Euler Substitution: MIT Integration Bee (3)

[@nicholasleclerc1583]

1:37 “And I *encourage* you guys to [*do it*] [...], but I *recommend* you *do not* [do it]” 😕

[@GreenMeansGOF]

For the trig substitution(1:23), multiply by cos^2(θ)*(1-sin(θ))^2 to numerator and denominator and you get the integral of (sin^2(θ)-2sin(θ)+1)/cos^4(θ) which equals the integral of tan^2(θ)*sec^2(θ)-2tan(θ)*sec^3(θ)+ sec^4(θ). These three parts can be integrated using appropriate u-substitutions.

[@jonasdaverio9369]

I found the substitution by u=sinh^(-1)(x) even simpler. Then you have integral of e^(-2u)*cosh(u)*du between 0 and +infty and it is veery simple

[@glydon-w2w522]

Jonas Daverio What is sinh? Or cosh? Ar they similar to sin or cos?

[@jonasdaverio9369]

Look up on Wikipedia. sinh=1/2(e^x-e^(-x)) and cosh=1/2(e^x+e^(-x) and thus sinh(x)+cosh(x)=e^x

[@glydon-w2w522]

Jonas Daverio thanks

[@Roth-kana]

yeah, it's pretty straightforward that way, instead of whatever solver he's looking up.

[@jonasdaverio9369]

Is that ironic?

[@lakshaymd]

Actually the trig sub one isn't bad at all. You have sec²θ/(secθ+tanθ)² Multiply and divide by (secθ-tanθ)² You get sec²θ(sec²θ-2tanθsecθ+tan²θ) =sec²θ(1+2tan²θ)-sec²θ(2secθtanθ) Use u=tanθ in the first term and u=secθ in the second term.

[@darcash1738]

I never know when to multiply and divide by 1 like this. I've seen it a few times and I don't quite understand all of the cases. What is the main "give-away" exactly, if you can call it that? Eg, are the bottom and top certain kinds of trig functions, is the bottom more complex than the top, etc.

[@lakshaymd]

@@darcash1738 There are several different things that can tip you off, but often, as in this case, it's when you have a sum or difference in the denominator that would look better if it had squares. Sums and differences in the denominator are bad for most purposes and usually you wanna get rid of those first, so that's where you first look. What I mean by "look better if it had squares" is that if you have a+b or a-b at the bottom, you should check if a² - b² is something that you can work with. This is most often the case when a and b are trig functions (because trig identities like cos² + sin² = 1 and sec ² = 1 + tan²) or when a and b have square roots in them (like 5 + 3√x, because then a² - b² is 25 - 9x). In these cases, you multiply and divide by the conjugate of the denominator (ie if the denominator is a + b, you use a - b) which leaves a² - b² in the denominator and pushes all the messy stuff to the numbers where it is easier to handle.

[@darcash1738]

⁠@@lakshaymd epic, I see now. Bc of those sec and tan identities, it does make a lot of sense to try the conjugate, even though it isn’t the classic example with square roots in the denominator. I guess the fact that the whole thing was squared threw me off. Yooooo bro I just realized you can do conjugate from the very beginning I’m having trouble though. How do you evaluate this? 0 to inf: 2/3 x^3+x-ln|x+sqrt(x^2+1)|-xsqrt(x^2+1)

[@lakshaymd]

@@darcash1738 I have no clue. What prompted the question?

[@darcash1738]

@@lakshaymd i did the conjugate without trig sub

[@coolclips101]

Thanks so much for these videos. I'm really enjoying them! :)

[@vorldveace646]

There is a way to use trigonometric substitution using x = tan(v) and have it come out in a nice form. dx / [x + sqrt(1 + x^2)] ^ 2 = sec^2(v)dv / [sec(v) + tan(v)]^2 sec(v) + tan(v) = 1 / cos(v) + sin(v) / cos (v) = [1 + sin(v)] * sec(v) So, sec^2(v)dv /(sec(v) + tan(v))^2 = dv/[1 + sin(v)]^2 We know cos(2v) = cos^2(v) - sin^2(v) = 2 * cos^2(v) - 1 We also know that sin(v) = cos (v - pi/2) = 2 * cos^2(v/2 - pi/4) - 1 substituting that into the equation gives us dv/(1+sin(v)) ^ 2 = dv/(1 + 2 * cos^2(v/2 - pi/4) - 1) ^ 2 = dv / (2*cos^2(v/2 - pi/4))^2 = sec^4(v/2 - pi/4) dv / 4 = [tan^2*(v/2 - pi/4) + 1]*sec^2(v/2 - pi/4) dv / 4 now, let y = tan (v/2 - pi/4) and dy = sec^2(v/2 - pi/4) dv / 2. our new lower bound then becomes tan (-pi/4) = -1, and our new upper bound then becomes tan (pi/4 - pi/4) = tan (0) = 0 [tan^2(v/2 - pi/4) + 1]*sec^2(v/2 - pi/4) dv / 4 = (y^2 + 1) dy/ 2 This definite integral then evaluates to y^3 / 6 + y / 2 from -1 to 0. Plugging in the bounds then gives (0^3 / 6 + 0 / 2) - [(-1)^3 / 6 - 1 / 2] = 0 - (-1 / 6 - 1 / 2) = -(-2/3) = 2/3. Great video as always. Thanks so much for introducing Euler substitution in this video! It's a method I never heard about before and it was really exciting to see you apply it.

[@rodolfodamianchapagonzalez5971]

Damn, I've used that substitution a lot of times, but never had the idea that it has that name

[@anurodhkumar2943]

How handsome the equation looked at the end!!! Superb

[@angelmendez-rivera351]

The easiest method is to simple notice that 1/[sqrt(1 + x^2) + x] = sqrt(1 + x^2) - x, antidifferentiate after the simplification, and evaluating the resulting limit, which is trivial if you use a Taylor series.

[@muhammadmazhari6900]

We could use x=tan(t) and then simplify it to : integral from 0 to pi/2 of 1/(1+sinx)^2 Then use tan(x/2)=u

[@green0563]

I did it using trig sub in 5 lines, got 2/3. In your face! This is how I did it: x=Tanθ, then u=Secθ +Tanθ. du/dx = uSecθ, multiply u above and below, one Secθ is used up. Now from u=Secθ +Tanθ, we get (u-Secθ)^2=(Secθ)^2-1, using (Secθ)^2= (Tanθ)^2 +1 after subtracting Secθ from and squaring both sides. Simplify to get Secθ=(u^2+1)/2u. Put in the original to get the integral of 1/2[1/u^2+1/u^4] with u going from 1 to infinity.

[@adolfhitler7394]

great job using wolfram alpha, you must be sooooo smart

[@manishkumarsingh3082]

I worked out with, secx+tanx=y ....{1} sec^2(x)-tan^2 (x)=1 (secx+tanx)(secx-tanx)=1 (secx-tanx )=1/y .....{2} Subtracting 1 and 2 we get tanx={y/2-1/2y}.......{3} Differentiating { 3 } Sec^2(x)dx=(1/2+1/2y^2)dy Substitute in the integral we get (1/2+1/2y^2)dy÷y^2 It becomes just power form and answer comes out to be 2/3☺ Thx

[@green0563]

I did a similar thing.

[@aksjeofmel]

genius man😳😳😳

[@maf9120]

I did a normal trig sub, just like this guy at the start, and since both the numerator and the denominator were squared terms I "extracted" the square from the fraction and converted to sines and cosines. Then used Weierstrass sub and solved an easy partial fractions problem. Didn't take me very long even though I am rusty and had no idea what I was doing. Not difficult. But as others pointed out an hyperbolic substitution works better.

[@buxeessingh2571]

I don't know where to find a proof, but every Euler substitution can be found -- much more labouriously -- by first using a trig substitution followed by the z = tan(theta/2) substitution.

[@nessfrommother2147]

I think it's called Weierstrass substitution, and Euler substitution does look awfully similar to that.

[@othman31415]

The substitution t=arshinh(x)=ln(x+sqrt(1+x^2)) works like a charm! We have, x =sinh(t) so dx=cosh(t)dt, and that's not all! 1/(x+sqrt(1+x^2))^2 becomes exp(-2t). Overall the integral is now: Integral from 0 to infinity of exp(-2t).cosh(t)=1/2(exp(-t)+exp(-3t)). This integral is just 1/2(1+1/3)=2/3

[@douglaslee6731]

Great explanation

[@fabiansalinas946]

I think perhaps an easier method is to use a sinh substitution for x.

[@LetsSolveMathProblems]

That is also an excellent approach, but I feel the resulting integral may be still difficult to evaluate. If substitution u = sinh x is used, we arrive at integral from 0 to infinity of (cosh u)/(sinh u + cosh u)^2. I do not see any quick way of evaluating this, but perhaps I am not seeing an obvious path. In the video, I wanted to stick with algebraic manipulation, not trig or hyperbolic trig substitutions. Nevertheless, thank you for pointing out an alternate path! =)

[@calcul8er205]

LetsSolveMathProblems sinhu+coshu =e^u :). Expressing the coshu in the numerator in terms of the exponential definition allows you to combine and evaluate

[@glydon-w2w522]

LetsSolveMathProblems brother if we factorise then solution of this problem coming to be 1

[@leif1075]

@@LetsSolveMathProblems Can,you PLEASE ANSWER what,other,more,,intuitive way could,you solve,it..why,not,cos x or,sin,x..it,doesn't have to,be,hyperbolic does it?

[@leif1075]

@conacal rubdur I never said he did..nit confused..just frustrated he ddint use a more intuiitve approach

[@wanghuiyuan7930]

The Euler substitution used is equivalent to using half angle formula to tan x/2 after you first use trig substitution.

[@chinmaydeshmukh9776]

it is very similar to substituting x=cota and then using half angle formula, then substitute tana/2=t.

[@krabkrabkrab]

Whenever I see that denominator I remember 1/(sqrt(1+x^2)+-x)=sqrt(1+x^2)-+x. From there it's simple. You get with x=sinh(u), that it's just integral of cosh-sinh, all squared times cosh. Replace cosh and sinh with their exponential definitions, and finally end up with integral of (exp(-u)+exp(-3u))/2. 2/3

[@dipenthumar3727]

Hey just chill the first method that he dropped due to a scary trigonometric integral can be solved further sec^2x/(tanx+secx) dx in this put tanx as sinx/cosx and secx as 1/cosx. By doing so the sec^2x will get cancelled and then put sinx= 2tanx/2 over 1+tan^2x/2...

[@clownprince6609]

Yeah

[@mdjwy]

I substituted like this, u = x+sqrt(1+x^2). It is similar to the method in this video but simpler to caculate.

[@lostwisdom8900]

Cool video, thanks! I was just asking myself: isn't at 7:07 an indeterminate form and we should study the limit in a deeper way? The result is correct, the limit as x approaches positive infinity of that expression is 0; but usually we should study it carefully, right?

[@martinepstein9826]

"usually we should study it carefully" that depends on your level of experience. Life is too short to always justify your steps as though you're in an intro calc sequence. The reason these classes make you show so much work is that if you don't have a lot of experience your intuition for calculus problems is extremely unreliable. But someone like OP looks at sqrt(x^2+1)-x and immediately thinks "sqrt flattens out so finite differences go to 0" or something like that, and it's not worth it for them to analyze the situation further.

[@vishalmishra3046]

Use y=sqrt(1+x^2). Then [1/(y+x) = y-x]^2 = 1+2x^2-2xy. On integration: F(x) = x + 2/3x^3 - (1+x^2)^(3/2)/(3/2). F(inf)-F(0) = (0)-(-2/3)=2/3=ANS.

[@okiltex]

Thanks! This is new for me!

[@user9287p]

I thought I was ready to tackle this problem. I was not Cx. I'm a high school student and I'm curious to know what kind of Calculus level is required to have the skills to approach this problem?

[@naif277]

you only need the substitution rule for this problem

[@green0563]

There's a trig sub method too, I'm in high school and could solve it because I've seen the derivation of the integral of Secθ, and it uses the fact that if u=Secθ +Tanθ, du/dx=uSecθ. I saw that it might be useful here as well.

[@green0563]

This is how I did it: x=Tanθ, then u=Secθ +Tanθ. du/dx = uSecθ, multiply u above and below, one Secθ is used up. Now from u=Secθ +Tanθ, we get (u-Secθ)^2=(Secθ)^2-1, using (Secθ)^2= (Tanθ)^2 +1 after subtracting Secθ from and squaring both sides. Simplify to get Secθ=(u^2+1)/2u. Put in the original to get the integral of 1/2[1/u^2+1/u^4] with u going from 1 to infinity

[@green0563]

I think it's the not just knowledge of certain techniques, but also that gained by doing many different integrals that helps you approach a new one.

[@informationparadox387]

1:15 Well I solved it by this method with some clever substitution , which didnt even took that much time! But whatever its quite common to intigrate some function with many methods!😙👍

[@AliVeli-gr4fb]

great, thank you

[@rajtagore6047]

Trig substitution gives an answer, its not impossible, its actually quite easy

[@ratnadeeppatra4252]

i got ans just by multiplying and dividing (x - √1+x²)² (but later i had to solve a limit) but I didn't knew this method, so thx.😁😁

[@skwbusaidi]

More obvious subsititusion u = x + sqrt(1+x^2) and isolate x ti get dx . Or much easier let x = sinht Then replace sinht and cosht with thier exponential equivalent

[@lovelygirl3677]

Why is trig sub not going to work? When (sec x+tan x)^-n appears, a useful trick is to rewrite as (sec x-tan x)^n. In this case, it will give the expression (sec^2 x)(sec x-tan x)^2, which can be integrated nicely.

[@ElusiveMind22]

Just tea, thank you!

[@LetsSolveMathProblems]

Of course! I'm glad you enjoyed the video. =)

[@holyshit922]

My advice if we want to remember this substitution 0:39 Lets divide angle complementary to theta with angle bisector then side with length one will be divided into 1 - y and y moreover we wll get another triangle then calculate ratio x/y in this new triangle

[@thecaptainindia9790]

Using trig substitution it is much more easy but you have to do a Lim theta tend to pi by 2 which will be 0....when you did x equals tan t you can multiply divide by tant minus sect whole squared and tant. Square minus sect square is 1 and just all the terms will be easily integrable you can try

[@matheusurbano7045]

Euler substitution?! I never heard of it before!

[@federicovolpe3389]

This is an example of an integral than can be solved much more easily by hyperbolic trig sub rather than by standard trig sub.

[@rylanbuck1332]

When u didn’t flip the bounds with the-1/2 I wanted to scream

[@socialintrovert6790]

very nice problem

[@souravjain2932]

Why can't we just rationalize the denominator.. help me plz @LetsSolveMathProblems

[@bigbrain296]

Yeah I did it that way

[@TONIO-ru4iu]

Yoy can integrate it by using trigonometric substitutions and Weierstrass substitution

[@holyshit922]

To cover all cases we need Euler substitution with real roots when discriminant is grater than zero

[@LetsSolveMathProblems]

Thank you for an important remark. Yes, it is very true that we should watch out for the mindless restrictions on the domain of our function as we attempt to integrate by making substitution. However, in our case above, we are integrating from 0 to infinity, so our substitution is one-to-one and onto with the given function; hence, we do not have to worry about the potential trouble the quadratic substitution may potentially engender. I believe we have covered all the cases in the given explanation. Still, I may be missing a crucial piece of logic. If that is the case, please comment (with a cogent argument) on where exactly I failed to produce a comprehensive explanation, and I will add the content to the video description. Thank you again! =)

[@holyshit922]

I wrote about integral int{R(x,\sqrt{ax^2+bx+c})dx} Yes we have to substitute function one to one and if we have not such function we can try to split interval of integration

[@bazejfiderek8038]

One could also observe that multypling num and den by (x - sqrt(1+x^2)^2) leaves 1 in denominator and then after using formula for (a-b)^2 the only problem is integral of 2xsqrt(1+x^2), which is not that hard

[@blazejfiderek5229]

Sorry, obviously i mean (x - sqrt(1+x^2))^2

[@Frank-xc8ys]

Ahi para integrar solamente se racionaliza el denominador nada mas confundes con mucho proceso

[@anishmathew7593]

I think the the substitution x= sinh(u) , dx= cosh(u) is simple. Sinh(u)= 1/2(e^u - e^-u), Cosh(u )= 1/2(e^u + e^-u) Sinh(u)+cosh(u) = e^u

[@faresberarma3349]

Hi Blackpen Redpen, Great job and funny integral, personally i use the hyberbolic substitution, it lead to the result quickly let x=sinh(t) te denominator become (sinh(t)+cosh(t))^2=exp(2t) the integral become the laplacien transform of cosh(t) wich is equal to p/(p^2-1) here p=2 the result is 2/3 have a nice day

[@LetsSolveMathProblems]

I am actually not blackpenredpen, contrary to a somewhat popular conjecture. =) For this integral, hyperbolic substitution is, as you illustrated, an efficient and straightforward method of finding the answer. Although Euler Substitution is still an excellent problem-solving tool, I personally did not consider employing hyperbolic functions when I created this video.

[@RehanKhan-si1oi]

😂😂😂 @blackpenredpen

[@raphaelmillion]

Ain't hyperbolic trig substitution much easier?

[@royphilip3964]

Great vid, even better accent...but could u help a guy out with a double integration of something that looks like a gaussain function?

[@kartiksharma7166]

At 1.22 you said "it's not going to work out ". You are wrong .. l evaluated this integral using this method ... After substituting x=tant We have. (sec^2t. dr)/(sect + tant ) Then assume sect + tant to be p. sect + tant =p sect - tant =1/p sec^2t-tan^2t=1 And solve like substitution method and you will get the answer ...

[@mortezamodarres2470]

The denominator is (sec(t) +tan(t))^2. This method is very difficult to get to the result

[@3P141592651]

Not very difficult

[@manishkumarsingh3082]

@@mortezamodarres2470 it works with secx+tanx=y substitution I got 2/3 answer

[@manishkumarsingh3082]

Yes bro it works I did the same and got answer 2/3 it's very easy

[@smithpereira459]

Just use (X+sqrt(1+x^2))=t (X-sqrt(1+x^2))=1/t Subtract: sqrt(1+x^2) = (t-1/t)/2

[@ManishKumar-sw8yk]

Sir what is the integration of 1/(a^2cos^2x+b^2sin^2x)^2 ?

[@abdulalhazred5924]

no u

[@xpjs]

Actually, when you do the x=tan substitution you can actually perform a weirestrass substitution :)

[@sunritroykarmakar4406]

PJS you are correct my friend So easy that way

[@txikitofandango]

I did it very similarly and without a sinh sub. Instead of t = sqrt(1+x^2) - x, I used t = sqrt(1+x^2) + x. Square both sides and some magic happens.

[@txikitofandango]

With x = u/2 - 1/(2u), dx = (u^2+1)/(2u^2). Your new integral is the 1/2 times the integral from 1 to infinity of t^-2 + t^-4. Not too bad!

[@xaxuser5033]

use t=tan(x/2)

[@oneinabillion654]

I shall force myself to do normal trig substitution

[@clownprince6609]

Just rationalise it, denominator vanishes...

[@lalitverma5818]

Nice question

[@Frank-xc8ys]

Que pasa si sustituyo t - x = sqrt(x^2 +1)

[@abhijeetpanda4523]

I just multiplied both numerator and denominator with (x-√1+x^2)^2 and I solved the integral but the major problem was the upper and down limits. Can you help me in this?

[@ananyashirakatsi9754]

I did the same thing you did. You quickly find an antiderivative of (2t^3/3+t)-(2/3)(1+t^2)^(3/2). At the lower limit of t=0, this is clearly equal to -2/3. The problem is the upper limit. You need to show that the limit as t-> infinity of this antiderivative vanishes. That way, the result will be 0-(-2/3)=2/3. To see that the upper limit of the antiderivative vanishes, you can multiply and divide the antiderivative by its conjugate expression (2t^3/3+t)+(2/3)(1+t^2)^(3/2). After some cancellation, the numerator of the resulting expression will be a quadratic polynomial, while the behavior of the denominator in the asymptotic limit of large t will be cubic, and so the limit is easily shown to vanish.

[@almanahulzilnicdesuceava5379]

Or you could multiply by[ x - sq(1+x^2)] and it.s easier

[@glydon-w2w522]

david shinigamigt doing that your ans should came -1 as it came in my case ..

[@almanahulzilnicdesuceava5379]

Hmm that means smthing is wrpng

[@kamarinelson]

david shinigamigt first of all, very elegant solution method to rationalize the denominator; seriously. The indefinite integration is easier than when x=sinhu from there but evaluating requires manipulating the indeterminate form (difference of infinities) which I did not bother to do. The answer still comes out to 2/3 tho. My n-spire cas had no trouble verifying once the integral was rewritten.

[@ernestschoenmakers8181]

@@glydon-w2w522 Nope the denominator becomes +1.

[@suleymanmercan389]

At 5.37 isnt that 4t^2

[@meiwinspoi5080]

not very convincing about the bound of the limit at x = infijity

[@rileydoan1194]

I think you got the bottom of the integral wrong at 8:04

[@thegolddog5583]

I think you are correct

[@Guru_Joe_Praise2023]

Euler substitution can work too Sir. ❤

[@bigDjake-t4o]

substituting x to (e^t+e^-t)/2 is a lot easier way to solve i guess!

[@cipherunity]

Terrific

[@Spacetime356]

We can just rationalize it and apply lim (tends to infinity) and approximate using binomial expansion, we land with the same answer..

[@randomname7918]

You have neat handwriting, assuming you're writing with a mouse

[@نعمللوحدة]

The greatest mathematician

[@thermodynamics458]

Yeah, this is crying out for a x = sinh(theta) substitution.

[@utkarshverma1]

Instead of using x=tan(theta), it is much better to use 1/x=tan(theta) since it allows us to factor out an x^2. If we use this sub alongwith double angle identity in trigonometry, we end up with an easy integral. I prefer the trig way to algebra because it is much neater and was faster for me to solve!

[@NotThatSocially]

why can't you just do u=x+sqrt(1+x^2) ? (u-x)^2=1+x^2 u^2-2xu+x^2=1+x^2 (x^2 cancel out, then solve for x) x=(u^2-1)/2u dx = (u^2+1)/(2u^2) du The resulting integral is then very easy to solve :)

[@giuseppemalaguti435]

Bastava moltiplicare sopra e sotto per (x-.....) ^2 molto più semplice

[@Mawang_Jisu]

try t=x+(1+x^2)^(1/2)

[@arvindjha6464]

Just can,t control my laughter You are trying to make this question difficult It is just an one minute question substitute x= cot(b) and u are done

[@snim9515]

Plz explain.

[@sunritroykarmakar4406]

Please shut up lol

[@نعمللوحدة]

U R absolutely right

[@نعمللوحدة]

He's the king of math

[@marvels3011]

Have you lost it?I think you are telling that denominator has (1+cosb).It is (1+cosb)^2

[@colorfulcalculus4526]

I think easiest way is to rationalise the fraction...

[@PhotonAvogadro]

The easiest solution is substituting x+√(1+x^2)=t Trust me, there isn’t a simpler way.

[@rollercoaster9719]

Is ans 2/3.. commenting before watching hope I am correct

[@nootums]

OILer

[@lambdamax]

You are cool

[@uva1312]

math is nuts

[@varadkousadikar9424]

Wow

[@andrewneedham3281]

I definitely disliked your weak justification of "infinity minus infinity is zero." That's not always the case, and there are multiple examples of using various forms of "infinity minus infinity" to prove that one number is equal to a completely different number. You should have shown how, by multiplying your root by a form of 1 equal to a numerator and denominator of the form of its conjugate, root (1 + x^2) + x. When you do, the quantity reduces to 1 divided by stuff in terms of x. When x goes to infinity, now you can justify the expression going to zero. Otherwise, an interesting method with this Euler substitution. I probably would have gone a trig substitution route myself, not having seen this technique before.

[@marvels3011]

substitute x=cot(b) and you are done

[@thapakaji8579]

how about wiesterass substitution? Sorry if i didnt speel his name correctly..

[@alainrogez8485]

For the bound infinity, you have an indeterminate form inf- inf. You should use another way to find out 0.

[@mathforbem]

Very heard😭😭😭😂

[@MrAman47]

I think the easiest way to integrate this is to use Wolfram Mathematica/Alpha

[@kakosullinho302]

it's foda! KKKK

[@leif1075]

You never explain how to think of this in the firdt place!! How? Its,not intuitive or logical in that sense at all! Please elaborate and correct this!