It doesn’t matter if it’s ILATE or LIATE. Since if you have an integral with both log and inverse function, then it’s most likely not doable in the first place 😆
I believe that by inspection we can easily see which one is easier to differentiate and which one to integrate. This idea or sense helps students in the long run. Tricks might help for short term, but not for the long run. This also helps students to recognise different patterns and get familiar with mathematics.
AMEN. Preach it. Ok, but seriously, these are the exact same words I had in mind as soon as I started the video, and I am glad I am not the only one, and I am glad you beat me to it.
Define formally and mathematically what "easier" to integrate or differentiate means. Otherwise it is bringing some pseudoscience argument into the mix. As far as I can tell there are dozens of patterns which work by trial and error, and a few of them are very common. But the patterns should ultimately be well defined and not rely on intuition. A computer program should be fed thousands of these problems applying pattern rules until a "perfect" pattern is found at least for those in the dataset
@@gregorymorse8423 The patterns need not be well-defined. LIATE is merely mnemonic, not a mathematical theorem. You are reading too much into this, and nearly everyone else knows what exactly is being meant by "easier".
My calc 2 teacher taught us the LIATE method but didn't make us use it if we didn't want to. Same with the DI method. I personally never used it but for some people it did help. IMO it's just about teaching the tools, not telling the students which ones to use.
Yeah same. Looking back at it, I think it was better that I never used methods like LIATE as this forced me to think more about which function to integrate and which to differentiate when using integration by parts. After some practice, I could calculate the second term quickly without a pen often and I could see if there was any way to integrate that term or if it was similar somehow to the original integral (like in case of sec³(x)) or if I had to do repeated integrations by parts which were getting even more complicated etc.
I agree in not using LIATE, it's not about memorising a mechanic, I like having the openness of realising a mechanic doesn't always work. When I teach integration by parts, I always teach find the function to integrate first
Excellent video. Before I retired, I told my students that LIATE is a nice "rule of thumb" but, when integration by parts is applicable, LIATE does not work 100% of the time. The example I used to demonstrate this fact is the integral of (xe^x)/(x+1)^2. in this case, the factor to be differentiated is xe^x and the factor to be integrated is 1/(x+1)^2.
Another way to deal with sec^3 is: sec^3(x)= 1/cos^3(x)= cos(x)/cos^4(x)= cos(x)/(cos^2(x))^2=cos(x)/(1-sin^2(x))^2 and with a substitution: sin(x)=u ; cos(x)dx=du we have the integral of 1/(1-u^2)^2 wich is a rational funcion, so quite easy to integrate. Is a method that works for 1/cos^n(x)and 1/sin^n(x) for every odd n.
I feel like most Mathematicians do this anyways subconsciously. And it is a good first instinct to have to quickly solve problems. But there are cases that don't quite work, and I think it's up to the student to discern for themselves.
My first class of calculus was in 11th grade in India, that's Junior Year in America. We were always taught ILATE, explanation was "Choose whichever is the harder to integrate" and like you said, it works in almost all scenarios. Wherever there was an exception we were given the solution and were told the specifics. It's the first time I'm hearing of DI method. Pretty amazing!!
The way I learned the LIATE method was that you should use it when first attempting the question. They said it won't always work, but it can help you start working out the problem.
In spanish we say ALPES: A=arcsin, arccos, arctan, etc. L=logs P=polynomios E=exponential S=sin, cos, tan, etc. It's said that not always work but it has always work for me with this set.
For the integral of xsin(x^2) there is no need of integration by parts. Just U-substitution let u=x^2 so du=2xdx It becomes integral of-1/2cos(u) and the final answer is-1/2cos(x^2)+C
Yep, and in fact, if you do LIATE integration by parts on this integral, then you'll have to differentiate x and integrate sin(x^2), the latter of which is impossible!
for guys lived in morocco 🇲🇦, they have been using a technic called "ALPES" istead of laatte A -Arctan ,arcsin,arcos L - ln P - polynomial E - Exponential S -Sin , cos , tan
This video is great. There's always that tension between "tools that are very helpful most of the time" and "concepts that work nearly all the time," and you balanced this excellently in this video with examples to both sides. 💪
For integral of xsin(x^2)dx in the end card: its easier to do u-sub then by parts u=x^2 , dx=du/2x integral of xsin(x^2)dx =1/2 * integral of sin(u)du =1/2 * -cos(u) =-1/2 * cos(x^2)+c
Moral of the story is just do whatever sounds like less of a nightmare and fuck around if you’re stuck. Liate is just a way to get you moving if you’re a hesitant student. Only way to solve a hard problem is to try solving it, even if there’s no clear solution.
My teacher didn’t like the idea of restricting us to LIATE so we instead subscribed to a some rules of thumb. These are the ones I remember, still use, and are almost always enough to make the right choice. A) If one of the functions has cyclic (sines, cosines, exponentials, etc.) or terminal derivatives (polynomials), let that one be the one you differentiate. Unless the other function’s integral is unknown or unsolvable by you, in which case… B) Let the most difficult of the functions that you CAN use integrate be the one you integrate.
I just discovered your videos last week but have only seen ones from about 5 years ago. Having seen them and just coming across this one and the change from no beard to beard is the best thing I've seen in a while
It's fun, in France we have ALPES, for arccos/arcsin ; log ; polynomial ; exponential ; sinusoidal. At the time I learnt integration, my teacher said like you that it is better to search which part is better to integrate than use this tip.
I never seen the LIATE method before. Like you said, I tell the students that they must figure out what to integrate first since finding the derivative is the easiest (sometimes). Unfortunately, many students of mine have trouble doing so.
In the bonus example, just substitute u=x^2 I know the question is what if I try to integrate by parts but: - maths is about getting to the right answer and spotting the quickest way is always a valuable skill - according to the rules of logic ("if today's Sunday I'm the pope" according to the late Doctor Richard Maunder, or "if my granny had wheels she's be a bike" according to Gino Cappucino), any answer to your question is correct if I'm not going to integrate by parts. If I integrate that by parts then I'll climb Nelson's Column naked.
For most of our IBP problems, when we have a transcendental function and a power function, if the derivative of the transcendental function becomes algebraic, then it is u, if it stays transcendental, then it is with dv.
9:06 I tried this integral by first substituting u = 1 + ln(x). I then ended up win an integrand that looked like (ue^u - e^u)/u^2 * 1/e And at that point I had the brilliant idea use the quotient rule in reverse And then it's just (e^u)/u * 1/e or just x/(1 + lnx) THE QUOTIENT RULE IS USEFUL FOR INTEGRALS!!
Yeah exactly. I was looking for this comment. You can also do it without a substitution. lnx/(lnx+1)² = ((lnx+1)-1)/(lnx+1)² = ((lnx+1)(d(x)/dx)-x(d(lnx+1)/dx))/(lnx+1)² Which is the result when the quotient rule is applied to differentiate x/(lnx+1). Hence the answer for the integral is x/(lnx+1)
I never thought LIATE as a rule or as a method. It's more like an advice from experienced mathematicians, "In most of the integrations, you'll find yourself integrating in this pattern."
I completely agree, you can't exactly impose strict rules for calculus. It always matters the type of problems you are solving and thinking a few steps ahead is the key. I always hated that rule
4:49 we just took this in class today and a student kept integrating by parts till he had about 10 terms because he didn't add the integration of sec^3 x on both sides 😂
It’s actually funny that in my education in Dominican Republic we use the method but the order is different, for us is ILATE, all the letters stands for the same.
I am lucky that I got a good Mathematics Teacher . When we came across this stuff , he taught us this method but suggested not to use it . He told us to integrate by parts by choosing which expression is "easier to integrate" by our intuition. And for real , in some questions even this rule fails(as in takes longer time ). And you sometimes can use different methods to integrate a function that seems harder to integrate but is easier than the other one.
for the 4th intergral, I can write down the numerator as 1 + lnx - 1, which is the same as (x)'(lnx + 1) - x(lnx + 1)' therefore, the whole fraction becomes (x/(1 + lnx))', which then nicely cancels out with the intergral sign
I love the multiply by 1 trick for #4. Adding zero and multiplying by one are like the invisible ninja assassins of mathematics. They come out of nowhere and defeat the problem quickly. One of my faves is to integrate 1/(1+e^x) by adding zero.
I often tell my students, "We artificially adjust things to fit into standard forms we can use." Especially with situations we are integrating constant/quadratic or linear/quadratic.
Your videos are so easy to follow, not sure why, it must be a combination of your natural ability to explain the subject matter, along with the use of your whiteboard to write down equations in real time.
another way to deal with the integral at the end is you can make u = 1+ln(x) then x = e^u-1 then its basically just two almost similar integrals that cancel with each other
we can substitute for 1+lnx as t differentiating we get dx=xdt and we know x=e^(t-1) there or getting the integral as e^(t-1)(t-1)/t^2 which is e^(t-1)(1/t-1/t^2) as -1/t^2 is derivative of 1/t we get the result as e^(t-1)/t+c which is x/1+lnx+c after substitution (i used the result int e^x[f(x)+f'(x)]dx = e^xf(x)+c, which we can prove by doing another substitution)
In my college, they had taught us about the ILATE rule but they also mentioned that it is not always necessary to follow the order and gave us the examples. So we just had to understand which function was easily integrable and which wasn't. I guess it is more sort of, an intuitive method of solving.
I really like the way physicists think about integration by parts -- in direct analogy with (y, Ax) = (A*y, x) ---- this is really the power of integration by parts -- not in finding particular integrals -- like Integral of x^2 Log(x) dx, but in manipulating the action (integral) to obtain deep insight into the physical solutions. This sort of manipulation come up in math classes in ODEs, PDEs and Calculus of Variations.
My teacher teaches me both ways. First is to use the LIATE, but more importantly, she stresses that we need to choose correctly in order for the IBP to work, because she used the same example of integrating sec^3 x in our homework
I never heard of this LIATE method and maybe it was for the best. Instead of giving us another "trick" to memorize, they just told us to choose which to integrate and which to differentiate.
The LIATE method is essentially good enough to be worth a try, like a kind of "let's see if this works" approach, but it is also important to develop a feel for which parts that would be better to differentiate.
Another easier way to do #4 is with reverse quotient rule. Since the denominator is squared take it to be the denominator of the antiderivative function and by some guess and check, x is a perfect candidate for the numerator.
The sec^3 problem, and in fact all rational functions of trigonometrics of x, can be solved using Weierstrass’ substitution. You wouldn’t have to guess or experiment which factorization is convenient for an integration by parts, since you wouldn’t need any. In general, Calculus teachers would benefit from having seen Risch’s algorithm. In particular, that in integration, trigonometric function don’t really exist. They are just exponentials and logarithms. Another general mantra from the algorithm is that you can integrate all integrals (that have elementary primitives) by eventually reducing to the rational case.
I tried to use LIATE in the bonus example. So I chose sin(x^2) to be integrated. When I tried to do that found out it becomes soooo easy if you just use the form 1/2∫2xsin(x^2)dx.
I have been having no luck learning calc 2, but this method actually makes sense. (I am a bit biased to the tabular method though, we use it in quine-mcklusky models for circuit design)
For example #4, my first thought was to be suspicious of the denominator being squared and that the numerator had parts of the denominator in it. That immediately reminds me of the result when differentiating a quotient. So, instead of trying to actually do the integration by parts, I would have just played around to try and build a fraction that would give the original integrand when differentiated. You already know the denominator, so it's just a matter of working out what the numerator could be. This example is quite easy.
I don't mention LIATE. Let students explore on their own. If their u fails, then go back and try a different u. My tip for students is to find the u-term first. The u-term in the original integrand is usually the term you want to change or get rid of through differentiation.
Sneaky fact from complex numbers, trig functions and exponential functions are actually the "same." So in an ordering like this, swapping them makes just as much sense as swapping logs (inverses of exponentials) and the other inverses.
Let me tell you another approach to the same problem we assume lnx=t dx=e^xdx integral reduces to e^t[ t/(t+1)²]dt e^t [ 1/1+t - 1/(1+t)²]dt this becomes d(e^t/1+t) using the product rule and the derivative and integral cancel out. so we get the same answer substituting t back
We indians learned ILATE rule this also we don't use maximum cases We apply integration by parts By just checking which is more preferable to integrate
I guess I should have realized that it was supposed to be the other one when I got very close, but simply conflicting signs that didn't allow me to cancel when i chose lnx as "I"
There are two ways for this triple product of x, sin(x), and e^x. One method is to use complex numbers, to rewrite sin(x)*e^x as a sum of two complex exponential functions, which will be i/2*e^([1 - i]*x) - i/2*e^([1 + i]*x). Pull the i/2 out in front as a constant, and assign constants N = 1 - i and P = 1 + i, to simplify our writing. Now you can integrate i/2*x*[e^(N*x) - e^(P*x)], with only one IBP table, after pulling i/2 out in front as a constant, and it is a simple ender. Then convert it back to the real world. Another way you can do it, is to assign J and K such that: J = integral sin(x)*e^x dx, and K = integral cos(x)*e^x dx Suppose we've previously solved these two integrals. The results are: J = 1/2*[sin(x) - cos(x)]*e^x K = 1/2*[sin(x) + cos(x)]*e^x We'll use J & K, as we construct our IBP table. Let sin(x)*e^x be integrated, and x be differentiated: S _ _ D _ _ I + _ _ x _ _ sin(x)*e^x - _ _ 1 _ _ J + _ _ 0 _ _ integral J dx Construct our result: x*J - integral J dx Since J is a linear combination of sin(x) and cos(x), this means integral J dx is a linear combination of J & K: integral J dx = 1/2*J - 1/2*K Thus, our result is: x*J - [1/2*J - 1/2*K] = (x - 1/2)*J + K/2 Fill in J & K: (x - 1/2)*1/2*[sin(x) - cos(x)]*e^x + 1/4*[sin(x) + cos(x)]*e^x Consolidate and simplify, and add +C: 1/2*[x*sin(x) - x*cos(x) + cos(x)]*e^x + C
This is why IBP is the last choice in "which integral technique should I use" since the others have tells or signs to look for. IBP is like "meh nah, just figure it out"
Bonus example, you get stuck at integral sin(x^2)dx, but if u=sin(x^2), integral v du is just integral x cos(x^2)dx, which is easy by substitution! I'm stealing your last example for when I teach Int. by parts next.
I have encountered integral sec^3(x) multiple times in all of science subjects. Mostly in physics. Here and there in Chemistry. Most of the times during maths calculus 2
Never heard of LIATE. Though for a sec that was an weird English abbreviation for “Integrating by parts” because I didn’t learn math in English (also never had an equivalent to LIATE taught to me). So I thought this video was about why you didn’t teach integrating by parts. Lmao
OMG the integral at 9:05 I just didn't solve it with LIATE I solved it using partial fractions can't believe it 😋😋😋😋 wish u can see my solution it gave the final answer directly 😂😂
It doesn’t matter if it’s ILATE or LIATE. Since if you have an integral with both log and inverse function, then it’s most likely not doable in the first place 😆
Yup 👍
What's your qualifications
Are you a PhD holder?
In india it is called ilate 🤔🤔
@@AbhishekKumar-jg7gq ya..you ar right..
@@AbhishekKumar-jg7gq mere ko LIATE sikhaya hai lol
I prefer the LATTE method. If the integral looks hard, go and make yourself a coffee.
Haha good one 😂
XD
😂😂😂😂😂😂😂😂😂😂😂😂😂👏
Noice
Gigantic heart palpitations, here we go!!!!!!!
I believe that by inspection we can easily see which one is easier to differentiate and which one to integrate. This idea or sense helps students in the long run. Tricks might help for short term, but not for the long run. This also helps students to recognise different patterns and get familiar with mathematics.
I agree. It’s better to understand why you shouldn’t do a certain thing, instead of just learning a rule to say you shouldn’t do it.
AMEN. Preach it. Ok, but seriously, these are the exact same words I had in mind as soon as I started the video, and I am glad I am not the only one, and I am glad you beat me to it.
Yes this helps a lot in integrating products of similar functions. Like in sec³(x), one can easily say that we will integrate sec^2(x)
Define formally and mathematically what "easier" to integrate or differentiate means. Otherwise it is bringing some pseudoscience argument into the mix. As far as I can tell there are dozens of patterns which work by trial and error, and a few of them are very common. But the patterns should ultimately be well defined and not rely on intuition. A computer program should be fed thousands of these problems applying pattern rules until a "perfect" pattern is found at least for those in the dataset
@@gregorymorse8423 The patterns need not be well-defined. LIATE is merely mnemonic, not a mathematical theorem. You are reading too much into this, and nearly everyone else knows what exactly is being meant by "easier".
My calc 2 teacher taught us the LIATE method but didn't make us use it if we didn't want to. Same with the DI method. I personally never used it but for some people it did help. IMO it's just about teaching the tools, not telling the students which ones to use.
Yeah same. Looking back at it, I think it was better that I never used methods like LIATE as this forced me to think more about which function to integrate and which to differentiate when using integration by parts. After some practice, I could calculate the second term quickly without a pen often and I could see if there was any way to integrate that term or if it was similar somehow to the original integral (like in case of sec³(x)) or if I had to do repeated integrations by parts which were getting even more complicated etc.
I had a question?
Is there like a proof to the ilate rule
I agree in not using LIATE, it's not about memorising a mechanic, I like having the openness of realising a mechanic doesn't always work. When I teach integration by parts, I always teach find the function to integrate first
12:50
The smile of cancellation
best cancelling i've seen in a while
0 and 1 are so beautiful!
Excellent video. Before I retired, I told my students that LIATE is a nice "rule of thumb" but, when integration by parts is applicable, LIATE does not work 100% of the time. The example I used to demonstrate this fact is the integral of (xe^x)/(x+1)^2. in this case, the factor to be differentiated is xe^x and the factor to be integrated is 1/(x+1)^2.
I think the answer is e^x/(x+1)+c
@@clementfradin5391 You are correct!
we said U times V - the integral of V du. ... How math geeks make poems
Thanks @richardryan5826 ! This problem helped me!
Another way to deal with sec^3 is:
sec^3(x)= 1/cos^3(x)= cos(x)/cos^4(x)= cos(x)/(cos^2(x))^2=cos(x)/(1-sin^2(x))^2
and with a substitution: sin(x)=u ; cos(x)dx=du
we have the integral of 1/(1-u^2)^2 wich is a rational funcion, so quite easy to integrate.
Is a method that works for 1/cos^n(x)and 1/sin^n(x) for every odd n.
But is that really simpler?
∫ sec³x dx = ∫ 1/cos³x dx = ∫ cos x/cos⁴x dx = ∫ cos x/(1−sin²x)² dx → sin x = u, cos x dx = du
= ∫ 1/(1−u²)² du = 1/4 ∫ ( 1/(1+u)² + 1/(1−u)² + 1/(1+u) + 1/(1−u) ] du
= 1/4 ( −1/(1+u) + 1/(1−u) + ln|1+u| − ln|1−u| ) + C
= 1/4 ( 2u/(1−u²) + ln|(1+u)/(1−u)| ) + C
= 1/2 ( sinx /(1−sin²x) + 1/2 ln|(1+sinx)/(1−sinx)| ) + C
= 1/2 ( sinx/cos²x + 1/2 ln|(1+sinx)²/(1−sin²x)| ) + C
= 1/2 ( secx tanx + 1/2 ln|(1+sinx)²/cos²x| ) + C
= 1/2 ( secx tanx + ln|(1+sinx)/cosx| ) + C
= 1/2 ( secx tanx + ln|secx + tanx| ) + C
The solution above doesn't even include the work required to find partial fraction decomposition.
@@MarieAnne.Hey who are you.
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you really an god gifted child.
Demn bro nice answer.. I will use this sht in my upcoming exam 🗿
I feel like most Mathematicians do this anyways subconsciously. And it is a good first instinct to have to quickly solve problems. But there are cases that don't quite work, and I think it's up to the student to discern for themselves.
I agree with you of doing these subconsciously.
My first class of calculus was in 11th grade in India, that's Junior Year in America. We were always taught ILATE, explanation was "Choose whichever is the harder to integrate" and like you said, it works in almost all scenarios.
Wherever there was an exception we were given the solution and were told the specifics.
It's the first time I'm hearing of DI method. Pretty amazing!!
By parts ke kaam asan ho jaata hai...
Hame
LATEC padhaya Gaya hai
The way I learned the LIATE method was that you should use it when first attempting the question. They said it won't always work, but it can help you start working out the problem.
In spanish we say ALPES:
A=arcsin, arccos, arctan, etc.
L=logs
P=polynomios
E=exponential
S=sin, cos, tan, etc.
It's said that not always work but it has always work for me with this set.
For the integral of xsin(x^2) there is no need of integration by parts. Just U-substitution let u=x^2 so du=2xdx
It becomes integral of-1/2cos(u) and the final answer is-1/2cos(x^2)+C
Yep, and in fact, if you do LIATE integration by parts on this integral, then you'll have to differentiate x and integrate sin(x^2), the latter of which is impossible!
Small brain: memorize this rule
Big brain: recognize the easier integral like a boss
for guys lived in morocco 🇲🇦, they have been using a technic called "ALPES" istead of laatte
A -Arctan ,arcsin,arcos
L - ln
P - polynomial
E - Exponential
S -Sin , cos , tan
the same as ILATE
Bro think exponential are more hader than sin, cos. Think integrating expon: that make terrifying.
This video is great. There's always that tension between "tools that are very helpful most of the time" and "concepts that work nearly all the time," and you balanced this excellently in this video with examples to both sides. 💪
For integral of xsin(x^2)dx in the end card: its easier to do u-sub then by parts
u=x^2 , dx=du/2x
integral of xsin(x^2)dx
=1/2 * integral of sin(u)du
=1/2 * -cos(u)
=-1/2 * cos(x^2)+c
dont forget the C 😅
welldone, it's a right ans. I got it too
@@thekingdragon6660 damn how did i forget the c 😰
Thanks lol
@@thekingdragon6660😂 bro never missing c your are amazing ❤
Moral of the story is just do whatever sounds like less of a nightmare and fuck around if you’re stuck. Liate is just a way to get you moving if you’re a hesitant student. Only way to solve a hard problem is to try solving it, even if there’s no clear solution.
My teacher didn’t like the idea of restricting us to LIATE so we instead subscribed to a some rules of thumb. These are the ones I remember, still use, and are almost always enough to make the right choice.
A) If one of the functions has cyclic (sines, cosines, exponentials, etc.) or terminal derivatives (polynomials), let that one be the one you differentiate. Unless the other function’s integral is unknown or unsolvable by you, in which case…
B) Let the most difficult of the functions that you CAN use integrate be the one you integrate.
I just discovered your videos last week but have only seen ones from about 5 years ago. Having seen them and just coming across this one and the change from no beard to beard is the best thing I've seen in a while
We learned "choose one, and check if you've reduced the mess. Back out if you have not." Would a substitution on u = lnx not help on the 4th case?
It's fun, in France we have ALPES, for arccos/arcsin ; log ; polynomial ; exponential ; sinusoidal. At the time I learnt integration, my teacher said like you that it is better to search which part is better to integrate than use this tip.
Wow that's such a cool acronym!
Well I guess this acronym really… alp-ed you out!
@@goodplacetostop2973 lol
@@goodplacetostop2973 you’re a celebrity on this part of RUclips.
In Mexico we use ALPES too and it's dope
I haven't finished the video, but I'm so much satisfied by the way you change markers (so Smooth dude) 😂❤️
YAY ! Just took membership ! (both channels)
I never seen the LIATE method before. Like you said, I tell the students that they must figure out what to integrate first since finding the derivative is the easiest (sometimes). Unfortunately, many students of mine have trouble doing so.
In the bonus example, just substitute u=x^2
I know the question is what if I try to integrate by parts but:
- maths is about getting to the right answer and spotting the quickest way is always a valuable skill
- according to the rules of logic ("if today's Sunday I'm the pope" according to the late Doctor Richard Maunder, or "if my granny had wheels she's be a bike" according to Gino Cappucino), any answer to your question is correct if I'm not going to integrate by parts. If I integrate that by parts then I'll climb Nelson's Column naked.
For most of our IBP problems, when we have a transcendental function and a power function, if the derivative of the transcendental function becomes algebraic, then it is u, if it stays transcendental, then it is with dv.
-cos(x^2)/2+C will be the answer to the last question. No need to apply LIATE or ILATE. It can easily be done by substitution by substituting x^2=t.
K genieass
Whenever i see x^2 and x together , always do it
@@Katoto112 why do you even care lol
9:06 I tried this integral by first substituting u = 1 + ln(x).
I then ended up win an integrand that looked like
(ue^u - e^u)/u^2 * 1/e
And at that point I had the brilliant idea use the quotient rule in reverse
And then it's just (e^u)/u * 1/e
or just x/(1 + lnx)
THE QUOTIENT RULE IS USEFUL FOR INTEGRALS!!
Yeah exactly. I was looking for this comment. You can also do it without a substitution. lnx/(lnx+1)² = ((lnx+1)-1)/(lnx+1)² = ((lnx+1)(d(x)/dx)-x(d(lnx+1)/dx))/(lnx+1)²
Which is the result when the quotient rule is applied to differentiate x/(lnx+1). Hence the answer for the integral is x/(lnx+1)
I never thought LIATE as a rule or as a method. It's more like an advice from experienced mathematicians, "In most of the integrations, you'll find yourself integrating in this pattern."
I completely agree, you can't exactly impose strict rules for calculus. It always matters the type of problems you are solving and thinking a few steps ahead is the key. I always hated that rule
4:49 we just took this in class today and a student kept integrating by parts till he had about 10 terms because he didn't add the integration of sec^3 x on both sides 😂
It’s actually funny that in my education in Dominican Republic we use the method but the order is different, for us is ILATE, all the letters stands for the same.
I am lucky that I got a good Mathematics Teacher . When we came across this stuff , he taught us this method but suggested not to use it . He told us to integrate by parts by choosing which expression is "easier to integrate" by our intuition. And for real , in some questions even this rule fails(as in takes longer time ). And you sometimes can use different methods to integrate a function that seems harder to integrate but is easier than the other one.
The last line was a twister lmao.
Integration by parts felt impossible until i found this video, thanks!
for the 4th intergral, I can write down the numerator as 1 + lnx - 1, which is the same as (x)'(lnx + 1) - x(lnx + 1)'
therefore, the whole fraction becomes (x/(1 + lnx))', which then nicely cancels out with the intergral sign
I love the multiply by 1 trick for #4.
Adding zero and multiplying by one are like the invisible ninja assassins of mathematics. They come out of nowhere and defeat the problem quickly.
One of my faves is to integrate 1/(1+e^x) by adding zero.
They way he solved the hard integral blew my mind! I've never heard of LIATE, so this entire video is just me learning.
Nice!
I often tell my students, "We artificially adjust things to fit into standard forms we can use."
Especially with situations we are integrating constant/quadratic or linear/quadratic.
In my 6 years of knowing integral calculus this is my first time that I've heard of LIATE. At this point I would trust my intuition more than LIATE.
The way i teach it, its usually usable if there is no composite function involved in the integrand.
At 10:00 assume lnx to be t. dx equal to e^t dt and then apply by parts
THAT'S IT
7:15 how did you know to stop there?
Your videos are so easy to follow, not sure why, it must be a combination of your natural ability to explain the subject matter, along with the use of your whiteboard to write down equations in real time.
For the past 3 years before I started my own RUclips channel, I ever called you Father of Calculus 💪
In the integration of secant cube we can assume there is a x to the power 0 which is ultimately 1 and then do the D I method
when you put +C it makes me happy
another way to deal with the integral at the end is you can make u = 1+ln(x) then x = e^u-1 then its basically just two almost similar integrals that cancel with each other
we can substitute for 1+lnx as t differentiating we get dx=xdt and we know x=e^(t-1) there or getting the integral as e^(t-1)(t-1)/t^2 which is e^(t-1)(1/t-1/t^2) as -1/t^2 is derivative of 1/t we get the result as e^(t-1)/t+c which is x/1+lnx+c after substitution (i used the result int e^x[f(x)+f'(x)]dx = e^xf(x)+c, which we can prove by doing another substitution)
I did something similar.
In my college, they had taught us about the ILATE rule but they also mentioned that it is not always necessary to follow the order and gave us the examples. So we just had to understand which function was easily integrable and which wasn't. I guess it is more sort of, an intuitive method of solving.
I really like the way physicists think about integration by parts -- in direct analogy with (y, Ax) = (A*y, x) ---- this is really the power of integration by parts -- not in finding particular integrals -- like Integral of x^2 Log(x) dx, but in manipulating the action (integral) to obtain deep insight into the physical solutions. This sort of manipulation come up in math classes in ODEs, PDEs and Calculus of Variations.
6:24 secx=under root(1+tan²x)
So eqn becomes underroot(1+tan²x)×sec²x
Let t=tanx then subsitution you also get answer
The feeling being ready before diving to Calculus 2 this upcoming 2nd sem.😌
My teacher teaches me both ways. First is to use the LIATE, but more importantly, she stresses that we need to choose correctly in order for the IBP to work, because she used the same example of integrating sec^3 x in our homework
In fact (on 14:32), I didn't really know about LIATE. But, I am fully agree with Bprp's opinion: "Choose which part is to be integrated first"
for the sex^3(x) qn we can use the result of int root(a^2 + x^2) dx = x/2 root(1+x^2) + a^2/2 ln |x + root(1+x^2)| + c to get quick ans
The integral of the end can be done with a u-sub
I never heard of this LIATE method and maybe it was for the best. Instead of giving us another "trick" to memorize, they just told us to choose which to integrate and which to differentiate.
liate is typical in this question . So,Without using liate answer is comming --(1/2) cos(x^2 ) +c
The LIATE method is essentially good enough to be worth a try, like a kind of "let's see if this works" approach, but it is also important to develop a feel for which parts that would be better to differentiate.
Very cool about your number 4. I remember that video, glad to still be watching!
Another easier way to do #4 is with reverse quotient rule. Since the denominator is squared take it to be the denominator of the antiderivative function and by some guess and check, x is a perfect candidate for the numerator.
We were taught the ALPES stands for
A for the arcsin arc cos etc
L for log
P polynome
E exponential
S sin and cos etc
The sec^3 problem, and in fact all rational functions of trigonometrics of x, can be solved using Weierstrass’ substitution. You wouldn’t have to guess or experiment which factorization is convenient for an integration by parts, since you wouldn’t need any. In general, Calculus teachers would benefit from having seen Risch’s algorithm. In particular, that in integration, trigonometric function don’t really exist. They are just exponentials and logarithms. Another general mantra from the algorithm is that you can integrate all integrals (that have elementary primitives) by eventually reducing to the rational case.
in France we sorta use ALPES which stands for Arctan, Log, Polynomial, Exponential and Sinus/Cos
5:23 actually it helps u need to know an identity of integration of root of x^2 + a^2
For the bonus problem, i think we can use u substitution u=x^2
The third integral came in handy when i was computing distance traveled by a projectile. Glad I did it right
Thank you for the integration by parts techniques and solving problems in calculus.
I tried to use LIATE in the bonus example. So I chose sin(x^2) to be integrated. When I tried to do that found out it becomes soooo easy if you just use the form 1/2∫2xsin(x^2)dx.
thank u so much. you have all my respect. this is what maths teachers should be like; positive, fun and informative.
Another (UK) who hasn't seen this before
As I understand it, it helps in a lot of occasions, but you can't always see when it *won't* work
I have been having no luck learning calc 2, but this method actually makes sense. (I am a bit biased to the tabular method though, we use it in quine-mcklusky models for circuit design)
My teacher told me that DI method doesnt work with logarithms... but I know when to use the 3 stops
For example #4, my first thought was to be suspicious of the denominator being squared and that the numerator had parts of the denominator in it. That immediately reminds me of the result when differentiating a quotient. So, instead of trying to actually do the integration by parts, I would have just played around to try and build a fraction that would give the original integrand when differentiated. You already know the denominator, so it's just a matter of working out what the numerator could be. This example is quite easy.
I don't mention LIATE. Let students explore on their own. If their u fails, then go back and try a different u. My tip for students is to find the u-term first. The u-term in the original integrand is usually the term you want to change or get rid of through differentiation.
Right LIATE rule is (3rd example) assume secx a and assume sec^2x b and solve [a (integral b)-[integral (derivative a).(integral b)]]
K genieass
My Calc teacher taught us LIPET. No idea why trig and exponentials are swapped, but that's what I was taught.
Sneaky fact from complex numbers, trig functions and exponential functions are actually the "same." So in an ordering like this, swapping them makes just as much sense as swapping logs (inverses of exponentials) and the other inverses.
The 4th is amazing !
Let me tell you another approach to the same problem
we assume lnx=t
dx=e^xdx
integral reduces to e^t[ t/(t+1)²]dt
e^t [ 1/1+t - 1/(1+t)²]dt
this becomes d(e^t/1+t) using the product rule and the derivative and integral cancel out.
so we get the same answer substituting t back
My math teacher was very simple about this, "try to find what would be the most convenient way to solve it"
7:06 hi how did you know to stop here? Thanks
In France we learn the ALPES method which means Arctan, log, algebraic, exp and sinus (and cos)
We indians learned ILATE rule
this also we don't use maximum cases
We apply integration by parts
By just checking which is more preferable to integrate
d/dx (x cos x) = cos x - x sin x, which is pretty close to the integral as well.
Very insightful. I like the DI method for IBP.
I guess I should have realized that it was supposed to be the other one when I got very close, but simply conflicting signs that didn't allow me to cancel when i chose lnx as "I"
The hardest integral I solved with integration by parts was x*sin(x)*e^x - I did it with a double D-I-method, idk if it could have been easier
There are two ways for this triple product of x, sin(x), and e^x. One method is to use complex numbers, to rewrite sin(x)*e^x as a sum of two complex exponential functions, which will be i/2*e^([1 - i]*x) - i/2*e^([1 + i]*x). Pull the i/2 out in front as a constant, and assign constants N = 1 - i and P = 1 + i, to simplify our writing. Now you can integrate i/2*x*[e^(N*x) - e^(P*x)], with only one IBP table, after pulling i/2 out in front as a constant, and it is a simple ender. Then convert it back to the real world.
Another way you can do it, is to assign J and K such that:
J = integral sin(x)*e^x dx, and
K = integral cos(x)*e^x dx
Suppose we've previously solved these two integrals. The results are:
J = 1/2*[sin(x) - cos(x)]*e^x
K = 1/2*[sin(x) + cos(x)]*e^x
We'll use J & K, as we construct our IBP table. Let sin(x)*e^x be integrated, and x be differentiated:
S _ _ D _ _ I
+ _ _ x _ _ sin(x)*e^x
- _ _ 1 _ _ J
+ _ _ 0 _ _ integral J dx
Construct our result:
x*J - integral J dx
Since J is a linear combination of sin(x) and cos(x), this means integral J dx is a linear combination of J & K:
integral J dx =
1/2*J - 1/2*K
Thus, our result is:
x*J - [1/2*J - 1/2*K] =
(x - 1/2)*J + K/2
Fill in J & K:
(x - 1/2)*1/2*[sin(x) - cos(x)]*e^x + 1/4*[sin(x) + cos(x)]*e^x
Consolidate and simplify, and add +C:
1/2*[x*sin(x) - x*cos(x) + cos(x)]*e^x + C
sir for 4:55 we can use reduction formula of secx wiz::In=integral sec^nx where n>=2 thenI=(((sec^n-2x)(tanx))/(n-1))+((n-2)/(n-1))In-2
How'd you derive that? The same way he used. Using the standard way is better than remembering these hell long of formulas, I think.
In Spain we have the keyword ALPES:
Arc(sin,cos,tan...)
Logarithm
Powers (x^n)
Exponential
Sin, cos, tan...
on the third example you can do reduction formula right
?
This is why IBP is the last choice in "which integral technique should I use" since the others have tells or signs to look for. IBP is like "meh nah, just figure it out"
Bonus example, you get stuck at integral sin(x^2)dx, but if u=sin(x^2), integral v du is just integral x cos(x^2)dx, which is easy by substitution! I'm stealing your last example for when I teach Int. by parts next.
Someday I want to learn english totally to learn about Calculus 2. It's too difficult for me. TY for your videos
Thx.Ill use it,but you really made me thinking about the 3th and the 4th integral
Thank you professor, for helping me earn my engineering degree, keep up the great work, we are so grateful to you. God bless you sir!
There is somerhing called the reduction formula in integration so reduction formulae is a special case for integration by parts
2:48 integrating with sin in a single go, wow, looks like i will have to learn the DI method ( i already saw this in related vids, but ignored)
I have encountered integral sec^3(x) multiple times in all of science subjects. Mostly in physics. Here and there in Chemistry. Most of the times during maths calculus 2
Never heard of LIATE. Though for a sec that was an weird English abbreviation for “Integrating by parts” because I didn’t learn math in English (also never had an equivalent to LIATE taught to me).
So I thought this video was about why you didn’t teach integrating by parts. Lmao
OMG the integral at 9:05 I just didn't solve it with LIATE I solved it using partial fractions can't believe it 😋😋😋😋 wish u can see my solution it gave the final answer directly 😂😂