Sharpen your integration skills!

[@blackpenredpen]

I got so hyped when I saw u getting into the backflip position at 4:00!!

[@goodplacetostop2973]

69 likes, STOP THE COUNT!

[@Xenon7170]

100 people ruined it.

[@mathematics_and_energetics]

Yes, Michael's backflips are AMAZING. I look forward all the time to seeing him doing backflips! 🌺

[@aarryasaraf]

The back flip! It's back!!!!!!!!!!

[@nicolocantaluppi5572]

It caught me off guard!

[@goodplacetostop2973]

4:04 YO, THE BACKFLIP IS BACK!!! 8:32

[@ZeonLP]

im waiting for the double backflip

[@captainsnake8515]

YES

[@aril_archive6518]

Instantly push like button

[@ruanramon1]

Penn's BACKFLIP is actually the BESTFLIP

[@tonyhaddad1394]

Can any one do a double backflip (Its possible phisicaly ??)

[@rohitraut5925]

just completed fist solution: instead more easily: cosine squared x - sine squared x = cos2x whose integration is: sin2x/2

[@poissonsumac7922]

Beat me to it! That's way more simple imo, tho what he did was neat too

[@Tiqerboy]

Exactly how I did it.

[@changjeffreysinto3872]

beat me to it +1

[@Kanbei11]

To be fair he did say you could use a double angle identity to solve

[@cyberpunk_edgerunners]

the same i thought

[@aviralsood8141]

4:04 BACKFLIP CUTS ARE BACK!

[@manucitomx]

Oh, oh, oh, what a nice early Xmas present at 4:05. Thank you professor.

[@adityamohan7366]

I learned today that cos of double angle can also be written as a difference of fourth powers of cos and sin of single angle..just as it can be of difference of squares of both which is by definition.

[@arduous222]

Another way to evaluate first integral is to use (cos^2 x - sin^2 x)=(cos x - sin x)(cos x + sin x). If we let u=(cos x+sin x), then du/dx = (cos x - sin x). So, (up to constant), int (cos^2 x - sin^2 x) dx = int u du = (1/2) u^2 = (1/2)(1+2 sin x cos x)= sin x cos x +1/2.

[@holyshit922]

0:06 Difference of two squares , and double angle for cosine 4:18 By parts it may be necessary to do it twice

[@mathematics_and_energetics]

Since I love calculating in C, I solved the integral over cosx*coshx in C with cosh(x) = cos(ix) and sin(ix) = -i*sinh(x). It took me a little bit longer, but I didn't need to integrate by parts... Of course, Michael's ways to solve the intgrals are the best ones... 🌺

[@ntvonline9480]

Got my “OK. Great” shirt and “Do Hard Math” sticker. Will be inspirational for grad school. Thanks!

[@rickshaw1641]

Your backflip is so amazing, so great, so powerful!

[@rhettolsontime8835]

You could have also set cos^2x-sin^2x equal to cos2x, by the double angle theorem.

[@kennethandrew8878]

He explicitly says he didn't want to do it that way, but yes you could

[@mathunt1130]

There are two ways you can deal with the second integral. You can note cosh x=(e^x+e^-x)/2, and then use integration by parts on e^\pm x\cos x, or that cos x =(e^ix+e^-ix)/2 and then use complex integration. It's simple either way.

[@_GOUTHAM]

This shows besides maths class you soon create gymnastics channel too😁

[@tomkerruish2982]

Two other ways for the second integral: 1. Guess it's a linear combination of cos x sinh x and sin x cosh x and figure out the coefficients. 2. Recall that cosh x = cos ix and use cos x cosh x = ½(cos(x + ix) + cos(x - ix)).

[@giacomocervelli1945]

you could also see that the integral was just the real part of integral(coshx*exp(ix)dx). Then expand the coshx as 1/2[exp(x)-exp(-x)]

[@ruanramon1]

4:04 THE EPICNESS

[@abhimanyu88872]

Sir you are really a genius.🛡️‼️

[@mihaipuiu6231]

Very smart solution. Good for you Penn!

[@dodokgp]

why go through all the splitting and integration by parts when you can simply replace cos^2(x)-sin^2(x) by cos(2x) ?

[@RieMUisthegoaT]

i think the point of this video is to show the technique of canceling two integrals. pretty cool if you ask me

[@tomatrix7525]

That would have been easier. I think the purpose of his approach was for demonstration purposes. Nobody would bother to watch cos2x being integrated it is too basic

[@ghanshamchandel1854]

because it cannot be applied to cos(x)cosh(x)dx. He was building up for that integral.

[@tonyhaddad1394]

Back flip is incredible good job in math and in sport 💓💓

[@kumailabdulaziz4096]

One of the best mathematics channels; Go ahead !!

[@beneustace3709]

I liked just for the backflip!

[@andreivila7607]

Hello Mr. Penn. Instead of separating integral of cos^2(x) - sin^2(x) in two integrals, couldn’t you just say that cos^2(x) - sin^2(x) = cos(2x) and integrate from there?

[@eugenekim3012]

He mentioned double angle formula, he's doing this just for the fun of overkilling.

[@andreivila7607]

@@eugenekim3012 Sorry, I didn’t watch carefully. I just speed ran through the integral 😅

[@ghanshamchandel1854]

@@eugenekim3012 he was also building up the technique for the later example.

[@PubicGore]

Dr. Penn*

[@ArjunBhanap]

Another method for the cos(x)cosh(x) integral is by integrating the function cos[x(1+i)] and taking the real part of the answer, using Euler's formula to express cos(ix) and sin(ix) in terms of real and complex hyperbolic functions with real arguments

[@skwbusaidi]

The second question can be done using integration by part twice or DI method D I + cosx coshx - -sinx sinhx + - cosx coshx Integeral of cosx coshx = cosx sinhx +sinx coshx - integeral of cos x coshx So integeral of cosx coshx = 1/2 (cosx sinhx +sinx coshx ) + c

[@MichaelPennMath]

I like to think of my solution as a sort of "final draft" of either of those methods.

[@polychromaa]

The first one is pretty simple. I was able to do it in my head

[@JM-us3fr]

I just got finished watching the 2006 MIT integration bee a couple days ago. I gotta say (as a master's student) it's nice to see I was able to keep up with those undergrad math majors.

[@danielmirandacastro7161]

Imo this actually looks so much better than using double angle formulas os writing stuff as exponentials

[@josephmartos]

You are a 10/10 man! Congrats!!

[@justacutepotato2945]

I prefer to use the exponential expression for cosh. That makes the integral extremely easy.

[@ShivKumar-im3pm]

But that would be boring

[@michaelgolub2019]

The difference of squares in the first example may be expressed as a cosine of a double argument, then no integration by parts is required; as to the second example, we can express both trigonometric and hyperbolic function via exponential function and integrate easily in complex domain, then recombine as trigonometric and hyperbolic functions. The latter may be not so fast, but straightforward.

[@liyi-hua2111]

i like the way he did at second part. It's like doing a problem in a knee observation instead of doing the problem like a software, and also a bit inspiring. (But not the first part)

[@SKARTHIKSELVAN]

Thanks for teaching us. I like your flip also.

[@joeg579]

HELL YEA BROTHER THE FLIPS ARE BACK 🤘🤘🤘 nice integrals by the way

[@particleonazock2246]

I would like to express my gratitude toward the honoured and venerable mathematics professor, Michael Penn, for his remarkable video on integration skills.

[@MaromonLup]

Thank you for helping me recall my first year of Uni again

[@jamincywong2159]

For the first question, just use cos^2 x - sin^2 x = cos(2x) That would be much faster

[@hamzahamdain9162]

Thank you for the new technique 😎

[@profamitgupta]

The backflip was awesome

[@sarthakanand4515]

Now whenever I will be solving a paper, as I go from sum to sum, in my head, you, sir, would be backflipping.

[@simaomoreira20]

4:04 dammmmmmmmmmmmmmmmmmmmmmmmmmm

[@abrahammekonnen]

The idea of taking the function, adding another copy and multiplying by 1/2 is an interesting idea. I wonder how it was motivated.

[@CM63_France]

Hi, For fun: 1 "ok, great", 1 "let's may be go ahead and", 4:13 : perilous jump, it's been a long time.

[@replicaacliper]

Can you just replace cos and cosh with their exponential representations

[@MrRyanroberson1]

i'll just add something very subtle you pointed out: cos^4 - sin^4 = cos^2 - sin^2. very strange identities pop out of that. now if only cos^4 + sin^4 had a solution...

[@hi-bw6yv]

Didn't realize you could do a flip so hard that you break space and time.

[@lazaroychasoperez8882]

I think than the first integral will be easier if you had use cos^2(x)-sin^2(x)=cos(2x)

[@tomkerruish2982]

Yes, you can do a backflip, but can you snap the bad guy's neck when you do it? It'd actually be super easy, barely an inconvenience.

[@mohammedsalouani7672]

Yeeeaaah the back flip

[@flux4162]

BACK FLIP IS BACK!!!!!!

[@theimmux3034]

My guess is that you just integrate by parts till you have the integral we started with, add on both sides and divide by two.

[@abdallahmohamedelhady6668]

Thank you so so much

[@mahmoudalbahar1641]

Many thanks for this good video

[@MGoebel-c8e]

Did anyone try to substitute x for (-x)? The integral then equals its negative which would mean it’s zero (in a way cos(x)*sin(x) is a result “similar” to zero). I do not see how this is not valid. I also don’t see how this is less fascinating than people doing random acrobatics.

[@rain2001]

what is that thing falling behind him at 4:46?

[@Te4mRyouko]

Looks like dust, maybe from doing that backflip 😅 isn't he filming this in his basement?

[@mariannanezhurina7399]

cos^2x-sin^2x=cos2x so it makes integration a lot easier

[@jameskeegan00]

I was not expecting the flip

[@marclinlin2799]

讚 back flip yes!

[@respectmath95]

Another way to cancel out the integral INT (cosx)^4 - (sinx)^4 dx =INT (cosx)^3 d(sinx) + INT (sinx)^3 d(cosx) =(sinx)(cosx)^3 + INT 3(sinx)^2(cosx)^2 dx + (sinx)^3(cosx) - INT 3(sinx)^2(cosx)^2 dx =sinxcosx + C

[@lorentzbear]

Black flip 😍😍😍

[@berzerksharma]

The backflippp🔥🔥

[@JSSTyger]

How did that backflip erase the board?

[@Tiqerboy]

You should get on Ninja Warrior as a side gig.

[@granaro8236]

I don't get it. Once you got to cos^2-sin^2, that is cos(2x). the integral was immediate

[@MohammedAli-ig2nu]

Will you consider the question of how to decide if the anti-derivative can be expressed in elementary functions?

[@arvindsrinivasan424]

Goddamn.... I wish I had enough motivation to become this athletic

[@muckchorris9745]

THE BACKFLIPS ARE BACK!!!!!!!!!!!!! YEAAAAAAAAAAAAAAAAAAAAAHw

[@ruanramon1]

Here's a Michael Penn's backflip compilation kzbin.info/www/bejne/raGcl3SinbZ9nck

[@karolakkolo123]

But cos^2(x)-sin^2(x) is just cos(2x)

[@DavidCorneth]

I added a bunch of links to Michaels videos to the Online Encyclopedia of Integer Sequences (OEIS). See oeis.org/search?q=%22Michael+Penn%22&sort=&language=&go=Search. Some of his video's contain integers sequences, some give a constant which is in OEIS. I got word from Neil Sloane (the founder) saying "the more we have the better!". Can you maybe help by saying which of his video's could be in OEIS to which sequence?

[@jounik]

And that's a good place to flip.

[@royalmate9919]

Sir, please solve questions which are involving mod

[@hassanalihusseini1717]

Amazing backflip.... but can you next time multiply it by (-1)? :-)

[@dkravitz78]

Cos^2-sin^2 is sin+cos times sin-cos. Easy u sub from there.

[@prabkiratsingh4846]

It would have been easier with cos 2x identify

[@freddysilva1938]

Amazing. Mind blowing haha

[@DrChess8]

Beautiful maths, thank you!

[@nelprincipe]

you could use the identity cos2x = cos^2(x) - sin^2(x) to simplify steps on the first example.

[@lukaobradovic9629]

Sharpen your integration skills: buy a calculator

[@djvalentedochp]

back flips yay

[@MelonMediaMedia]

For the first one you can use cos^2-(x)sin^2(x)=cos(2x) Edit: and for the 2nd one I think you can use complex definition for cosine, and the e^x thing for the hyperbolic cosine and do some stuff with it

[@drsonaligupta75]

The first one could have been done by cos2x=cos^2x -sin^2x

[@anjaneyasharma322]

Trigo calculus the way it is being taught is nonsense. Draw the graph for the function. Find out the average value of Y Multiply it by the base in radians. U get the area under the curve Integration finished I have crossed 60 reading such mistakes. The instructor has not yet realized. As for t shirts they are fine so no need of this type of selling.