life changing integration by parts trick
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- Опубликовано: 5 июл 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
same here!!!!!
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.
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
3:11 “Isnt that nice!” Nice ? NICEEE ??? SIR THAT IS STRAIGHT AMAZING my life is never the same after this
"Arent you glad I found my video? 😁" Yes! So happy!
Glad to see you so happy about math too :)
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.
As a physic student I can say, this is really lifechnaging, hours of pain will be joy instead. Thanks 🙏🏼
blessed to have him as my differential equations professor, couldn’t be luckier
Awwwww thank you ❤️ It’s symmetric!
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.
How have I never seen this before? This is amazing!
Omg hi Mu Prime Math !!!
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 don't even know integration 😂
There are two basic techniques which are pretty easy to understand in isolation. One is integration by parts (Multiple interations of it are done with the tabular method, or DI method) the other is U-Substitution, which is like anti-chain rule. You learn those two bad boys, and if you are good with algebra and can also think geometrically, then you have a solid foundation to learn calc I-II ideas
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 gotta say I love learning math from people who talk about its tricks and properties with such passion. Thank you for your video!!
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 😉
I've learnt the +1 -1 trick. on integration but this takes it to a new level - can't thank you enough!
Me too
Beautiful tricks never cross one's mind until someone with enough insights reveal them. Thank you Dr. Peyam!
This is mind-blowing I love it, never thought of taking advantage of the constant like this. Nice!!
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 ❤️❤️❤️❤️
I never even considered this. Makes it so much easier!
My lord, such a simple yet elegant trick and it's so convenient.
Many thanks for sharing it with us!
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 !
Just can't explain how grateful I am to you for explaining this miraculously magical method
Thanks alot ❤️
Absolutely stellar expedient method!
You're the man!! Usually it takes more than one videos to get me to subscribe.
He knows how amazing this trick is and he is absolutely right, very glad I watched, thank you so much for the integration skills
What an amazing trick !!!
Thank you for sharing.
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
Incredible. Thank you so much for this trick!
It s amazing how something so important and so simple ,isn't used often
Truly exciting and imaginative!
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
Just impressed!!!
I have learned to integrate ln(x), arcsin(x) etc and looking for arctan(x)...... And found your method!!!
Can't express how happy I have become!!! 😁😁
My life has truly been changed. Thank you , Dr Peyam!
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
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)
This is dope
Please keep making integration easy and fun!! I need it badly
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!
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
this is some serious math hack territory we are treading in
Im almost crying how good this is , Thank you very much for sharing
This video came in my feed many times, glad I finally watched it.
Love you Dr peyam!
Thank you!!!!
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 😁
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)?
Wow. This just blew my mind! That‘s crazy cool!
OHH GODD! I HAVE BEEN TRANSFORMED AND WILL NEVER LOOK AT ANY INTEGRAL THE SAME WAY AS I DID EARLIER!!!
This is just ingenious and astoundingly simply, yet so effective!
@@robertveith6383 OK
Life changing man, for real 💯
Nice!!!
This is so awesome!
totally agree
Pretty cool method. Never thought of it that way. This is pretty helpful. Thanks so much for sharing this
This helps a lot with both definite and indefinite integral. It saves lots of calculating time.
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.
I don't know why, but this makes me really happy 🤤🤗🖤🖤
Me 2😚
Solving integrals is allways fun!!!
Wow!! I did not see that coming. Very powerful trick. And I like the "marker toss" at the end. "Marker drop" will be the new "mic drop". Subscribing now.
Truly life-changing. Was absolutely mind-blown
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
Solving integrals is allways fun!!!
this trick is awesome, as is your presentation!! loved the video
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.
Mann that boom boom 3:01 !!! was so energetic 😅🎉
what do i say except THANK YOU! and i absolutely love your energy! makes one want to study math with enthusiasm. :)
this was very helpful indeed and i wish my teachers taught this way. all they did was give us a sheet of results they wanted us to memorise, explain first few of them and then give practice problems. :/
Keep up the good work. I love it. That trick is awsome!
Dr. Peyam: Thanks for watching.
Me: Thanks for changing my life!
I can't believe my eyes. Why didnt we know this before?
You rock, man! What a useful pattern. Thank you!!
Thanks so much!!!
Totally blown my mind! Shared to all my calc homies
This feels illegal
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.
This is helpful! Somebody might say that you still get the answer by the old way, but this new trick saves one step at least.
Waiiiittt why haven't anyone thought of this, It's simple yet genius. Thank you so much Dr Peyam
5:13
My favorite part of this video 😎
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 video was recommended to me at midnight the day of my advanced integration unit test which is basically just a bunch of integration by parts and partial fractions. I’m both grateful and very concerned that youtube recommended me this
Totally life changing trick , Thank you!!
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 is some omega brain shit lmao
Incredible. I never thought so.... extremely good idea. Thank you.
That was actually incredible.
Who's here for CBSE class 12th boards 2024?
broooo me after struggling from integration
Very useful trick i wish every maths content creator should come up with trick like that to make the life of students easy and also it creates interest in maths
You have helped me achieve a new awareness in life. Thank you.
This is such a powerful clever trick!
I’m just now learning integrals and I’m so happy I found this
Appreciate this approach! Thanks for posting the video
Great technique! I like the marker-drop at the end.
my teacher taught me this trick before and I was like wow, thank you. Because of your video now I remember a useful trick to apply in my life
BRO this helps me so much with speeding up my solving time omg
thank you so so much, you're the goat frfr
You’re welcome :)
a neat concept that clarifies the +c as well👍👍👍
This is truly life changing!!!
I found this when I’ve been suffering the most with Calc 2 thank you kind sir
Wonderful! Wish I'd known about this technique sooner.
Wow. Genuinely impressive.
This sort of reminds me of 7th grade, when we learned about "completing the square."
This trick is really smart and I will use it in the coming years. Thank you :)
absolute game changer!! Thank you.