what tricks unlock this integral??

[@unturnedd]

could have mentioned in the end that the result is equivalent to pi/sinh(pi)

[@SuperSilver316]

I’m glad I didn’t need to scroll far for this one, Hyperbolic trig always

[@CM63_France]

And that is only the value of the function. But the integral of a constant from 0 to 1 is the value of that constant multiplied by the length of the interval : 1 .

[@joydeepc1]

You deserve more subs. Although I am below my age i am learning integrals.. one of the best videos on this topic

[@jeffreywood4759]

9:20 For those wondering, use the substitution u = (1-t) for the second part of the integral, you swap the bounds of integration, but pick up a minus sign from the derivative, so the minus becomes positive and the bounds swap back to 0 to 1. Then just do the trivial substitution of u -> t, and you'll see the two parts of the integral are the same.

[@krisbrandenberger544]

Yes, or you could just use king's Rule.

[@59de44955ebd]

Footnote: the final result can also be expressed as pi * csch(pi), where csch is the hyperbolic cosecant

[@xinpingdonohoe3978]

Forget that. That's boring. It's equal to |i!|²

[@karnetik]

i see a michael penn integral, i click

[@dang-x3n0t1ct]

Why not use cosx= Re(e^ix) ?

[@xizar0rg]

Because that wouldn't show the same techniques he's demonstrating here. It's a pedagogic choice rather than a pragmatic one.

[@eiseks3410]

I agree, it was an overkill

[@danielc.martin]

Better

[@SuperSilver316]

Agree

[@ruilopes6638]

Deep Michael Penn lore on this video. Many recalls

[@TheNiplesslyNippled]

I know right?! Glad to be here 🤓

[@goodplacetostop2973]

12:55

[@jesusthroughmary]

And here I thought I was first

[@benardolivier6624]

It was a good place to st... 😉

[@IoT_]

Not necessarily, but the answer could have been written as pi/sinh(pi)

[@wolfmanjacksaid]

3:27 not a big fan of setting "u" equal to two different things

[@karnetik]

u=1/u😵

[@noxfortes]

Also noticed it. Should've used 'v', maybe.

[@lythd]

its color coded and with an arrow specifying which is for which integral. once on the inside of integral they are the same thing, an integration variable over the specified bounds. so it doesn't really matter how they related to the previous integration variable. so its not really two different things, it just relates to the previous integration variable differently in each integral, and its very clearly defined which is which

[@forcelifeforce]

@@lythd Please write in sentences.

[@lythd]

@@forcelifeforce i did

[@alipourzand6499]

At some point, I was affraid to see a complex result for the integral, but there was a happy end!

[@bassboy14110]

Michael is basically Sean Evans who can do math.

[@angelosettanni559]

i can' t understand the final result. can someone help me to understand?

[@amritlohia8240]

Which step do you need help with?

[@billycheung5114]

Still curious why he can let u be different things of x and add them up to become an integral from 0 to inf 🤔🤔🤔

[@ClaudioCabrera-d4g]

Hi Michael, Claudio from Chile

[@士-x7e]

π/Sinh π

[@BrunoMoreiraTorres]

Funny things always seem to happen when you get an integral with bounds at 0, 1 or infinity and applies u=1/x 🤔

[@ClaudioCabrera-d4g]

How can I send you a similar problem so that you have a try in the channel?

[@amritlohia8240]

If you look in the video description, right at the bottom, there's a link to "Suggest a problem".

[@celestialmath4797]

Isn't the answer 2π/(e^(-π)-e^(π))

[@fhffhff]

$(0;1)cos(lnx-ln(2-x))dx=1+|(0;1) En=1;≠(-1)^nEj=0;2n-1C(2n-1;j)((lnx) ^(2n-1-j)(-1)^j(ln(2-x))^(j+1)/(j+1)-$dx (2n-1-j)ln^(2n-2-j)xx-¹ln^(i+1)(2-x)/(j+1))/(2n-1)!*2=π/shπ

[@قاسمشبرنگ]

In the last step, you forgot to divide by Gamma (1+i+1-i)=gamma(2)=1

[@meurdesoifphilippe5405]

I love this world where there is no definition problem in 0

[@jardozouille1677]

So complicated ... wouldn't it be interesting to start by noting that ln(x/(2-x)) = 2 argth(x-1) ?