Integral of x^2*cos(x) and why the DI method works better than udv integration by parts
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- Опубликовано: 3 сен 2018
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In this video, we will do the integral of x^2cos(x) with the traditional u-dv setup and the DI setup. We will also discuss why the DI method works better than udv integration by parts. This method will help you tremendously with your Calculus 2 class. Once again, the DI method is just Integration by parts, and sometimes it's called the Tabular Integration
DI method, all 3 stops: • integration by parts, ...
This method be like:
- Hey you silly integral, you’re going to DI!
Why isn't the DI method used commonly? It would solve so many difficult and long int by parts questions
But isn't DI method coming from ibp and if you prove it and show why it works the students don't need to prove it everytime since they understand why it works already
It's commonly used by the teachers here
Because professors wanna do it the " hard way " first before this method. When i took calc 2 we did integration by parts until the final exam. Then i went to diff equations and prof was surprised we did not know DI method lol 😅😅😅
We were taught both the methods because sometimes while solving problems we are meant to solve linear differential equations that involve byparts integration and to save time we just use DI
In Japan, this method is called ”瞬間部分積分(Shunkan Bubun-Sekibun)”, which means “instant integration by parts”!
The DI method improved my life. Honestly proving and calculating Fourier series coefficients becomes so much easier with DI hahah
Yee!! fourier an bn coeficient integrals are not that tedious with DI method.
All my years taking math and I was never taught this. This method is amazing! It's more impressive (scary to most people) to do it the original way, but the DI method really cleans it up and makes integration by parts seem a lot less daunting. I wonder if my professors would have docked me points if I had known and used the DI method on a test?
I teach both to my students and I don't mind which one they use. The main part is still to know the idea for IBP is "one part to be differentiated, the other part to be integrated" (then hope for the best)
@@blackpenredpen
I love how you add "hope for the best" hahaha
I can scarcely believe how easy and well-organised this DI method is. How wonderful! Thanks BPRP.
When I went into calc AB junior year, I watched the video on the DI method before we got to integration by parts. By doing that, not only did I understand the lesson better than everyone else in my class, but I also was able to make my own connections between the DI method and the traditional formula, allowing me to safely use the DI method for all integration by parts problems, even though my teacher was only aware of the "first stop." By explaining this method to my friends in ways they could understand, they too grasped the concept much better, and I still use the DI method to this day, 3 years later.
since you first uploaded the DI method on this channel, things became way more simple and easy to understand. and now a more detailed explanation as to how this works.. wow, it's great
[yay][yay][yay][yay][yay][yay][yay]
I am very happy to hear! Thanks for the comment!!
LOVE your vids (I've watched well over 100 of them I am sure.) Thanks for your work!
Gary Tugan wow!! Thank you!!
Playing around with this integral, I think I found another approach for solving it. As integrals and derivatives are inverse operations, the basic idea is to write the solution as a function with variables, then take the derivative of this function and solve for the variables, as follows:
All derivatives of sine and cosine functions are other sine and cosine functions, and all derivatives of polynomials are polynomials of a smaller degree. So, let's write the antiderivative in the following format, where a, b, c, d, e, and f are variables to be worked out:
Integ{x^2 cos(x)} = (ax^2 + bx + c)(sin (x)) + (dx^2 + ex + f)(cos (x)).
Take the derivative of both sides of this equation (use the product rule on the RHS), and you get:
x^2 cos (x) = (ax^2 + bx + c)(cos (x)) + (2ax + b)(sin (x)) + (2dx + e)(cos (x)) - (dx^2 + ex + f)(sin(x))
I know this looks a little messy right now, but it's not as bad as it appears. Matching up terms on the RHS, we get:
x^2 cos (x) = (ax^2 + (b + 2d)x + (c + e))(cos (x)) + (-dx^2 + (2a - e)x + (b - f)(sin (x))
Since we have to end up with x^2 cos(x) with no sin(x) term, we know:
a = 1
b + 2d = 0
c + e = 0
d = 0
2a - e = 0
b - f = 0
Now finding the values for all six variables is easy:
a = 1
d = 0
b + 2d = 0, therefore b = 0
b - f = 0, therefore f = 0
2a - e = 0, therefore e = 2
c + e = 0, therefore c = -2
And so our integral is (x^2 - 2)(sin(x) + 2x(cos(x). Final answer!
Simplicity and Elegance.....It's a beautiful thing.
Why isn't the DI method used commonly? It's like the calculation becomes a lot easier.
when i was starting on calc (i learned calc by myself), i was a bit confused on the udv method until i see ur pi function vid, so i learned di instead
Excellent!
Thank you
Great method, thank you!
Wow! Really elegant!
You are the best teacher ever!
Wonderful DI method of diagonalisation.
you're amazing at explaining this dude
This is the most complicated, lengthy explanation I've seen for the tabular method. It really shows you what's going on inside it though.
thank YOU!!
I learn more here than in school, thank you
Thank you!
I'm old, and we used the 'uv'-method for integration by parts. I really love this DI-method. I'm curious, would there ever be a time when using the DI-method when you might need to switch the D and I columns after a few steps?
if you switch the columns you're basically going backwards and so you'll just undo the progress you made with your previous step.
the point is usually to have something that gets simpler when you differentiate until eventually it disappears, which you pick as the thing that you differentiate, and the something that hopefully stays about the same level of complexity when you integrate so that overall each time you do IBP the bit in the integral is simpler until eventually you solve it.
So tl;dr i don't think you should ever have to reverse it. but maybe there's some really weird special case where it makes sense to do so?
Elegante explicación. Thanks you.
Now that's efficient !
老师你的讲得特别好！
Sir, you are so funny and your method is really useful. Thanks a lot.🙏🙏🙏
Love you from INDIA.
I just... I love u Hahaha i'm in my last math course in university and fourier an & bn coeficient by doing parts was making those integrals so tedious. Thanks a lot. Greetings from Argentina. Future electronic engineer here hahahahah. Peace ✌🏼
Thank you bro from India
thank you
The last part where it is integral of -2cosx you can do that as you normally would your doing extra work
tnx alot
Legend!
It is literally enlightenment
What would change/happend if insted of taking the derivetive of u I would integrate, and same with v?
wish he could be my professor , his smile is adorable tf
wow
so easy ID METHOD
#yay DI method
Everybody gangsta until blackpenredpen uses a blue pen
How will I integrate xlogx by D-I method???
nice
Awesome :))))
6:35 that yay soundtrack
9:02 Is there any weird case where you gotta switch the columns? Like you pick the I part to differentiate and pick the D part to integrate?
Even during the traditional method switching what you are differentiating and what you are integrating causes you to undo your last step and end up with what you started with.
Good Job BPRP!
Unathi Gxarisa good job to you as well!
Love u
Thank you!!
Cool 😎
YAY
I am confused about choosing D PART AND I PART (7:10) in the video. Why can't we choose cosx in D PART and x^2 in I PART?
because then x^2 will get to 0 by taking the diretetive, and cos(x) will not get to 0.
for the answer it does not matter where you put it, but it is really simpler to do it so that you get to 0 at your D colom
*you can put cosx in the D part, but then the problem wont get easier (try it yourself)
Hey sir, how can the di method be applied in int(lnx dx)? I can’t get it right. Please help. Thanks’
u=ln x and dv=1 dx. You just stop at the 2nd line.
did you actually come up with the DI method yourself? I can't wait to tell my maths teacher about it though we haven't even begun learning about integral calculus in school yet :D
t. gobold No he did not. It is just a superb method that he is showing to us.
And, of course, Box The Answer!
What is the integral of x.tan(x) ?
So we can make choice which we have to take under D and which we want to take under I ? As per ILATE rule
Plz answer
Yes. You still have to make a choice which term to integrate, and which term to differentiate. The ILATE or LIATE strategy works about 80% of the time, but there are some cases where it works against you.
What to look for, is which function is easiest to integrate, and prioritize assigning that to the integration column. Another thing to look for, is which function will become simpler as it gets differentiated, and assign that to the differentiation column.
Great video! How would you use the DI method for (e^x)(sin x)?
James Dirig that's the 3rd stop. Check description
Thanks! I really love your videos. I tell my students to visit your channel.
Hey there..... Ohayogozaimqsu (as I think you are japanese) If we calculate ln^3x then upto which point will we integrate 1 bcoz one is considered as the first function in this question....
Can you do a video using the D I method to integrate (ln x)^2 it seems to not work! I had to abandon it and use the old method :-(
Given:
integral ln(x)^2 dx
Multiply by x/x, so the derivative of the inside function appears in it:
integral x/x * ln(x)^2 dx
Set up IBP table:
S ____ D _______ I
+ ___ ln(x)^2 ___ 1
- ___ 2*ln(x)/x __x
x*ln(x)^2 - 2*integral ln(x)/x * x dx
Cancel x's:
x*ln(x)^2 - 2*integral ln(x) dx
Recall integral ln(x) dx = x*ln(x) - x
Thus our result is:
x*ln(x)^2 - 2*x*ln(x) + 2*x + C
#YAY !!!
Can the DI method be used for all equations ?
Depends on what you mean by "can"?
Can it always work? No. Many integrals have no closed-form elementary solution, even if they might initially seem like they are a straight forward IBP problem.
Can it be effective at solving ALL integrals that integration by parts can solve? Absolutely. I don't know of a single counterexample, where the traditional method works, but the DI method doesn't. It's just a reformatting of the traditional method.
Will a teacher accept it? Depends on how fixated they are on the traditional method.
How do i know which term of the original integral i have to differantiate and which one to integrate?
Follow the ILATE RULE
I is inverse
L for logarithmic
A for algebraic
T for Trigonometric
E for exponent function
There are some exceptions
Like e^x*sinx
So you may just try another one if you get stuck
@@vishalbhatia6652 what does this rule say? If one of the functions in the integral is one of the ones you listed than what... i didnt quite understand it😅
Now, what if the side cannot be differentiated to zero for the DI method?
If it can't then you can only B.I.P
🙏🙏🙏🙏please reply🙏🙏🙏🙏.
One of my sir claim that he discovered this method in 2003. Please tell me who is the actual discoverer of this tabular form and when it was discovered.
it is just intergration by parts( multiple times), only written very smart and simple.
Is this how mathematicians integrate? This is really fast method
Thanks from Russia!
This reminds me a lot of the Russian algorithm for multiplication.
Can you talk about quaternions like 3b1b
Well, I don't see why quarternions are a thing.
The complex numbers set is algebraically closed, so how would you define quarternions?
Quaternions would be numbers in a fourth dimensional plane, similar to how complex numbers are in the second dimensional plane.
@@Mot-dh5sx Quaternions are really built for 3-dimensional applications. We just have four parameters to define them, because three of the four are set up to be normalized to represent direction as a unit quaternion. The fourth term is for magnitude, and contains no unit quaternion of i, j, or k to go with it. The terms in front of each of i, j, and k in any given quaternion quantity are all real numbers, between -1 and +1, and add up as vectors to have a magnitude of 1. The unit quaternions of i, j, and k are each a form of the sqrt(-1), such that when you square any one of them, you get -1. Combinations of unit quaternions either multiply together to -1 or +1, depending on the patterns within the right-handed coordinate system and a corresponding table.
This is the reason we use i, j, and k to represent coordinate unit vectors, even before thinking about quaternions. They are alphabet neighbors to i for the imaginary unit. Hamilton, who coined quaternions, thought that they could replace vectors in general, but it didn't catch on. The notation still caught on, with other mathematicians and physicists.
Try getting the formula for x^1/n that has no calculus inside that...
Can you explain the error function? ....
Yes. The bell curve function, whose simplest form is exp(-x^2) cannot be integrated in closed form. We know the total area under the curve is sqrt(pi), from a polar coordinate substitution and double integral over the full domain. But we cannot integrate the original function in closed form.
In order to evaluate the integral of the bell curve function, as is commonly needed in probability, the error function erf(x), and error function complimentary erfc(x) were defined as the solution. The bell curve is normalized by dividing by sqrt(pi) as the leading coefficient, so that the area adds up to 1, because for the application of probability, the probability the random variable being any real number is 1. erf(x) indicates the improper integral from -infinity to the value of x, and efrc(x) indicates the remainder of the integral from x to +infinity. By definition, erf(x) + erfc(x) = 1. erf(x) indicates the probability that a continuous random variable that can be any real number is less than x. erfc(x) indicates the probability that a continuous random variable that can be any real number is greater than x. That is, for the given probability function that is normally distributed, with a mean of zero and a standard deviation of sqrt(2)/2
What is ultimately behind evaluating this function is an infinite series. Just like what is behind sine and cosine. The original function exp(-x^2) has been expanded to an infinite series like a Taylor series, and each term of the sum is then able to be integrated in closed form. This generates the summation that defines erf(x) and erf(c).
damn, that is some legit black magic there
Integral u*dv = u*v - integral v du
u*v = ultraviolet
v*du = voodoo
#yay
so, you just do intergration by parts over and over, until there is a 0 in the intergral part ( and then the intergral is also 0) so you end up with the answer??
my teacher did not know this method and he did not like it. i can only use it as a check, not as proof.
#YAY
This kind of method should only be used to check your work. I'd have to look into it more to see where possible pitfalls are, but I don't think students should depend on this too much.
this is just integration by parts but with a more convenient format.
#YAY Did you create this method?
Nop
This is lebnitze rule for integration
This is all very nice, but it only works when one of the functions is a polynomial, correct? In other words, you have to get to a derivative equal to zero, which can only happen with a polynomial.
Nope, works for every ibp even if the d/dx never hits 0 e.g integral of e^xsinx, there’s another stop for it
You're right. As is clear from another post I made for this video, I must have written this comment before I had a complete understanding of integration by parts.
u playing pupg ?
Is it just me, or did I hear the Doraemon theme song in the beginning of the video?
Can't you also stop DI method when you know how to integrate the row?
Yup, that's the 2nd stop. I have the full video in the description : )
Wait a second, integration with u; dv ended at half of the video. That mean...
OK, what's the origin of this method? Who invented it?
Mathematician Brook Taylor, the namesake of the Taylor Series, discovered integration by parts, first publishing the idea in 1715.
i don't get it
AndDiracisHisProphet
U must be joking right?
yes
AndDiracisHisProphet
Called it! Lol
I didn't get the DI method, I can't understand him.
How old are you?
Does this work for any, ANY integral of f(x)g(x) type? #yay
No. For instance, if you have a product of sine and cosine, with a different coefficient in both functions. Using integration by parts will put you in an infinite loop. You have to simplify with trig identities.
I cannot use this on test NOOOOOOOOOOO
Wait so Integration has a product rule now?
you can derive integration by parts by starting with the product rule, rearranging a little and integrating both sides.
(uv)' = vu' + uv'
vu' = (uv)' - uv'
integrate both sides
integral vu' = integral (uv)' - integral uv'
integral vu' = uv - integral uv'
@@khajiit92 Sounds like ultraviolet voodoo to me.
Integral u*dv = u*v - integral v du
u*v = ultraviolet
v*du = voodoo
Why tf am I paying for uni again?
DI stands for don't inhale
Cut the Bs, video starts at 7:01