounds cooler though, hahaha

[@dp121273]

@@drpeyam: It's totally nasty too, urban language wise, haha :D

[@dp121273]

@@no-one-in-particular: Good explanation .. Thanks!

[@azmath2059]

my old maths professor pronounced sinh as "shine", cosh as "cosh", tanh as "than", sech as "sheck", cosech as "cosheck", and coth as "coth"

[@bandamkaromi]

brilliant solution.😃😃

[@D-Bar]

That voice crack in the 3 2 1 intro though 😅

[@chengchen9032]

you are wearing the same shirt that you wore in yesterday's math 121A class!

[@drpeyam]

Yep :)

[@leonardromano1491]

Is this a female saying 'ge' at 1:40?

[@jameswilson8270]

I enjoyed it :)

[@newtonnewtonnewton1587]

Nice video D peyam thanks a lot السلام عليكم

[@gamedepths4792]

Now find a facebook or instagram integral!

[@drpeyam]

Hahahaha

[@blackpenredpen]

Hahahha

[@skeletonrowdie1768]

myspace integral, it converges to 0 if you get what i mean

[@drpeyam]

Skeleton Rowdie Burnnnnnn 😂😂😂

[@BRORIGIN]

The CSCH republic

[@drpeyam]

Hahahahaha

[@bandamkaromi]

hahaha. I don't know what "csch^2 x" and "coth^-1 x" are pronounced. lol. :D

[@shandyverdyo7688]

Ok

[@adityavij667]

I request you to try some jee adv questions if possible 😁

[@pbj4184]

Couldn't you just u-sub u=t*sec(alpha)? You don't have to use hyperbolic trig sub then...... unless you're Dr. Peyam 😎😛

[@ricardoalexreyvenegas6454]

Doctor con mucho respeto y cariño sus videos son extraordinarios, pero la verdad que no se ve muy bien. Le aconsejo mejorar la visión de nosotros para así poder apreciar de forma más clara sus desarrollos. No se ve nada de bien. Saludos cordiales.💪💪

[@drpeyam]

Yo sé

[@ricardoalexreyvenegas6454]

@@drpeyam sería genial ver sus videos desde una óptica distinta. Gracias y saludos.