Estonian Math Olympiad

This integral can be computed using elementary techniques but the goal is to find a way to not integrate by using the odd property of the function. It is not immediately obvious that the function is odd, until we apply the properties of logarithms and then investigate.

[@randomjin9392]

For proper integrals: ∫(a, b, x) = ∫(a, b, b + a - x), aka "King's property". So we can replace x with (-1 + 1 - x) = -x. This means ∫ = ∫ln(√(1+x²) - x)dx. Adding this to the original ∫ we get 2∫ = ∫ln(1+x² - x²)dx = ∫ln(1)dx = 0. So ∫ = 0

[@syedahmed8335]

bro 2∫=∫1/2.log(-1)dx with limits -1 to 1 so at the end we get answer as log(-1) so I don't know if it have any value but log (x) domain is x>0

[@randomjin9392]

@@syedahmed8335 re-read the comment.

[@KarlFredrik]

This is a very nice solution

[@Grecks75]

Nice. "King's property"... Didn't know it had a name... 😅

[@nanamacapagal8342]

BEFORE WATCHING: I recognized the inside as some inverse hyperbolic trig, I just didn't know which. I assigned the integrand = u and solved for x, and got x = sinh(u), or u = arsinh(x). Because sinh(x) is an odd function and the region being integrated over is symmetrical around 0, the integral is equal to 0

[@Grecks75]

I guess what you mean is: "Because the integrand arsinh(x) is an odd function and ...". Elegant and efficient answer. ❤ I solved the integral by substitution of x = sinh(t) and integration by parts. Got the same answer and was wondering if the result was by coincidence. 😅

[@nanamacapagal8342]

@@Grecks75 yep. typo on my part, my bad

[@l_a_h797]

@@Grecks75 True, but if f is odd, then the inverse of f is also odd, right (within the applicable domain)? Still, an important detail.

[@Grecks75]

@@l_a_h797 Yeah, it's true in this case (f=sinh, f^-1=arsinh), and also more generally whenever f is bijective so that it has an inverse function in the strict sense (i.e. one that is both a left and a right inverse). The same statement is NOT true when f has only a right inverse. Example: f(x) = sin(x), g(y) = pi - arcsin(y). Now: f is an odd function, and g is a right inverse of f (i.e. (f°g)(y) = f(g(y)) = y for all y in [-1, +1]). But g is certainly NOT an odd function! Therefore, one has to be a bit careful with that statement.

[@Jeremy-i1d]

Thank you for another wonderful and blessed video. I liked your discussion of conjugation. A related point though is that the need for it is avoided by re-expressing the condition for an odd function in the form f(x)+f(-x)=0. Here f(x)+f(-x) = ln(x+root(1+x^2)) + ln(-x+root(1+x^2)) = ln(-x^2+1+x^2) = ln 1 = 0 which establishes the that f(x) is odd and hence that the integral is 0. Again, thank you for your lovely videos and i look forward to seeing the next one. God bless you ❤

[@l_a_h797]

I kind of did the same thing, once the presenter set the goal of showing f was an odd function. But I would say we are still multiplying the argument of ln by its conjugate; we just didn't have to set it up that way intentionally.

[@the_circleguy]

so detailed and precise! I need more of this ! You earned a sub ❤

[@EdiukuMatematica]

I'm from, Angola and i Love you so much, as mathmatic i'm, i used to learning so deep when i watch your vídeos!

[@jay_sensz]

First you should check that f(0) = 0. If that's not the case, you're not dealing with an odd function and checking anything else is a waste of time. In this example, it's easy to check that indeed f(0) = 0. After that, you can take the derivatives of both functions `ln(-x+√(1+x²))` and `-ln(x+√(1+x²))`, which are both equal to `-1/√(1+x²)`. Since the two functions have the same derivative and have the same value at x=0, they must be the same everywhere.

[@Grecks75]

I like your argument about the two functions having the same derivative and the same value at one point.

[@anthropicandroid4494]

Just... wonderful to see an evaluation of a reciprocal unlock that pine knot; thanks for what I'm sure will be good dreams!

[@tessfra7695]

Congratulations on reaching 200k subscribers, sir!

[@EzekielBrockmann]

You're a very clear teacher. Thank you.

[@maxgoldman8903]

If we change this question to an indefinite integral, we can still manage to get a closed-form solution,i.e., the anti-derivative of ln(x+sqrt(1+x²)) exists in terms of elementary functions by doing u= x+sqrt(1+x²) and integration by parts.

[@taito404]

4:49 there is. there actually is an algebraic manipulation we can do and it's really simple: multiply the top and bottom by its conjugate

[@taito404]

oh, that's what you actually did lol

[@nol2521]

If you do x = tan(theta), you also can solve the indefinite integral, you get sec^2(theta)*ln(sec(theta) + tan(theta)) and integration by parts is doable because sec^2 integrates to tan and ln(sec + tan) differentiates to sec

[@fedorkochemasov4533]

Trig sub could be easier in terms of noticing the solution: consider that x = tan t, then sqrt(1+x^2) = sec t and dx = sec^2 t dt. All in all we get an integral from -pi/4 to pi/4 of ln(tan t + sec t) * sec^2 t dt. We know that even*odd is odd, and since sec^2 is even, then we need to prove that the function ln(tan x + sec x) is odd (I won't use t from here, but the idea is the same). Next we note that tan x + sec x = (sin x + 1)/cos x, then by definition of an odd function we need ln((-sin x + 1)/cos x) = ln (cos x / (sin x + 1)) take what's on the right side under the natural log, and multiply the fraction by 1 - sin x, in the denominator we'd get 1 - sin^2 x which is cos^2 x, and in the numerator we'd get cos x *(1 - sin x), cosines divide and we are left with (-sin x + 1) / cos x which is exactly what we have under the natural log on the left side hence the function is indeed odd. This is basically the same approach as yours, but it might be easier to see because of trigonometric identities.

[@Mediterranean81]

5:35 for this We just need to show that ln (sqrt (1+x^2)-x) + ln (sqrt (1+x^2)+x) =0 Using log property (sum to product) We get ln [(sqrt (x^2-1)-x)(sqrt (x^2+1)+x)] Expand = ln (x^2+1-x^2) = ln (1) = 0 So f(-x) + f(x) =0 So f(-x) = - f(x) So the inside of the integral is an odd function That means the whole integral is 0

[@bogydan4223]

I recognized fast that the argument of the integral is just arcsinh, which is an odd function

[@AlexandreRibeiroXRV7]

Another way to see this: let x = sinh(t), the hyperbolic sine of t. We know that: * d/dt (sinh(t)) = cosh(t) * cosh²(t) - sinh²(t) = 1 * sinh(t) + cosh(t) = e^t The integrand becomes ln(sinh(t) + √(1 + sinh²(t))) * cosh(t) dt, which simplifies to ln(sinh(t) + cosh(t)) * cosh(t), which becomes ln(e^t) * cosh(t) and finally simplifies to t*cosh(t) The domain of integration changes from [-1, 1] to [arcsinh(-1), arcsinh(1)], which due to the properties of the hyperbolic sine (i.e. it's an odd function) can be rewritten as [-arcsinh(1), arcsinh(1)]. Finally, given that f(t) = t is an odd function and g(t) = cosh(t) is an even function, their product produces an odd function, which if integrated over any interval of the form [-a, a] where a is a real number yields zero as a result.

[@braindamage7311]

Hyperbolic functions. Be x = sinh(u), solves it easily. Note that cosh(u)^2 - sinh(u)^2 = 1

[@wolfix20021]

Thank you!

[@JoachimFavre]

Here's another method using one of my favourite integration method, the King property (int_a^b f(t)dt = int_a^b f(a+b-t)dt). Applying it here we get that I = int_-1^1 ln(x + sqrt(1+x²)) = int_-1^1 ln(-x + sqrt(1+x²)). Adding the two forms of I we get, 2I = int_-1^1 [ln(x + sqrt(1+x²)) + ln(-x + sqrt(1+x²))] = int_-1^1 ln(sqrt(1+x²)² - x²) where we used ln(a) + ln(b) = ln(ab) and (a+b)(a-b) = a² - b². But then simplifying it, 2I = int_-1^1 ln(1 + x² - x²) = int_-1^1 ln(1) = 0, and so I = 0. Note that this method is more general, it allows to find that int_-1^1 ln(x + sqrt(a + x²)) = ln(a).

[@GreenMeansGOF]

My thought was to replace x with -x and add the integrals. Once adding, the logs combine and we multiply conjugates. It follows that the integral is 0.

[@hardamlimit427]

I tended to freak out but made one step back n got it due to your perfect explanation!!! Thx man 🙏

[@marcinmierzicki1746]

Very nice problem and nice showing 😊

[@arnavthesceintist1149]

Thanks for helping jee students

[@chrisbarrington108]

Thank you… I had a vague recollection, and it is true that the integrand is in fact a standard formula for the inverse sinh(x) (solve (e^y - e^(-y))/2=x)… Since sinh is an odd function, then the integral over a symmetric domain must be zero. I can’t imagine that many 12th graders would see that, but you never know.

[@RezaNamakiyan]

بسیار متشکرم استاد گرامی

[@celinelefevre5056]

Many thanks for all your Videos I am looking from France . A french fan Thierry

[@mathunt1130]

If you know your mathematics, and in particular hyperbolic trig, you'll notice that ln(x+\sqrt(1+x^2))=asinh(x). Sinh(x) is an odd function, and so the inverse function is also odd. As it's integrated over a symmetric interval, this integral is just zero.

[@vivic2939]

Nice video! I also came up with this way using trig sub The expression √(1+x²) made me remember the identity tan²u+1=sec²u, so I tried the sub u=arctan x => x=tan u, such that dx=sec²u du and this gave ∫ln(x+√(1+x²))dx=∫ln(tan u+sec u)sec²u du with bounds from -π/4 to π/4. Now using integration by parts I made w=ln(tan u+sec u) and dv=sec²u du, such that dw=sec w dw and v=tan u, and using the formula we have ∫ln(tan u+sec u)sec²u du=ln(tan u+sec u)tan u|_(-π/4)^(π/4)-∫tan u sec udu The latter integral is of an odd function over a symetric interval, so it's zero, and we can compute that ln(tan u+sec u)tan u|_(-π/4)^(π/4)=ln(√2+1)+ln(√2-1)=ln(2-1)=ln1=0 and this concludes that the integral is 0

[@naakatube]

5:18 MUCH EASIER: bring the right hand negative logarithm to the left, apply logarithm of the sum rule, do the multiplication, and you get ln(1)=0 which is obviously true

[@gurjotdosanjh7828]

4:45 From here, I just set the two expressions equal to each other. I thought that if they are the same expression just altered (like you explained later) I wouldn’t find a solution for x. Instead, I’d get something like 0=0 or 1=1. And that’s what I got. I guess it’s not a formal way but yeah. It worked.

[@mekbebtamrat817]

Substitute x = sinh u. The integrand is the inverse sinh function.😊 The inverse being odd, the integral vanishes

[@makehimobsessedwithyou6412]

how can integrate the function if not proving it is odd?

[@Grecks75]

Substitute x = sinh(t) and perform integration by parts on the result. I did it this way. Easy. 😃 Alternatively: Look up the antiderivative of arsinh(x) in a table of integrals of your choice. It is a standard integral.

[@KrishnaAtri08]

Sir, what comes after Graham's Number. Can you make a video on it please 🙂

[@OlavoLeopoldinodaSilvaFilho]

Just substitute x=sinh(u). the integrand becomes proportional to u.cosh(u), which is odd.

[@mathematical902]

We know that d(ln(x+rootunder(x^2+1)))/dx = 1/rootunder(x^2+1), And we know that derivative of odd function is even, And here we can see 1/(rootunder(x^2+1)) is even so we can say that ln(x+rootunder(x^2+1)) is odd And hence the integration of ln(x+rootunder(x^2+1)) from -1 to 1 is 0.

[@CalculusIsFun1]

Hers how I did it. define A(x) = integral from -1 to 1 or ln(root(1 + x^2) + x)dx B(x) = integral from -1 to 1 ln(root(1 + x^2) - x)dx A(x) + B(x) = integral from -1 to 1 of ln(1) = 0 so A + B = 0 additionally this means A(x) = -B(x) but notice that if A(x) = root(1 + x^2) + x A(-x) = root( 1 + x^2) - x = B(x) so B(-x) = -B(x) for this situation. and since the integral from -a to a of a odd function is zero, B = 0. So A + 0 = 0 so A = 0 so integral from -1 to 1 ln(root(1 + x^2) + x)dx = 0

[@Jason-ot6jv]

This is one of those questions where you say in your head "If this is not an odd function then im screwed" xD

[@adw1z]

It can be noticed that the integrand arsinh(x) is an odd function, and so integrating across a symmetric bound gives 0. But out of interest, we might want to find the area under this function anyways. arsinh(x) = ln(x+sqrt(x^2+1)) If we define: I = ∫[0 to 1] arsinh(x) dx Let u = arsinh(x) ==> x = sinh(u) , dx = cosh(u) du arsinh(1) = ln(1+sqrt(2)) = a arsinh(0) = 0 ==> I = ∫[0 to a] u cosh(u) du = u sinh(u) - cosh(u)]a,0 by parts = a sinh(a) - cosh(a) + cosh(0) = a + 1 - cosh(a) . Indeed: cosh(a) = 0.5exp(ln(1+sqrt(2))) + 0.5exp(-ln(1+sqrt(2))) = 0.5(1+sqrt(2)) + 0.5((1+sqrt(2))^-1) = 0.5 + 0.5sqrt(2) + 0.5sqrt(2) - 0.5 = sqrt(2) ==> I = ln(1+sqrt(2)) + 1 - sqrt(2)

[@muhammadkumaylabbas8513]

Great video!

[@nedmerrill5705]

Very nice!

[@zairon1192]

By doing f(x)+f(-x) , you'll get 0 by using simple logarithm properties and prove f(x)=f(-x)

[@surendrakverma555]

Thanks Sir

[@tonyhaddad1394]

just take e on both side -x + sqrt(1+x^2) = 1/(x+sqrt(1+x^2) 1= (a+b)(a-b) 1 = (1+x^2) - x^2 1= 1 so this equation always hold

[@mihaibriscan3348]

You can also just integrat : S ln[x+√(1+x^2)]dx =x*ln[x+√(1+x^2)] - -S x/√(x^2+1)dx= =x*ln[x+√(1+x^2)]-√(x^2+1) and now if we add 1 and -1 it will pe also equal to 0 but I find this solution easier

[@johnplong3644]

I didn’t get this until college Calculus 11 grade was trigonometry 12 grade was pre-calculus I only got to Algebra 2 in High School.I took Trigonometry my freshman year in College along with College Algebra I took Calculus 1 next semester.All together I had 20 credits in Calculus

[@Shishir27]

We could have done a trigonometric substitution x=tanθ, and then integrate by parts

[@zmaxic]

1. Just used ln(a) + ln(b) = ln(a * b) ln( sqrt(1 + x^2) - x) =? - ln ( sqrt(1 + x^2) + x) ln( sqrt(1 + x^2) - x) + ln ( sqrt(1 + x^2) + x) =? 0 ln( (sqrt(1 + x^2) - x) * (sqrt(1 + x^2) + x) ) =? 0 ln( 1 + x^2 - x^2) = ln(1) = 0 so true, it is odd function. 2. arsh x= ln(x+ sqrt {x^2+1} ) and it is odd function.

[@tgx3529]

I tried to use substitution x= sinh y, there is1= sinh ln(1+ sqrt2), then there Is the integrál ( y* cosh y), by party IT Is zero.

[@SALogics]

Nice explanation

[@antosandras]

int tan(2x/pi) on [-,1,1] is not 0. You should check also that the integral exists.

[@Modo942000]

For such questions, I always like to graph it after to confirm. The fact that the function has Ln made me really skeptical of it too, even though it was just proven to be odd. I tried graphing it and lo and behold, it really is an odd function. Now this is making me wonder how to actually manipulate Ln to make it have similar effects. Is there an even ln function out there?

[@bjornfeuerbacher5514]

Lots of possibilities. E. g. try ln(x²+1).

[@Modo942000]

@@bjornfeuerbacher5514 ok when you actually state it that way, I guess it makes a lot of sense. Guess it was just me being fixated on normal ln(x).

[@world669]

Make 1 as 2nd function and proceed further

[@pojuantsalo3475]

I am VERY surprised the integrant is an odd function since ln(x) isn't even defined for x ≤ 0. That is an odd function in disguise if I ever saw one!

[@82rah]

Very nice Mr. Prime!

[@koonwong8582]

Sub x=sinh(y) will solve the integral too.

[@davidwhiteley3879]

Wow if I learned this in high school advanced program calculus or in college calculus 1, 2, 3, or advanced calculus classes I sure don't remember odd and even functions. I'm not a spring chicken and I have had near fatal brain swelling but still!

[@abdellatifmounsif5352]

Let determinate f(x)+f((-x)=log sqr( x+sqr(x2+1))+ log sqr[-x+ sqr(-x2*1)= log1=0 f(-x)=-f(x) Log a+ log b =log ab

[@tmrapper6378]

This question has also appeared on jee mains exam

[@laitinlok1]

This method would only work with limit and √(1+x^2) + x ≠ 0

[@khalidmahmoodaamir]

To show Ln(-x + sqrt(1+x^2))=-Ln(x + sqrt(1+x^2)) We show Ln(-x + sqrt(1+x^2)) + Ln(x + sqrt(1+x^2)) =0 LHS=Ln([-x + sqrt(1+x^2)] . [x + sqrt(1+x^2)]) =Ln(-x^2 +1 + x^2)=Ln(1)=0= RHS

[@RyanLewis-Johnson-wq6xs]

Integrate[Log[x+Sqrt[1+x^2]],{x,-1,1}]=0 It’s in my head.

[@edoardocabrini5997]

4:50 I think that you can do this ln(-x+(x^2+1)^(1/2))=-ln(x+(x^2+1)^(1/2)) => ln(-x+(x^2+1)^(1/2))+ln(x+(x^2+1)^(1/2))=0 => ln((x+(x^2+1)^(1/2))*(-x+(x^2+1)^(1/2))) => ln(-x^2+x^2+1)=0 => ln(1)=0 that is true for every value of x. Right?

[@herbertsusmann986]

Very sneaky!

[@nickmgls6523]

I did x=tanθ and then integrated by parts but apparently there are smarter ways to solve this 😅. I retried by proving f(x)+f(-x)=0 with some basic algebra which is a much better solution

[@dean532]

87% would’ve just dived into solving it. Hehe

[@childrenofkoris]

Hi sir, do you sell your T Shirt with infinity symbol under your brand?.. i like to have a dark color one, possible grey or dark blue 😊, i want to buy one

[@IslamDarbal]

we know if f is odd f(x)=-f(-x) then f(x)+f(-x)=0 ln(sqrt(x^2+1)+x)+ln(sqrt(x^2+1)-x)=ln(sqrt(x^2+1)+x)(sqrt(x^2+1)-x)] =ln(x^2+1-x^2) =ln(1) =0 Then f is odd so the integral is equal to 0

[@АнатолийЩведов]

Проще взять по частям. Получим ln(2^0.5+1)+ln(2^0.5-1)=0 и интеграл от -1 до +1 от f(x) = x/(1+x^2)^0.5 = 0, т.к. подинтегральная функция нечетная.

[@shreyashkokare2072]

Anyway log function is always odd.

[@yba7586]

I would have just added -ln( x + sqrt(1+x^2)) on both sides. This makes it ln( (sqrt(1+x^2)-x) (x + sqrt(1 + x^2))) = 0. Simplifying with difference of squares makes it ln( 1 + x^2 - x^ 2) which is just ln(1) or 0. Therefore f(-x) = -f(x)

[@giuseppemalaguti435]

I=arcsh1+arcsh(-1)=ln(1+√2)+ln(-1+√2)=ln(1)=0.... comunque i calcoli non servono, è una funzione dispari,perciò I=0

[@mistermudpie]

But what clues you on from the beginning that you should pursue the odd function avenue?

[@sausge6887]

i assume just the fact that the bounds are equal and opposite

[@mistermudpie]

@@sausge6887 that could be misleading, if the problem authors are particularly devious.

[@sausge6887]

@@mistermudpie oh boy, that’s some evil stuff right there

[@nicolasb11]

Wow good job Prime Newton 😂❤

[@holyshit922]

Olympiad ? Try to check if f(-x) = -f(x) otherwise by parts

[@สุธีร์แซ่ฮุ้น]

0

[@Somerandomdood-gk2we]

Please do calculus tutorial i can't do these problems since i am in 9th grade i dont know calculus please taech

[@algoboi]

Well he has many videos on his channel regarding Calculus and Pre-Calculus. Make sure to check out his older videos and other resources if you get stuck while learning. Enjoy your journey learning Calculus.

[@Somerandomdood-gk2we]

@@algoboi from where?

[@jensdittrich8178]

-f(x) = - ln(x + sqrt(1+x^2)) = ln(1 / (x + sqrt(1+x^2)) = ln(1 / (x + sqrt(1+x^2) * (x - sqrt(1+x^2))/(x - sqrt(1+x^2))) = ln((x - sqrt(1+x^2))/(x^2 - (1+x^2))) = ln(-x + sqrt(1+x^2)) = f(-x)

[@MrJPI]

ln(f(x))+ln(f(-x)) = ln(f(x)*f(-x)) ln((x+sqrt(1+x^2))*(-x+sqrt(1+x^2))=ln(-x^2+1+x^2)=ln(1)=0.

[@AmlanSarkar-wr2pr]

First one to comment.

[@amritpatel3794]

Excellent

[@cycklist]

Americans always say "12th grade" or whatever without explaining what that means. You just assume your system is universal. It's not. Can't you just say what age it is instead?

[@m3nny125]

Sorry but coming from a non american, it basically is universal, if anything id argue non-american systems are less universal. because to understand the american system, its simple. if someone is in the Nth grade, then they are in their nth school year in your system (Excluding kindergarten). ie here in egypt we call a 9th grader "middle 3" but you can easily conclude that 9th grader means middle 3 as an egyptian by thinking "hmm ninth grade means someone in their ninth year of school, and a person in middle 3 is in their ninth year of school, so 9th grader must mean middle 3" so its objectively the most universal system since its just counting and doesnt include terms/names (keep in mind im not glazing america or anything i hate that country, i just dont like to hide the truth) if anything age is MUCH harder to figure out since im a 7th grader but 13, but i know that NINTH graders are 13, so same age 2 grade difference. using grades just makes it easier for everyone mate

[@rhvre]

Estonian here. 12th grade is the final grade of high school (i guess in US that would be called a senior year?), students age is around 17-18 at that time.

[@sHiNcHaN12850]

Thank you ! 🎉