This is one of those integrals that just gets way out into the weeds. Multiple substitutions, hyperbolic trig functions. Very challenging. Great job explaining the steps. Especially the ones where someone can easily get lost on.
I love this problem. And, of course, I’ve seen it before. How would a student who has never seen it know what the first move would be? I used to tell my students, “now that you’ve seen me do it, remember the first move!” My students would ask, “how did you know how to do it?” and I would answer, “I saw my professor do it in college!” Lol Anyways, I love your very clear and detailed explanation of a great problem. As always, you amaze with your teaching skills!
There are people who can do it when they see it for the first time, without being taught! But these days as we have Wolfram Alpha, we don't have to manually do any integration😊
I did it like this: I=int(sqrt(tanx)) Now cosider a new integral J J=int(sqrt(cotx)) I+J=Int.(sqrt(cotx) + sqrt(tanx)) I+J=sqrt(2)*Int.( (sinx+cosx)/sqrt(sin2x)) we know that sin2x = 1-(sinx-cosx)² I+J=sqrt(2)*Int.( (sinx+cosx)/sqrt((1-(sinx-cosx)²) Now substitute sinx+cosx=t (cosx+sinx)dx=dt I+J=sqrt(2)*int.( dt/(sqrt(1-t²)) I+J=sqrt(2)*sin-¹(sinx+cosx)+c1 NOW I-J=Int.(sqrt(cotx) - sqrt(tanx)) I-J=sqrt(2)*Int.( (sinx-cosx)/sqrt(sin2x)) we know that sin2x = (sinx+cosx)²-1 I-J=sqrt(2)*Int.( (sinx-cosx)/sqrt(((sinx+cosx)²-1) Now sinx+cosx=t (cosx-sinx)dx=dt (sinx-cosx)dx=-dt I-J=sqrt(2)*int(-dt/sqrt(t²-1)) J-I=sqrt(2)*int(dt/sqrt(t²-1)) J-I=sqrt(2)*ln|x+sqrt(x²-1)|+ c2 J+I=sqrt(2)*sin-¹(sinx+cosx)+c1 Subtract them -2I= sqrt(2)*[lnx+sqrt(x²-1)-sin-¹(sinx+cosx))+c3
That form of the final solution is the most simplified and symmetric form, because you can also express the inverse hyperbolic tangent as a logarithm, and yet another form if you use partial fractions after 2t^2/(t^4+1)
Thanks - you always make it so simple and intuitive. ...A hallmark of a genius-teacher. 🎉❤ A small observation. The first term has + sign and the second term has -. ( I is inv tan exp and the second is hyp as derived. In the last step, by oversight you have inverted u and v. ( Happens to me always over the board😢)...
great solution and also a fan of the handwriting. But can we get answer in the form of natural log instead of inverse hyperbolic tangent. We could use the natural log substitution in form of 1/(x^2-a^2).
I remember once in school, one of us wanted to troll the teacher, so we asked, "what is the integral of e^(tan(x))". While it was a joke, I have sometimes wondered about it. Integral of e^(sin(x)) is a Bessel function of order 0. Integral of e^(tan(x)) shows some interesting, convergent properties. But I never get around to formalising it, only numerically studying it. Would be interesting if we could some day find an analytical expression for that, or just a "special functions" recursive series (I think I have that somewhere).
I had this setup on my exam and I was stuck, I just couldn't figure out what to do and wasted so much time. So after the exam I put this problem into symbolab, since nobody got the answer, and I couldn't beleve the result Thanks for the video : )
Hey sir i hope ur doing well can i ask a doubt after we get the integral as ∫2dt/(t²+1/t²) cant we factor the deno as {(a²+b²) = (a+b)² -(2ab)} SO WE GET 2∫dt/(t+ 1/t)² - √ 2² then just apply the formula so the final answer in terms of t will be 1/√2 {ln [(t+ 1/t)+ √2] / [(t+ 1/t) - √2]} + c
The maestro. Very inteligent your tecnic of solution. The same strategy of solution if the int \sqrt{\cot x}dx? And too \int \sqrt{\sec x}dx? The variable \phy and \theta not same? Here in the Brazil congratulation teacher
This is why they came out with books of tables of integrals! People doing real work want to look it up in a book and not try to derive it from first principles and probably get a sign wrong or something!
Cool but sqrt(tanx) +1/sqrt(tanx) is always >1 (ex: 1.46 for π/6) so you have to use coth−1 instead of tanh−1. It is always necessary to pay attention to the domain of definition of hyperbolic trigo. functions tanh−1 ∈ (-1;1) and coth−1 ∈ (-∞;-1)∪(1;∞)
Because the integral is more complicated than the derivative (the integrand). That is why integration is more difficult than differentiation. Differentiation is just mechanical/algebraic manipulation and simplification, and integration is an art. And many elementary expressions, functions, or integrands don't have elementary integrals/antiderivatives
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
This is one of those integrals that looks "simple enough" when you're taking an exam.
This is one of those integrals that just gets way out into the weeds. Multiple substitutions, hyperbolic trig functions. Very challenging. Great job explaining the steps. Especially the ones where someone can easily get lost on.
I do enjoy your patience and step by step breakdown. Too bad I'm retired and no longer have students to share this with.
Keep it going.
I love this problem. And, of course, I’ve seen it before. How would a student who has never seen it know what the first move would be? I used to tell my students, “now that you’ve seen me do it, remember the first move!” My students would ask, “how did you know how to do it?” and I would answer, “I saw my professor do it in college!” Lol
Anyways, I love your very clear and detailed explanation of a great problem. As always, you amaze with your teaching skills!
Very nice way of explanation nice n clear voice
I think our plan was to get rid of root
@@bravo2992getting to 2t²/(t⁴+1) is natural enough, but the steps after that just seem too complicated for any student to do in the first time imo
That's great !!!
There are people who can do it when they see it for the first time, without being taught!
But these days as we have Wolfram Alpha, we don't have to manually do any integration😊
My God, you're smart and have a gift for teaching. I plan to absorb all you have to give.
I used to do this stuff over40 years ago and it’s amazing to me how much I don’t remember. You just blew my mind.
I've never dived this deep into integrals before and this is probably the most complicated integral ive seen explained so succinctly
I did it like this:
I=int(sqrt(tanx))
Now cosider a new integral J
J=int(sqrt(cotx))
I+J=Int.(sqrt(cotx) + sqrt(tanx))
I+J=sqrt(2)*Int.( (sinx+cosx)/sqrt(sin2x))
we know that sin2x = 1-(sinx-cosx)²
I+J=sqrt(2)*Int.( (sinx+cosx)/sqrt((1-(sinx-cosx)²)
Now substitute sinx+cosx=t
(cosx+sinx)dx=dt
I+J=sqrt(2)*int.( dt/(sqrt(1-t²))
I+J=sqrt(2)*sin-¹(sinx+cosx)+c1
NOW
I-J=Int.(sqrt(cotx) - sqrt(tanx))
I-J=sqrt(2)*Int.( (sinx-cosx)/sqrt(sin2x))
we know that sin2x = (sinx+cosx)²-1
I-J=sqrt(2)*Int.( (sinx-cosx)/sqrt(((sinx+cosx)²-1)
Now sinx+cosx=t
(cosx-sinx)dx=dt
(sinx-cosx)dx=-dt
I-J=sqrt(2)*int(-dt/sqrt(t²-1))
J-I=sqrt(2)*int(dt/sqrt(t²-1))
J-I=sqrt(2)*ln|x+sqrt(x²-1)|+ c2
J+I=sqrt(2)*sin-¹(sinx+cosx)+c1
Subtract them -2I=
sqrt(2)*[lnx+sqrt(x²-1)-sin-¹(sinx+cosx))+c3
@@syed3344damn
That form of the final solution is the most simplified and symmetric form, because you can also express the inverse hyperbolic tangent as a logarithm, and yet another form if you use partial fractions after 2t^2/(t^4+1)
Maravillosa integral y maravillosa solución. Thanks from Spain. !!!!!
Outstanding explanation, you are the guy, I really enjoyed, best regards from Brazil.
This is amazing. Many thanks for this awesome "journey".
Glad you enjoyed it!
Nice clear writing for this interesting integral.
I solved that integral with two maths skills.
1. Using substitucion.
2. Completing the perfect trinomial.
i can finally binge ur videos, as i have just started integration. thanks
From India absolutely amazing sir
Mind boggling ,sir
Keep going your videos are the highlight of my day❤
Sir you are a genius at mathematics thank you
Thanks - you always make it so simple and intuitive.
...A hallmark of a genius-teacher. 🎉❤
A small observation.
The first term has + sign and the second term has -.
( I is inv tan exp and the second is hyp as derived.
In the last step, by oversight you have inverted u and v.
( Happens to me always over the board😢)...
Huh, what an integral. Thanks for sharing. Never stop learning or you not living 👍👌👍I have to watch this many times...
great solution and also a fan of the handwriting. But can we get answer in the form of natural log instead of inverse hyperbolic tangent. We could use the natural log substitution in form of 1/(x^2-a^2).
однозначно, красивое решение. Достойно похвалы
if this video is too long or slow for you, press F12 and type "document.querySelector(".video-stream").playbackRate = 3;" to konsol
Love your dedication BRO keep samshin integrals
Excellent Teacher, congrats
Good math go head for more thank you man 👍👍👍
Different level of problem. Very nice..
You made a difficult integral look easy.
Keep going man!
Love what you do ❤
Wow, that was intense, kinda a detective story to find the suspect (the integral) LOL
You are very smart. God bless you!
I remember once in school, one of us wanted to troll the teacher, so we asked, "what is the integral of e^(tan(x))". While it was a joke, I have sometimes wondered about it. Integral of e^(sin(x)) is a Bessel function of order 0. Integral of e^(tan(x)) shows some interesting, convergent properties. But I never get around to formalising it, only numerically studying it. Would be interesting if we could some day find an analytical expression for that, or just a "special functions" recursive series (I think I have that somewhere).
I actually watch all of your videos in 2x speed lol
your videos are the best!
Great Sir
Thanks for the video sir !
And that was perfect
Thank you for the lesson
I knew that the integral of 1/x^2+a = 1/sqrt(a) .arctan(x/sqrt(a)) + c but not why that was the case, thanks for the video!
I love your stuff man! Love maths. Maths is my "God Zero"!
You are my favorite ❤❤❤❤ bro
Thank you soooooo much!
I was helped a lot by this!
Great derivation, but when tanh instantly turns into tan for v/sqrt(2), at 23:48, you really should have mentioned that correction or edited over it.
I'll have to watch it again to see what you're referring to. Thanks for the feedback.
I had solved this question recently it kinda esy
If you are preparing for competitive examinations
Some integrals require more than 2 or 3 consecutive substitutions or methods to get a solution, and there may be equivalent solutions.
Greetings. Thanks for sharing.
I had this setup on my exam and I was stuck, I just couldn't figure out what to do and wasted so much time.
So after the exam I put this problem into symbolab, since nobody got the answer, and I couldn't beleve the result
Thanks for the video : )
Thank you for the save ❤
Amazing man!
Learned a lot. But why not let u be cos(X) . Then it's sqrt-(lncos(x)) . You can get ride of the negative as cos(-x) is also cos(x).
If I knew it was a better option, I would have used it.
That was wonderful ❤
Thanks this problem was in my text book
Actually this is a very famous question in our board(exam conducts) education system
Hey sir i hope ur doing well can i ask a doubt after we get the integral as ∫2dt/(t²+1/t²) cant we factor the deno as {(a²+b²) = (a+b)² -(2ab)}
SO WE GET
2∫dt/(t+ 1/t)² - √ 2²
then just apply the formula so the final answer in terms of t will be
1/√2 {ln [(t+ 1/t)+ √2] / [(t+ 1/t) - √2]} + c
The maestro. Very inteligent your tecnic of solution. The same strategy of solution if the int \sqrt{\cot x}dx? And too \int \sqrt{\sec x}dx? The variable \phy and \theta not same? Here in the Brazil congratulation teacher
This is why they came out with books of tables of integrals! People doing real work want to look it up in a book and not try to derive it from first principles and probably get a sign wrong or something!
Facts
I solved this question yesterday in my school in one try ✌️
Do you study in college or highschool...you might be genius
how can we write root 2 φ as the result of that integration?
as tanh^2 x+ sech^2x=1
Cool but sqrt(tanx) +1/sqrt(tanx) is always >1 (ex: 1.46 for π/6) so you have to use coth−1 instead of tanh−1.
It is always necessary to pay attention to the domain of definition of hyperbolic trigo. functions
tanh−1 ∈ (-1;1) and coth−1 ∈ (-∞;-1)∪(1;∞)
You are right
Piękny wywód.
Thanks, integral sqrt sen x
Sir can't we use The method of by parts to solve this problem??
Could you differentiate to prove there is no errors? BTW, great job!!!!
I did
Brilliant!
Your 📸 are most recommended
Wow
حلوة ولكن الطريقة طويلة
WOW!
((tanx)^2)/ 2(tanx)^0,5
Damn
O Bruv, why is the answer more complicated than the question itself 😅😅😅😅
Because the integral is more complicated than the derivative (the integrand).
That is why integration is more difficult than differentiation.
Differentiation is just mechanical/algebraic manipulation and simplification, and integration is an art.
And many elementary expressions, functions, or integrands don't have elementary integrals/antiderivatives
-ln(sinX)^0,5
Just another ordinary problem for Jee advance aspirants😂😂
What do you do for a living? Are you a teacher? You’d make a good one