But wouldn't you agree No one would EVER thinknofbthat substitution unless you knew that would work..nkt even Ramanujan or anyone else..you'd have to do it some other way first and somehow discover why that works.
@@leif1075 is that not the case for most subtle tricks? Some poor soul had to try it first and to their delight, find that it works. They can then pass down this knowledge so future people don’t have to spend a large amount of time on trial and error.
@@Mystery_Biscuits But then it's not a matter of intelligence mostly..just dumb luck or luck and maybe persistence. So you are saying he just did a bunch of examples and found a pattern..I don't see how anyone else could find it right? Not even Ramanujan or me or anyone else..
@@leif1075 I’m sure you could ask the same question of Ramanujan or Euler or Cauchy or Euclid or … Asking the greats how they came up with their most creative insights is something I think we’d all like to do but without the ability to time travel, we’re not going to get that chance. While some tricks may look like black magic to us, with the right person in the right time to have the right spark of imagination (and probably a little bit of luck too), some of the most surprising results can come to light.
My favorite substitution when I have a factored quadratic, I.e. sqrt((x-a)(x-b)) is to substitute t=sqrt((x-a)/(x-b)). This takes the quotient of the two factors. The integral reduces nicely to a rational function of t.
Two substitutions I like: first is hyperbolic trig substitution. Basically the same as trig substitution except you use hyperbolic trig functions. Does not really solve anything new, but it means you can avoid using reciprocal trig functions. There's also the weierstrass substitution, which is a crazy substitution I learned for the Putnam. This substitution is t=tan(x/2), which seems like it would be garbage, but it turns expressions with trig functions into fairly easy to work with rational functions.
One of the nicer substitutions I've seen is the "Möbius Transformation" - I think it has more of an application in complex analysis but it's basically a substitution of the form y = (ax + b)/(cx + d) which nicely simplifies some logarithm integrals. It works out nicely because the derivative ends up ad-bc/(cx + d)^2 which isn't difficult to work with with enough cancelling.
This is one of the main methods for solving root integrals in the swedish calculus book i used. Another very helpful substitution is tan t/ 2 = x, which after all simplifications looks very similar to this method and amazingly solves many trig functions
Wouldn't you agree I don't see anyone coming up with these substitutions unless told beforehand..nkt even Ramanujan or me or anyone else..they must have stumbled upon them by chance or done a lot of work to get to get them
Constant of integration forgotten at 13:07 until the last step, absolute values forgotten at 13:13 (but reintroduced in the next step), and the graphic at 13:23 is wrong - the substitution was t = √(x^2 + bx + c) - x. The final answer would also have been a lot neater if polynomial division was done on (t-1)/(t+1) to give 1 - 2/(t+1).
I know the video will be interesting every time Euler is involved. Apart from some mistakes, great video to make us understand how the substitution works. I wish Michael would explain to us how Euler found this substitution.
@@gp-ht7ug I mean once you have found that subtitution is actually helpfull in evaluating integrals you might as well plug whatever function is possible an specificaly the simple ones and see what happens also sometimes you go the opposite way lets say you want the integral to take a specific form then try using a fucntion as sub to get to that form
actually don't go for a trig sub at all in my opinion, if we start the way you did then do a Hyperbolic trig sub, we end up with the same result which is -2arctanh(of that beast).
Use Euler substitution In this integral all three of them are possible Although all three of them are possible the second one (this with constant term) will be best choice for this integral Try this substitution sqrt(x^2+3x+1)=xt±1 You are free to choose + or - sign This is also Euler's substitution
i dont live or study in the states, so I don't know if it is taught there, but I think the weierstrass substitution is a very useful tool. It may be time consuming but it really is helpful.
the answer is even nicer than this video shows cuz Micheal did the pfd wrong. this is a super nice substitution. it's all about that minus x which causes the x^2's to cancel
Hello Michael, a very nice presentation as always. I haven’t seen this Euler substitution before. Another method to solve this kind of integrals is a hyperbolic substitution with x=a*sinh(t) resp. x=cosh(t).
@@MichaelPennMath No problem, thanks for all your help on the channel by the way! These already excellent videos are made even better by your smooth editing :)
I've always wondered... suppose there exists a method to calculate a certain antiderivative that involves some kind of trigonometric substitution. Now if it happens to be the case that there are no trigonometric functions in the integrand, and no trigonometric functions in the antiderivative (so essentially the place where trigonometric functions appear is the substitution, but then they simplify away). Does this mean that there also existed a method that did not involve any trigonometric functions at all?
Hey! It would be really kind of you to listen to this and actually clear my doubt which could’ve been of many people but they just accepted it at school and forgot the why? My question is when we represent numbers like rt(3) on a number line we can easily do that by using pythagoras theorem.Its also very intuitive But, (Now comes the actual part , for the sake of conv. I’ve stated it here)pls make a video for this The things is when we encounter numbers like rt(9.3) oof that number takes a bit of construction like making a line then adding 1 unit to it then making circle marking the point on that circle then taking that point making another curve which finally marks the point rt(9.3) Hoping that you are familiar with it , kindly explain this strange magical seeming concept about circles. Thank you, sir
There are three Euler substitutions leading to integrals of the form ∫R(x, √(ax^2+bx+c))dx, where R(x, √(ax^2+bx+c)) is a rational function to integrals of rational functions. en.wikipedia.org/wiki/Euler_substitution. The lecturer outlined one of them. However, the situation is funny because, as a concrete example, he gave an integral that is easily calculated using an obvious trick without the first Euler substitution. ∫dx/[x*√(x^2+3x+1)] = {x>0}= -∫d(1/x)/√(1+3*(1/x)+(1/x)^2) = {t=1/x} = = - ∫dt/√(1+3*t+ t^2) = - ∫d(t+3/2)/√[(t+3/2)^2-5/4]= { ∫dz/√(z^2 +a)= ln[z+ √(z^2 +a)] +C } = - ln[t+3/2+ √(1+3*t+ t^2)]+C = - ln[(√(x^2+3x+1)+1)/x+3/2]+C.
Great content as always! To the editor, the editing is a bit intrusive (and was wrong near the end!). If there's a way to make it more subtle, it would add a lot more value to the video
@Michael Penn at 13:23, the slide-in of t should be sqrt(x^2 + bx + c) - x. I thought that the sound effects on that slide-in were over-the-top, and the sound effects on the cards at the beginning (like the bell) were a bit too loud imo. Just little things, but I think that could make the video better.
This should be taught in a complex analysis class, not an integrals class, because the substitution is relying on "unfolding the branch cuts" of the square-root, and can't be understood fully without this intuition (although perhaps Euler didn't know this, he probably had an intuition for it also).
Poor choice pf examples..all of them had a better easy substitution possible. Choose something only Euler's substitutions can do to tell its importance like 1/(x+√qudratic) form of things..
This substitutions are still trigonometric in nature, IMHO. The substitution t = (x^2...)^(1/2) -x looks like a transformation hyperbolic coordinate system.
DUDE!!! YOU WILL RIP MY EARDRUMS!! Don't make one video quiet, then another one super loud!!! Decide on the volume you want to use 😉 (ofc. I prefer the louder ones, as one doesn't need to turn it up all the time. And don't worry, I love your videos nevertheless! Peace ♥)
Missing a 1/2 in the partial fraction decomposition. Should end up as ln|t-1| - ln|t+1| + C
My favourite non-standard has got to be the Weierstrass substitution t = tan(theta/2) for integrals of a rational function of sines and cosines
But wouldn't you agree No one would EVER thinknofbthat substitution unless you knew that would work..nkt even Ramanujan or anyone else..you'd have to do it some other way first and somehow discover why that works.
@@leif1075 is that not the case for most subtle tricks? Some poor soul had to try it first and to their delight, find that it works. They can then pass down this knowledge so future people don’t have to spend a large amount of time on trial and error.
@@Mystery_Biscuits Well, sometimes tricks can be achieved by refining other tricks, but you are generally correct.
@@Mystery_Biscuits But then it's not a matter of intelligence mostly..just dumb luck or luck and maybe persistence. So you are saying he just did a bunch of examples and found a pattern..I don't see how anyone else could find it right? Not even Ramanujan or me or anyone else..
@@leif1075 I’m sure you could ask the same question of Ramanujan or Euler or Cauchy or Euclid or …
Asking the greats how they came up with their most creative insights is something I think we’d all like to do but without the ability to time travel, we’re not going to get that chance. While some tricks may look like black magic to us, with the right person in the right time to have the right spark of imagination (and probably a little bit of luck too), some of the most surprising results can come to light.
My favorite substitution when I have a factored quadratic, I.e. sqrt((x-a)(x-b)) is to substitute t=sqrt((x-a)/(x-b)). This takes the quotient of the two factors. The integral reduces nicely to a rational function of t.
The gnarliness goes away quite nicely.
Thank you, professor
Two substitutions I like:
first is hyperbolic trig substitution. Basically the same as trig substitution except you use hyperbolic trig functions. Does not really solve anything new, but it means you can avoid using reciprocal trig functions.
There's also the weierstrass substitution, which is a crazy substitution I learned for the Putnam. This substitution is t=tan(x/2), which seems like it would be garbage, but it turns expressions with trig functions into fairly easy to work with rational functions.
dam... Haven't heard about hyperbolic trig sub, now that's a fancy way to integrate
Actually i studied it at my college class
I like it when Michael shares some closing thoughts on the example at the end before ending the video.
i think thats a good place to stop
yep, saw it also... no big consequence.
At around 12:55, shouldn’t there be a 1/2 coming from the partial fractions decomposition?…which would cancel the 2 outside the integral?
Yes, looks like a mistake
I think u right.
It is in fact an error
Agreed. Too bad, it simplifies the final answer a little.
True
One of the nicer substitutions I've seen is the "Möbius Transformation" - I think it has more of an application in complex analysis but it's basically a substitution of the form y = (ax + b)/(cx + d) which nicely simplifies some logarithm integrals.
It works out nicely because the derivative ends up ad-bc/(cx + d)^2 which isn't difficult to work with with enough cancelling.
I have not heard this tricky substitution befor
I love it
Thanks for this new information
This is one of the main methods for solving root integrals in the swedish calculus book i used. Another very helpful substitution is tan t/ 2 = x, which after all simplifications looks very similar to this method and amazingly solves many trig functions
I would use other Euler's substitution
sqrt(x^2 + 3x + 1)=xt + 1
After this substitution partial fractions will not be needed
Wouldn't you agree I don't see anyone coming up with these substitutions unless told beforehand..nkt even Ramanujan or me or anyone else..they must have stumbled upon them by chance or done a lot of work to get to get them
Hi,
Very nice, I hope I will remember this because sometimes the classic method leads us to complicated formulas.
12:55 : missing 1/2
12:55 you forgot the 1/2
Shouldn’t the partial fraction decomposition have a 1/2 in it?
A/(t+1) + B/(t-1)
At-A+Bt+B
A+B=0
-A+B=1
2B=1
B=1/2
A=-1/2
I'd do x=1/t and the function inside is of the form 1/sqrt(x^2+a^2), which I remember the antiderivative is ln(x+sqrt(x^2+a^2))+C
We can substitute x=1/t to make it more simple
😂😂
Thank you for this great content, I have a small note that I think you forgot to remove the 2 multiplied by the integral in the last steps ❤
طبعا لقد نسي حذف 2
for rational functions in sin(θ) and cos(θ), the substitution u = tan(θ/2) turns it into a rational function in u alone.
That's the Weierstrass substitution. It's in Stewart's Calculus textbook.
I like the Lambert W function although I haven't seen many examples of its use in both differential and integral calculus.
Constant of integration forgotten at 13:07 until the last step, absolute values forgotten at 13:13 (but reintroduced in the next step), and the graphic at 13:23 is wrong - the substitution was t = √(x^2 + bx + c) - x. The final answer would also have been a lot neater if polynomial division was done on (t-1)/(t+1) to give 1 - 2/(t+1).
I know the video will be interesting every time Euler is involved. Apart from some mistakes, great video to make us understand how the substitution works.
I wish Michael would explain to us how Euler found this substitution.
Euler did every integral ever so I wouldn't be surprised if he just intuited it.
I am not so sure. There are three possible Euler substitutions. I don’t think he had three visions
@@gp-ht7ug I mean once you have found that subtitution is actually helpfull in evaluating integrals you might as well plug whatever function is possible an specificaly the simple ones and see what happens also sometimes you go the opposite way lets say you want the integral to take a specific form then try using a fucntion as sub to get to that form
I think it would be good to look into how people figure these things out to begin with.
this is absolutely beautiful. Thanks, Michael!
beautiful sir and very much understable.
actually don't go for a trig sub at all in my opinion, if we start the way you did then do a Hyperbolic trig sub, we end up with the same result which is -2arctanh(of that beast).
If we substitute t=1/x, we will get the form dt/sqrt(t²+3t+1) which now can be easily solved by completing square and taking t+3/2 = rt(5/4) coshθ
My favourite substitution is when symbolab substitutes me in having to solve the problem
Use Euler substitution
In this integral all three of them are possible
Although all three of them are possible the second one (this with constant term) will be best choice for this integral
Try this substitution
sqrt(x^2+3x+1)=xt±1
You are free to choose + or - sign
This is also Euler's substitution
i dont live or study in the states, so I don't know if it is taught there, but I think the weierstrass substitution is a very useful tool. It may be time consuming but it really is helpful.
14:28
the answer is even nicer than this video shows cuz Micheal did the pfd wrong.
this is a super nice substitution. it's all about that minus x which causes the x^2's to cancel
I think 2/(t^2-1) = 1/(t-1) - 1/(t+1)
Hello Michael, a very nice presentation as always. I haven’t seen this Euler substitution before. Another method to solve this kind of integrals is a hyperbolic substitution with x=a*sinh(t) resp. x=cosh(t).
I think you can write this as an inverse hyperbolic tangent at the end, provided you leave the negative on the outside.
Slight editing mistake at 13:30, the substitution had a minus x as well (doesn't change much though this was done correctly on the board)
thank you for the timestamp on that. I'll be more careful.
Stephanie
MP Editor
@@MichaelPennMath No problem, thanks for all your help on the channel by the way! These already excellent videos are made even better by your smooth editing :)
I didn't learn trig sub so i was creativ and played around. If u sub 1/z for x, most of the time the integral gets very simple.
I've always wondered... suppose there exists a method to calculate a certain antiderivative that involves some kind of trigonometric substitution. Now if it happens to be the case that there are no trigonometric functions in the integrand, and no trigonometric functions in the antiderivative (so essentially the place where trigonometric functions appear is the substitution, but then they simplify away). Does this mean that there also existed a method that did not involve any trigonometric functions at all?
Hey! It would be really kind of you to listen to this and actually clear my doubt which could’ve been of many people but they just accepted it at school and forgot the why?
My question is when we represent numbers like rt(3) on a number line we can easily do that by using pythagoras theorem.Its also very intuitive
But,
(Now comes the actual part , for the sake of conv. I’ve stated it here)pls make a video for this
The things is when we encounter numbers like rt(9.3) oof that number takes a bit of construction like making a line then adding 1 unit to it then making circle marking the point on that circle then taking that point making another curve which finally marks the point rt(9.3)
Hoping that you are familiar with it , kindly explain this strange magical seeming concept about circles.
Thank you, sir
Ahh, that's a LOT Eulier for sure.
This channel Is wonderful
Thanks!
there should be a 1/2 in the PFD part
There are three Euler substitutions leading to integrals of the form ∫R(x, √(ax^2+bx+c))dx,
where R(x, √(ax^2+bx+c)) is a rational function to integrals of rational functions.
en.wikipedia.org/wiki/Euler_substitution.
The lecturer outlined one of them. However, the situation is funny because, as a concrete example, he gave an integral that is easily calculated using an obvious trick without the first Euler substitution.
∫dx/[x*√(x^2+3x+1)] = {x>0}= -∫d(1/x)/√(1+3*(1/x)+(1/x)^2) = {t=1/x} = = - ∫dt/√(1+3*t+ t^2) = - ∫d(t+3/2)/√[(t+3/2)^2-5/4]=
{ ∫dz/√(z^2 +a)= ln[z+ √(z^2 +a)] +C } = - ln[t+3/2+ √(1+3*t+ t^2)]+C = - ln[(√(x^2+3x+1)+1)/x+3/2]+C.
Is there a nice Book with such exercises, that offer good solutions? That seems like quite the useful trick for integrals.
I just did an Euler substitution on my car...rotated the tires, too.
Very nice
Great content as always! To the editor, the editing is a bit intrusive (and was wrong near the end!). If there's a way to make it more subtle, it would add a lot more value to the video
How was the edit wrong "in the end" and what is the timestamp? Also, what was intrusive?
Thanks
Stephanie
MP Editor
@Michael Penn at 13:23, the slide-in of t should be sqrt(x^2 + bx + c) - x. I thought that the sound effects on that slide-in were over-the-top, and the sound effects on the cards at the beginning (like the bell) were a bit too loud imo. Just little things, but I think that could make the video better.
@@MichaelPennMath I thought that the reminder at this point was useful.
This should be taught in a complex analysis class, not an integrals class, because the substitution is relying on "unfolding the branch cuts" of the square-root, and can't be understood fully without this intuition (although perhaps Euler didn't know this, he probably had an intuition for it also).
You don't need to fully understand a substitution to appreciate its power. The algebra appears cumbersome but simplifies nicely.
that's crazy!
Can you please continue the olympiad series? love your content Micheal !!
13:30
Isn t there a minus x?
yes there should be
Weierstrass substitution could be usefull sometimes.
sir finally it it should have been 2*1/2*ln(t-1/t+1)= ln(t-1)-ln(t+1)
Why does no one use arcoth(x)?
How would you integrate this using trig sub? I am getting stuck at the integral of 2sec(θ)/(sqrt(5)sec(θ) -3)dθ.
Anybody knows a book where I can find this method detailed?
In Europe, we don't learn trig subs, we learn these Euler subs.
These substitutions are actually very common in jee type integration questions
That's why I couldn't understand it 😢
Favorites ..
Be careful with negative values because x^2 + 2xt +t = |x^2 + bx + c|
cool
Nice
HOW or WHY would Euler come up with these to begin with??
Poor choice pf examples..all of them had a better easy substitution possible. Choose something only Euler's substitutions can do to tell its importance like 1/(x+√qudratic) form of things..
I will be using trig substitution or i will be unable to complete the paper.
Nice :)
Is the character chalk in the descriptions all the different colors or what? Lol
Well obviously!
Stephanie
MP Editor
This substitutions are still trigonometric in nature, IMHO. The substitution t = (x^2...)^(1/2) -x looks like a transformation hyperbolic coordinate system.
Isn't hyperbolic the opposite of trigonometric (in a sense)? lol
@@TheEternalVortex42 both of them are aspects of conics, so, they are all related by coordinate transformations.
DUDE!!! YOU WILL RIP MY EARDRUMS!! Don't make one video quiet, then another one super loud!!!
Decide on the volume you want to use 😉
(ofc. I prefer the louder ones, as one doesn't need to turn it up all the time. And don't worry, I love your videos nevertheless! Peace ♥)
it's getting harder and harder to read the thumbnails. be careful not to clutter too much