so you want a HARD integral from the Berkeley Math Tournament

The first-place team wins $1000 bprp scholarship and I will be there!

Thank you!

0:00 we haven't done a hard integral for a while
2:33 the steps of Feynman's trick of integral
5:30 differentiate I(a) first, then do the integral
18:59 integrate I'(a) with respect to a to get I(a)

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[blackpenredpen]

The first-place team wins a $1000 bprp scholarship and I will be there!

[chinesecabbagefarmer]

And then multiply everything by BMT2020

[sabrinagiang]

Hello! You probably don’t remember me but I took one of your classes back in 2016. I was struggling with math and you advised me to refer to your YouTube channel. It’s crazy how much this channel has grown. Congrats!! 🎉

[aidankwok]

I have another way of doing it
I
= Integ (x cos x / sin x) dx
= Integ x/sin x d (sin x)
= Integ x d ln (sin x)

Applying integration by parts, I
= [pi/2 ln sin (pi/2) - 0 ln sin (0)] - Integ (ln sin x) dx

The integral from 0 to pi/2 of ln sin x dx is a classic integral (can be calculated by king’s property)

[limpinggabriel]

I actually solved this one on my own! I would never have been able to do that without guys like you teaching.

[wholepizzas]

20:30 by putting a=0 here and having a>=0 won't we have an issue while inputting the limits in tan(au) when u approaches infinity. If a can be equal to zero won't we have a possible 0×infinity situation

[silversleezy]

If you treat the integral as xcotx
Then apply the power series of expansion for cotx we have the integral as:
x ( 1/x−x/3−x^3/45−2x^5/945 - ⋯)
After multiplying the x into the expansion, we can then easily integrate and evaluate the function from 0 to pi/2 and see it converges to pi/2 ln2
(By factoring Pi/2 out and then consider power series of ln functions)

[seegeeaye]

A different way: integral = int of x cotx dx, then using By Parts, to get – int of ln sin x dx, by using the property of int of f(x) = int of f（pi/2 – x）, we have 2I = int of (ln sinx + ln cosx) dx. then follow some basic algebra and trig,

[gustavozola]

Excelent! You should post more challenging ones like this

[AriosJentu]

I've done this integral with a little bit different technique. First of all I've done IBP in first place to get "- int from 0 to pi/2 of ln(sinx) dx". Of course this integral also not so easy, and I've done separation of integration ranges from 0 to pi/4 and from pi/4 to pi/2. On second integral it was a u-sub like "x = pi/2 - u", to get from sine function - cosine. There is integral of sum of logarithms of sine and cosine functions from 0 to pi/4. Using log identities, we can get "int from 0 to pi/4 of ln(sin(2x)/2) dx", and with a u-sub "u = 2x". Assuming algebraic-integral equation "J = int from 0 to pi/2 ln(sin(x)) dx" and "J = -pi/2 ln2 - 1/2 J", we can get the same result as in this video. I've tried this first :D

[Happy_Abe]

Technically we can’t let a=0 at the end because then the argument from earlier doesn’t work so maybe we can take a limit as a approaches 0 to get the value for c

[bjornfeuerbacher]

I would have used the substitution u = tan(x) first, leading to \int_0^{\infty} arctan(u)/(u (1+u²)) du. Then in that integral, replace the arctan(u) with arctan(a*u), so you have an integral depending on a. Differentiating with respect to a then gives the same integral as in the video at about 12:10.

I think that's more intuitive than coming up with the trick x = arctan(tan(x)) first, and additionally, coming up with the idea that one should insert a factor of sec²(u).

[SpringySpring]

Just finished my first semester in Calculus 1, and now I finally understand a lot of the concepts presented in this video, as if it summarized my entire semester into one big challenge (the only thing I haven't really learned yet is the partial derivatives, but they seem to make sense from the video)
This was an AMAZING experience!!

[EmpyreanLightASMR]

I was just telling someone yesterday about Feynman "re-discovering" differentiating under the integral, but I'd never seen it in action. This was awesome!

[kaanetsu]

Another beautiful use of Feynman's technique!!

[Jack_Callcott_AU]

This was a very, very satisfying demonstration BPRP. I didn't know about the Feynman technique until recently. Thank you!😸

[rishabsaini]

Why so complicated?
Here are the steps I took:-
1) Do integration by parts taking x as the first function and 1/tan x as the second function as 1/tan x is easy to integrate over 0 to pi/2. (~20 seconds)
2) Find integral of 1/tan x over 0 to pi/2 by doing u-substitution, u = sin x, you'll get integral of 1/tan x as log(sin x) in like 2 sub-steps. (~1 minute)
3) When we put value of integral 1/tan x in our Integration by parts equation of step 1, we get

[x log(sin x) - integral (log(sin x) dx ] 0 to pi/2

Second term looks wild but It's easy to solve using properties of definite integrals (It was in my 12th grade NCERT book, I even remember the result). Int 0 to a F(A-x) = F(x) (~10 minutes)
We get -pi/2 log(2) for the second term.

Bam! We get pi/2 log(2) as the answer

[chennebicken]

I just love it, when pi and e show up together. Well, actually it makes sense… The pi comes from the trig-function and the log_e(2) from the integration.

[tgx]

You can use integral x* cotg x, then by parts, you have then - integral log( sinx), It's also on internet And some tric to use.

[Gabi-weff]

22:20 we can see the passion and appreciation this gentleman has for maths with just a look, he found his happiness and that's bittersweet af, nice video as always fella.