life changing integration by parts trick
ฝัง
- เผยแพร่เมื่อ 10 มิ.ย. 2024
- Let’s learn a life-changing integration by parts trick. Once you learn this integration technique for you calculus 2 class, many integrals will be much easier. The trick here is to choose a clever antiderivative with a smart constant.
0:00 Intro
0:15 Integral x arctan x
2:20 Integral ln x+2
3:30 Integral arctan square root x+1
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This technique is absolutely crazy. I feel like I now have the confidence to ask my boss for a raise and find a beautiful wife to marry. Thanks, Dr. Peyam.
lmao
but, be careful, if you watch the video backwards, it turns into a Country Western music video…
LMAOOO
LMAO
LMAOOOOOO
This is actually really cool 😆
Glad I actually watched it haha
Awwwww thank youuuu!!!
Please combine this with the DI method!😺
Nuuuuu! Now bprp will kill his students with these questions!
Master 😅
What is this crossover episode? 🧐
Now the +C is actually useful. Great video
Yup
So trueee
How did I not know this. My life is changed.
Greetings sir
Orz here
It's a very normal question in WBCHSE class 12
@@ayan7bhowmikNo one teaches this method in class 12
"Arent you glad I found my video? 😁" Yes! So happy!
Glad to see you so happy about math too :)
3:11 “Isnt that nice!” Nice ? NICEEE ??? SIR THAT IS STRAIGHT AMAZING my life is never the same after this
I've taught this method to my students for years and it's always made them much better at integrating in general
Hey, I have a doubt, Why do we have to choose the same constant at both the places though? for example at 4:47
symbolab gave me another result in his first integral
A Arjith look at the proof of integration by parts
@@aarjith2580 because we can't choose different ones. We are plugging in the antiderivative in both places you bum
@@aarjith2580 you integrate 1 and differentiate arctan(√(x+1)). Your derivative of that term is set, but the integral of 1 is x+c. This works for any fixed c. Hence, we fix c=2 as it's most useful to us. This means, each time we use this specific antiderivative of 1, our constant is 2.
I remember asking teacher, "can we add a constant to *inner integration* in the formula of integration by parts?"
He told me, "only a single +c can appear in a single integration", which feels so stupid to me now.
I am so glad I found this video. Kudos to you Dr. Peyam.
I think he probably thought you meant a general +C when applying the formula, which I'd also perhaps not recommend because it'll look too messy. But specific constants to make it easier, definitely do it!
I'm sure it was more a result of you asking your question unclearly
Probably your question wasn’t clear to him, so he meant you don’t add +C within each term.
@@skylardeslypere9909 that's still a bad answer from the teacher imo, especially without qualification. Of course many teachers won't get into the specifics of derivations and logic because of their effort,. communication skills, or rigid worldview
I think he meant you were asking if you can leave constants in the equation without absorbing them into +C
I love his attitude, man knows math and is happy about teaching it. I wish all my teachers were like him.
Thought the title was click bait.
It was NOT click bait.
Life changing, indeed!
Omg hi Floyd!!!
@@drpeyam Hey, I have a doubt, Why do we have to choose the same constant at both the places though? for example at 4:47
@@aarjith2580 When you're doing integration by parts, you integrate "dv", and you can add any constant to the v. But if the constants are different we would get v≠v
As a physic student I can say, this is really lifechnaging, hours of pain will be joy instead. Thanks 🙏🏼
The fact is (even if we just think up to calculus level): whenever we do antiderivates, 99% of the time we forget the "+C", and this "+C" is the reason why we can end up with different answers depending on how we choose. Moreover, this also shows that the operation of antiderivate is not even a function as we can pick anything in the codomain for C as we want. So as long as the "number" that we "need" belongs to the codomain we can add it freely.
Well, it is a function if you consider an equivalence relation between functions that differ by a constant.
(And then it is a linear isomorphism)
👆 put differently, Give it a couple years and you'll stop thinking of constants as really making them "different"
Wow, such a simple idea that's hidden in plain sight! Just in time for my calculus test this week.
Good luck for your exam bro.
blessed to have him as my differential equations professor, couldn’t be luckier
Awwwww thank you ❤️ It’s symmetric!
How have I never seen this before? This is amazing!
Omg hi Mu Prime Math !!!
This is one of the reasons why integration by parts is my favorite (and in my opinion, the most powerful) integration technique. Here's a really great (and difficult) integral for anyone interested: ∫√(1+x+x²)/(1+x)dx. Keep IBP in mind!
int (1+x+x^2)/(1+x) dx = int 1 + x^2/(1+x) dx, then use u = 1+x.
@@davida2810 no, that's the wrong function. (1+x+x²) should be under a square root. So the function was √(1+x+x²)/(1+x), which is much more difficult to integrate. Of course I wouldn't ask such a simple question that doesn't even involve IBP on a video about IBP.
It is unlikely that integration in parts will lead to a result in this case.
t=x+1 .
∫√(1+x+x^2)*dx/(1+x) =∫√(t^2 -t+1)dt/t=∫(t^2 -t+1)*dt/t*√(t^2 - t+1)=
=(1/2)∫(2t-1)dt/√(t^2 -t+1)-(1/2)∫dt/√(t^2 -t+1)+∫dt/t*√(t^2 -t+1)=(*)
In the last integral, replacing t=1/z leads it to an integral similar to the second.
(*)=∫d(√(t^2 -t+1)) - (1/2)∫dt/√[(t-1/2)^2 +3/4] - ∫dz/√[(z-1/2)^2 +3/4]=
=√(t^2 - t+1) - (1/2)* ln[t-1/2 +√t^2- t+1]- ln[z-1/2+√z^2-z+1]+C=
= √(1+x+x^2) - (1/2)*ln[x+1/2 +√(1+x+x^2)]-ln[1/t-1/2+√(1/t)^2-(1/t)+1]+C=
= √(1+x+x^2) - (1/2)*ln[x+1/2 +√(1+x+x^2)] - ln[(1-x)/2 +√(1+x+x^2)]+ ln|x+1|+C.
(I use the equality ∫dt/√(t^2+λ)= ln|t+√(t^2+λ) |+C,
a "long logarithm", and not a little informative expression through asinh).
@@Vladimir_Pavlov actually, after substituting 1/y=x+1, you can perform integration by parts with u=√(y²-y+1), dv=-1/y²dy. You actually did the same substitution (in two steps), but used a different method after that. Anyway, it's nice to see an alternate solution to mine. Reciprocal substitutions can be very powerful, along with ibp!
@@violintegral noicely done bro ... Yep we can use substitution.. but don't forget dy 😉
To be fair, the integral of x²/(1+x²) is relatively easy because that's just (1+x²)/(1+x²) - 1/(x²+1) which gives x - arctanx
But nonetheless, this is a really useful trick I've never thought about. Nice video
Yeah I ve thought the same thing
Me 2
yeah i agree but this does seem to be a nice tip
Nope. The 'trick' only works in cases where the algebraic simplification also works. This is not a method, it is nothing.
@@annaclarafenyo8185 the situations where OP's method work are precisely the situations in which Peyam's work. They're the same thing, just hidden behind algebra, like when an FB videos says to "take a number, double it, add one, half it, take away half the original number and boom you get 1/2 every time, spooky, right?"
I've learnt the +1 -1 trick. on integration but this takes it to a new level - can't thank you enough!
Me too
This blew my mind! So fascinating. I guess that's why in my courses it usually says "find the most 'general' antiderivative."
This trick is amazingly amazing, this really changed my life and how I see Integrals, thank you Dr. Peyam!
I don't even know integration 😂
For the ln(x+2)dx integral, you can substitute u=x+2 as well to bring it to a standard form
He is using the same method but not showing you the substitution. d(x+C) where C is a constant
Here's an example with ln(), that is more interesting.
integral ln(|x^2 - 4|) dx
I never even considered this. Makes it so much easier!
I gotta say I love learning math from people who talk about its tricks and properties with such passion. Thank you for your video!!
Absolutely stellar expedient method!
this is some serious math hack territory we are treading in
This is mind-blowing I love it, never thought of taking advantage of the constant like this. Nice!!
Beautiful tricks never cross one's mind until someone with enough insights reveal them. Thank you Dr. Peyam!
What an amazing trick !!!
Thank you for sharing.
Wow. As a math lover, I've fallen in love with this constant addition trick. As if, it was always there but we didn't see it ! Thanks a lot ❤️❤️❤️❤️
It s amazing how something so important and so simple ,isn't used often
My lord, such a simple yet elegant trick and it's so convenient.
Many thanks for sharing it with us!
Just can't explain how grateful I am to you for explaining this miraculously magical method
Thanks alot ❤️
Never thought of doing that, gonna call it the "Peyam Integration Method"
Something quite related of course is shifting the function,
so; ∫ f(x+c)dx = ∫ f(u)du letting u= x+c.
Ex: consider ∫ x/(x+1)dx
➙ ∫ (u-1)/u du with u= x+1, gives ∫ (1-1/u)du = u- lnu = (x+1)-ln(x+1)+const.
Be careful though to change limits in definite form,
so ∫ f(x+c)dx between x=a to b ➙ ∫ f(u)du between a+c to b+c
Yep this would also make the ln(x+2) integral easier too
@@robertveith6383 The area under 1.x yes, but simply ln(x) for integral, thats what Wolfram Alpha says anyhow.
It's inspired from Ross's analysis book.
isn't that just substitution?
@@coerciasink Indeed but works in general for any function f(x+c) -> f(u)
You're the man!! Usually it takes more than one videos to get me to subscribe.
I love his enthusiasm
Truly exciting and imaginative!
I’m in my first year in college studying physics AND OH MY GOD I’M SO HAPPY I FOUND THIS BEFORE MY ANALYSIS EXAM
That was actually incredible.
My life has truly been changed. Thank you , Dr Peyam!
Me: has used integration by parts for years.
Also me: never thought to use an antiderivative other than the C=0 version.
Thank you for teaching me 😁
At 0:50 he says "you start to cry necause this is very hard!"
Well, not really... x² ÷ (x² + 1) is easily written as
1 - 1 ÷ (x² + 1)
which integrates immediately to x - arctan x.
Sorry legend
Yea its easy
Im almost crying how good this is , Thank you very much for sharing
Truly life-changing. Was absolutely mind-blown
Mann that boom boom 3:01 !!! was so energetic 😅🎉
I don't know why, but this makes me really happy 🤤🤗🖤🖤
Me 2😚
Solving integrals is allways fun!!!
Totally life changing trick , Thank you!!
This video came in my feed many times, glad I finally watched it.
Love you Dr peyam!
Thank you!!!!
Very nice trick! It's always easy to forget the simple things if you're doing hard problems. This might save me half an hour sometime in the future :D
Very good, and very informative.
Integral of ln(x+c) is easy, by integrating the inverse y = e^x - c, and subtract from the total rectangle area (x+c)ln(x+c). So, in your case: (x+2)ln(x+2) - x + c
But don't forget to substitude the intervall endpoints. That's how I did it for years ^^
That's Laisant' method, quite smart really. Problem is you may not have the actual inverse function to work with (logs are easy of course) but what about 1/(x^3+x+1)?
Yeah that is pretty awesome. I definitely remember being taught this now that you mention it, but i didnt come up with it when i did the example so thanks for the reinforcement!
Wow. This just blew my mind! That‘s crazy cool!
This is so awesome!
totally agree
Life changing man, for real 💯
Nice!!!
This is amazing ! Definetly life changing in some ways, especially since i have my integration final soon :D
Also i just saw i wasn't subscribed yet but im litterally binge watching all of your content, so i clicked the button this time ! That's somme great content !
This is amazing, I always struggled with this but this helped so much!
Does this work with the D-I method of integration where you might have several levels of differentiation and integration? Here, if we are permitted to add an integration constant at each stage, we can build up a polynomial if required. That would be really cool and could really simplify some nasty integrals.
Why not? Any antiderivative is equally legit, isn't it?
Of course! When you choose the antiderivative in DI, just pick the right constant
@@drpeyam yah
It can work, but it generally helps you the most on the regrouper types of integrals that you usually can do in just one row.
Have never seen the method before... Really cool! I also remember than some functions (Ln, Arctan) are the same in the complex domain. It would be interesting to see similar simplification methods based on moving to the complex domain.
Another method is to simply add zero in a fancy way, to reconcile the fraction within the integral.
Starting with:
x^2/2*arctan(x) - 1/2*integral x^2/(x^2 + 1) dx
Add zero in a fancy way, to the numerator of the integral, by both adding 1 and subtracting 1:
integral x^2/(x^2 + 1) dx = integral (x^2 + 1 - 1)/(x^2 + 1) dx
Regroup:
integral [(x^2+1)/(x^2 + 1) - 1/(x^2 + 1)] dx
integral [ 1 - 1/(x^2 + 1)] dx = x - arctan(x)
Combine with the original expression:
x^2/2 * arctan(x) - 1/2*(x - arctan(x))
And simplify:
(x^2 + 1)/2 * arctan(x) - 1/2*x
He knows how amazing this trick is and he is absolutely right, very glad I watched, thank you so much for the integration skills
I’m just now learning integrals and I’m so happy I found this
Solving integrals is allways fun!!!
I can't believe my eyes. Why didnt we know this before?
Pretty cool method. Never thought of it that way. This is pretty helpful. Thanks so much for sharing this
This is actually amazing.
From my (limited) experience in math, if you just add and subtract that constant divided by the denominator in each case you should end up with the same result, still pretty fun though
Yes, by adding a zero, you can obtain the same result as the trick although you might need to do a couple more steps.
this trick is pretty cool, but after trying this in all my integration by parts questions, i can tell you that this trick only works for certain integrals and it's not something you should look out for all the time. just keep this trick in mind if you can cancel your antiderivative with a denominator later on
How do you figure out which integrals it does/doesn’t work for?
@@uwuifyingransomware im guessing by practice
@@Cjendjsidj yeah by practice. it's not worth it to intentionally leave a gap and write (x+_). i suggest using this trick only when you are stuck or happen to notice the perfect setup for this trick
@@uwuifyingransomware You'll figure it out by experience after dealing with a bunch of different functions in many integration problems. You must have substantial experience in functional analysis and how to manipulate those functions using several derived formula which will help simplify the problem. You may not be able to solve all integrals since not all integrals can be solved but most can be solved using the normal techniques you've learned i.e. elementary functions, algebraic/trigonometric/hyperbolic substitutions, powers of trigonometric/hyperbolic functions, recurrence relations, partial fraction decomposition for rational functions and of course, integration by parts.
absolute game changer!! Thank you.
This is dope
Please keep making integration easy and fun!! I need it badly
This feels illegal
Dr. Peyam: Thanks for watching.
Me: Thanks for changing my life!
Incredible. I never thought so.... extremely good idea. Thank you.
This is truly life changing!!!
5:13
My favorite part of this video 😎
so basically, this is taking the "integral(udv) = uv - integral(vdu)" and expanding it without loss of generality to "integral(udv) = u(v+C) - integral((v+C)du)"
Precisely!!
This helps a lot with both definite and indefinite integral. It saves lots of calculating time.
thank you! great trick
TBH without adding the constant term in the first place, I will still tackle the second term by adding 1 to the numerator which gives the same result with one more line, but this method is definitely more efficient and some thing I should keep in mind. Thx
Btw second one I will do with change of variable, same as third, but third one still need a constant to do neatly, so very clever examples for showing the power it does.
This is some omega brain shit lmao
Wow. Genuinely impressive.
Who's here for CBSE class 12th boards 2024?
broooo me after struggling from integration
this trick is awesome, as is your presentation!! loved the video
This is actually amazing
Totally blown my mind! Shared to all my calc homies
Keep up the good work. I love it. That trick is awsome!
Appreciate this approach! Thanks for posting the video
This is such a powerful clever trick!
What a nice trick! Thanks.
Wow. This is truly awesome.
You have helped me achieve a new awareness in life. Thank you.
u made my day ty!! xx
Genius, crazy technique, thank you guy
This was mindblowing thank you so much
This was great! Thank you!
You're a Life Saver Thank you Dr
Please make more such videos! Thanks!
Genuinely fantastic.
Thank you boss amazing trick
Wonderful! Wish I'd known about this technique sooner.
Great technique! I like the marker-drop at the end.
Thanks a lot. Really helpful