Sorry for the reupload. I actually had a mistake on the integral of 1/(x^2-a^2) in the previous video. I will make up to you guys by checking my answer by differentiation! That video will be done soon!
I think my professor summed up integration in a nice way. He said differentiation is all about technique. You see a scenario and have a set of rules you then follow. Integration is a form of art. It's much more intricate and delicate.
According to the question. For example, if the question is to find an original function for the next function, the answer without the constant is correct. But if the question is assigned to all the original functions. It must make +C
"Welcome to the Salty Spitoon, how tough are ya?" "How tough am I? I just integrated a trig function!" "Yeah, so?" "integral(sqrt(tan x))dx" "Uh, right this way..."
Me when finding this channel: "wtf is going on?!" Me rewatching 1 year later after having seen every bprp video: "alright easy didn't even need the DI setup"
I'm preparing a transfer exam for Korean universities and there was this question on my preparation problem set. Your solution was so helpful brother, thanks a lot!
I passed my semester of Calc 1! I did not fully understand everything but I believe I got a strong majority of it & I will be working on some of my weaknesses during break to prepare for Calc 2. For some reason when we got to U-substitution, everyone was confused but it seemed to make sense to me just based off the few example she gave in class and somehow that was all I needed. Anyway, so during last class before our final, she did a little review & answered questions. Someone asked about this sqrt(tanx) and she was like "oh you cant solve this with your current toolset, wait till next semester" and I was like "I can solve it" because I had all this false confidence from having understood the whole U-sub stuff. Well turns out this is a VERY difficult one lol. I guess if you are a student and want to be prepared for integrals, just spend a week studying just THIS one integral and you should be good to go lol.
Me too, I watched the whole thing wondering what was going to happen next! Really a great calculus problem. I'm going to show it to all of my students!
@@MelonMediaMedia im guessing they give you the space because with an A4 blank paper this would quickly turn messy for me as my handwriting is pretty big
Integration is just the easiest thing ever... I can integrate √tanx + e^x² in seconds: Set up the integral: ∫ √(tan(x)) + e^x² dt And then just use the "inverse" power rule: (t)√(tan(x))+ (t)e^x² And we're done... I didn't say that I'ld do it with respect to x...
I like this integral and its anti-derivative. I am impressed with your technique of using trigonometric and u-substitution along with algebraic manipulation to arrive at the answer.
Ok, so in the beginning we have a pretty simple mathematical expression while the result in the end is something horrible. Things tend to evolve naturally from more complex to simpler. I therefore consider it normal to leave this formula unintegrated. Thank you for all your thumbs up ! :D
I just differentiated this myself, it starts out ridiculously complex, but it slowly starts to fit in with everything, good luck on doing it! You might need 6 boards to do it, I managed to do it in 1 board, but I had to rub out a lot of the work out I did
I majored in physics in school and always preferred the more pure abstract mathematical parts of it. Watching this video is like taking a mental vacation back into the past. I'm happy that I was able to follow it through to the end on my first viewing :)
Slightly more straightforward (although much longer/messier): Factor x^4 + 1 = (x^2 + sqrt(2)x+1)(x^2 - sqrt(2)x+1), then do partial fractions, complete the square in the denominators and solve. This saves you from having to figure out the trick where you add 1/u^2 and subtract 1/u^2.
I think that the most monumental achievement humanity could ever hope to accomplish would be to find an intuitive understanding of how to go directly from the integrand to the antiderivative in one step...
@@blackpenredpen 4 years later you are still inspiring people. I'm only 14 right now, so don't have any solid plans for uni etc. (other than studying CS as I like programming) but now you got me interested in maths! I've been spending my afternoons just trying to learn maths for the past few weeks, and it's been really fun so far!
So... u substitute for fx Square u = sqrt (tanx). See that 2udu = sec^2(x) dx. Squaring u^2 = tanx gives us tan^2(x) = sec^2 (x) -1 = u^4 which we can see as sec^2 (x) = u^4 + 1. Thus dx = 2udu/u^4 + 1. Plug in original equation to have integral of u*(2u/u^4 + 1)du. Multiply top and bottom by 1/u^2 to get complex fraction with sum of squares in denominator. Complete the square to get (u + 1/u)^2 - 2, which has derivative of inside equal to 1 - 1/(u)^2. Now we want two integrals (why? it is not clear unless you see the tanh^-1 and tan^-1 option coming up), one with 1 - 1/(u)^2 in numerator and other with 1 + 1/(u)^2 in our numerator. Because the completed square can have two forms we can have the appropriate denominators to do two more substitutions, this time with t and say w. If we do the substitutions correctly we have two integrals, one being of 1/(t^2 - 2), and our other being of 1/(w^2 + 2), both in their respective worlds. A formula exists for these forms to be integrated neatly into tanh^-1 and tan^-1 forms. Substitute u back in for t, w, and sqrt (tanx) for u. Do this correctly, and then Add c and we're done. I did this mostly for my own understanding, but I'm fairly sure I didn't skip too much for it to act as a quick summary.
I passed calc 2 this summer and then I looked at this video because I couldn't figure this out on my own. I heard "Hyperbolic tan" and had a mental breakdown and screamed aloud, "SINCE BLOODY WHEN IS THERE A THING CALLED HYPERBOLIC TANGENT!? THEY'RE JUST MAKING NEW THINGS TO MESS WITH ME!" followed by expletives and crying.
I've seen this video a couple years ago but decided to comment only now. First of all, very nicely done! Without trying to diminish guy's effort and all the excitement of the viewing public, I just wanted to remark that from the view of pure math this is absolutely worthless. Here's why: In all applications (including physics, engineering, and even math) ALL integrals are definite. This integral would be of interest only if integral is from, say 0 to pi/2. In couple of steps now I'll solve the problem of integrability and the value of that integral. First, simple substitution v=pi/2-x translates this integral to the integral of \sqroot(cotv) over the same integral. Second, cotv is approximately inverse the of v for small v, so the question is \sqroot(1/v) integrable, and the answer is, of course yes. And we are done! If somebody needs the numeric value, just take integral of \sqroot(1/v) from 0 to 0.01 and then calculate the remaining part from 0.01 to pi/2 by whatever method of approximate integration.
Couldn’t you have just done tan inverse with the original u equation after you divided the fraction by u^2, or does it only work with one variable and one constant term?
Or you can do integration by parts! (i paused at 2:40 to try myself) First, use substitution to get integral of sqrt(tan(x)) = integral of u*2udu/u^4+1 Now do integration by parts to get, integral of sqrt(tan(x)) = u*(integral of 2udu/x^4+1) + integral of integral of 2udu/x^4+1 Use substitution again (set z=x^2) to get u*arctan(x^2) + integral of arctan(u) What's the integral of arctan(u)? it's u*arctan(u) - ln(1+u^2)/2 So we have integral of sqrt(tan(x)) = 2u*arctan(u) - ln(1 + u^2)/2 and substitute u = sqrt(tan(x)) back to get the answer: 2sqrt(tan(x))arctan(sqrt(tan(x))) - ln(1 + tan(x))/2 Now I am going back to the video to see how you did it and see if I'm right or not!
We can solve this in some other way too. We can write sqrt tanx as 1/2 {(sqrt tanx + sqrt cotx)+ (sqrt tanx - sqrt cotx)},and then break these into sin cos expressions,and then subtitute sinx + cosx = t and sinx-cosx = k in the first and second integrals respectively,and then apply the standard integral formula. Anyways love you process too!
If you plug in 0.5 for the integral function, then (sqrt(tan 0.5)+sqrt(cot 0.5))/sqrt(2)= 1.4793, but arctanh(1.4793) is undefined. The range of (sqrt(tan x)+sqrt(cot x)/sqrt(2)is alway greater than 1, which makes arctanh(sqrt(tan x)+sqrt(cot x)/sqrt(2)) always undefined for this integral function.
DougCube But integral from 0.2 to 0.5 of sqrt(tanx) is defined and real from the graph of sqrt(tanx). And all the result of this integral function is complex or undefined.
I did the integral in a slightly different way. I ended up with (sqrt(2)/4) * log(abs((1/2 + (sqrt(tan(x)) - sqrt(2)/2).^2) / (1/2 + (sqrt(tan(x)) + sqrt(2)/2).^2))) + (sqrt(2)/2) * atan(sqrt(2*tan(x)) + 1) + (sqrt(2)/2) * atan(sqrt(2*tan(x)) - 1). I also let u = sqrt(tan(x)) but in of multiplying through by 1/u^2 I added and subtracted 2u^2 from the denominator and then factoring the result using difference of squares. Once I had it in a factored form, I used partial fractions.
Finally! Loved the video! Is great! You could do the same integral for the 100k, but now you do it the hard way! (The one with partial fractions and with the factorization of u^4+1, and with the natural log result!)
At 9:59, I understand why you add the second fraction in, but doesn't the second fraction need the same denominator as the first to make it equal to the fraction on the line above?
I have a question: Does the identity x^2 - a^2 = x^2 + (ia)^2 in combination with the integral identities involving arctan(x) and arctanh(x) imply i•arctan(-ix/a) = -arctanh(x/a) ? Here is my reasoning: ∫1/(x^2 + a^2)dx = (1/a)arctan(x/a) + C, therefore ∫1/(x^2 + (ia)^2)dx = (1/(ia))arctan(x/(ia)) + C But, x^2 - a^2 = x^2 + (ia)^2, so ∫1/(x^2 + (ia)^2)dx = ∫1/(x^2 - a^2)dx = = (1/a)arctanh(x/a) + C Therefore (1/(ia))arctan(x/(ia)) = (1/a)arctanh(x/a) Multiplying the argument of arctan by i/i and multiplying both sides of the equation by ai^2 we get i•arctan(-ix/a) = -arctanh(x/a). Is this right? Am I wrong to cancel out the constants? Thanks for your help and thanks for the great videos!
Sorry for the reupload. I actually had a mistake on the integral of 1/(x^2-a^2) in the previous video. I will make up to you guys by checking my answer by differentiation! That video will be done soon!
One rather interesting thing about that expression is that (if we take C = 0) we'll generally get a complex number out of it. The imaginary part is always the same though. I think you get a nice expression if you introduce complex numbers (starting by factoring (u^4 + 1) into (u^2 +- i). But it seems non-obvious that 1) the manipulations I did after that are totally ok, since I kind of ignore principle value etc... and 2) how to rigouously show this is actually equal to the answer found with real math. Writing R = (1 + i) / sqrt(2) and u = sqrt(tan(x)) (and Re == real part) then I eventually got that Re[2R*arctan(Ru)] is an antiderivative of sqrt(tan(x)) w.r.t. x Writing out the def. of (complex) arctan => Re[(1/R) Log [(R - u) / (R + u)]] is an antiderivative. Numerically checking for some random values I find this does give the same values (up to a constant) as your solution. Though I did assume that tan(x) is non-negative here (that is how I ended up with "real part" in there).
There is simple method (I Feel) put x=(pi/4-t) dx=-dt and integral become sqrt(1-tan(t))/1+tan(t)) after rationalising numerator we get (1-tan(t)/sqrt(1-tan(sq)t) which is (cos(t)-sin(t))/sqrt(cos(sq)t-sin(sq(t)) now split the integrals. Cos(t)/sqrt(1-2sin(sq)t) other integral is sin(t)/sqrt(2cos(sq)t-1) by sqrt(2) manipulation we get u/sqrt(1-u(sq)) other u/sqrt(u(sq)-1) both forms are familiar having formulae
All depens on interval for x. If this integral is for x in (0.5;1) you can use sqrt(tg)=1/(sqrt (cotgx x) And use substitution y=sqrt(cotgx), than you have short solution
About the integral answer with inverse tanh and so on: is this integral defined anywhere as a function? A trip to Desmos seems to indicate that this function has no real values, and plugging in several of the usual trig values yields "undefined" no matter where I look.
What a gorgeous way to find the answer! Who is the genius first to find u+1/u and u-1/u pairs for this question? It really drive me crazy for such an integral. Thank BPRP.
Very symmetrical so much so with all those arctans and tan x's you could combine them except for that little h in the inverse hyperbolic tangent it stands for Hell in this integral bc you can't take tanh^-1(n) ||n|| > 1, ~|tan x| + |cot x| w/o screwing up the rest of it unless it's some kind of cubic solution with complex b coefficient ~ 1/3 ln (i) Point it seems so close to y = f(x); f^-1(y) = x, x could be 0-2π with no trouble you would run into complex numbers but they cancel themselves out ish
Hola, gracias por la explicación. Me parece que hay un problema con la solución: El dominio de la función arcotangentehiperbólica es el intervalo abierto (-1,1); por otro lado los corchetes que suceden esta función contienen una función cuyo recorrido o rango es algo así como el intervalo que va de 1.39... hasta infinito. De esta manera la antiderivada no se puede evaluar para ningún valor de x. ¿Es así?... Gracias. Hello, thank you for the explanation. I think there is a problem with the solution: the domain of the function hyperbolic arctangent is the open interval (-1,1); on the other hand, the square brackets that follow this function contain a function whose route or range is something like the interval that goes from 1.39... to the infinity. This way the antiderivative cannot be evaluated for any x value. is that so?... Thank you
Tell me if I’m wrong. When you had the choice beetwin partial fractions and tan-1, tanh-1 Wouldn’t you get 2 different results? One in ln, one in tanh ?
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
Sorry for the reupload.
I actually had a mistake on the integral of 1/(x^2-a^2) in the previous video.
I will make up to you guys by checking my answer by differentiation! That video will be done soon!
Np. it is fine
This one's more worth the storage area of google than a dozen 9 percent of the content of YT.
blackpenredpen Thanks
Did you have to redo the video? or just edit the definition that showed on screen
How do we factorize (u^4 + 1) ? plz
you know something is hard when blackpenredpen uses five colours
I only noticed 4 colours
@@seshnarayan7972 blue black red green purple!
@@spooky2526 Where is purple?
@@createyourownfuture5410 he mentioned he uses purple for the second part of observe section (10:39) although it doesn't look that diffirent from blue
I think my professor summed up integration in a nice way. He said differentiation is all about technique. You see a scenario and have a set of rules you then follow. Integration is a form of art. It's much more intricate and delicate.
TheDaltonGillespie I totally agree!!! Has he done this integral with u guys?
My Further Maths teacher calls integration Black Magic. Two kinds of people I guess
Integration is definitely more difficult, and thus satisfying
I agree. It takes a clever mind to do differentiation problems with ease. But it takes a creative mind to do integration problems.
AGMT further math? Are you international baccalaureate?
Imagine doing all this and then forgetting the +c
Hamish Blair lol
You would get zero marks in exam😂😂😂😂
Edit the video
It actually happened to me in an exam :(
According to the question. For example, if the question is to find an original function for the next function, the answer without the constant is correct. But if the question is assigned to all the original functions. It must make +C
"Welcome to the Salty Spitoon, how tough are ya?"
"How tough am I? I just integrated a trig function!"
"Yeah, so?"
"integral(sqrt(tan x))dx"
"Uh, right this way..."
Lol
"I just integrated 1/(x²+tan(x)) dx"
∫√t̅a̅n̅x̅ dx
@@jakepfeiffer6577 WHERE. DID. YOU. GET. THAT. INTEGRAL. SIGN. FROM?
@@createyourownfuture5410 on Mac it’s just option + b
Idk about windows but you can copy/paste it
Mr Math Man.
Math Me a Man.
Make him integrate the square root of tan.
stealing my joke, still love you though babe
"1+1 is 2, right?"
Calculus, everybody.
Yes. But in Z2, 1+1=0.
well 1+1 is 1 if you account for boolean algebra :)
You know you're high on math(s) when you forget what 1+1 is.
1+1=10 in binary
1+1=3
Me when finding this channel: "wtf is going on?!"
Me rewatching 1 year later after having seen every bprp video: "alright easy didn't even need the DI setup"
this is pure game. There are very very few maths teachers at level of you. Thank you.
Wow..you demonstrated this before 100k subs more than 3 years ago, today you have 687k. Wishing you for the next 313k
Thanks
@@blackpenredpen but what is special about the number 313
?
@@parkeryoung2471 it's prime
@@parkeryoung2471 to make it a million subscribers
I'm preparing a transfer exam for Korean universities and there was this question on my preparation problem set. Your solution was so helpful brother, thanks a lot!
And what you doing? In uni?
@@Awai_quotes Electrical and Computer Engineering, but I think I failed the exam. I might just quit and make indie game
@@프로틴요플레 So...how did it go for you?
I just saw you instagram reel, where you give credit question and searched for this integral on RUclips.
And I am so glad I found your video 😀!
😆
I passed my semester of Calc 1! I did not fully understand everything but I believe I got a strong majority of it & I will be working on some of my weaknesses during break to prepare for Calc 2. For some reason when we got to U-substitution, everyone was confused but it seemed to make sense to me just based off the few example she gave in class and somehow that was all I needed. Anyway, so during last class before our final, she did a little review & answered questions. Someone asked about this sqrt(tanx) and she was like "oh you cant solve this with your current toolset, wait till next semester" and I was like "I can solve it" because I had all this false confidence from having understood the whole U-sub stuff. Well turns out this is a VERY difficult one lol. I guess if you are a student and want to be prepared for integrals, just spend a week studying just THIS one integral and you should be good to go lol.
Oh man, I love this video. I watched the entire thing and enjoyed every second of it! Keep up the good work on your channel. :)
DGCubes what are YOU doing here??¿
What can I say, I like calculus. :P
thank you DGCubes!!
Me too, I watched the whole thing wondering what was going to happen next! Really a great calculus problem. I'm going to show it to all of my students!
BOI!
Evil integral to place on an exam ... :/
Would have to be like the only question or 1 of 2 questions.
I'm from India, and i am practising for this exam, I can assure you, there are more brutal questions.
@@MelonMediaMedia yeah for Indiana this is ez or normal
@@MelonMediaMedia im guessing they give you the space because with an A4 blank paper this would quickly turn messy for me as my handwriting is pretty big
This one is an easy problem
Integration is just the easiest thing ever... I can integrate √tanx + e^x² in seconds:
Set up the integral:
∫ √(tan(x)) + e^x² dt
And then just use the "inverse" power rule:
(t)√(tan(x))+ (t)e^x²
And we're done...
I didn't say that I'ld do it with respect to x...
Viktor Sundström LOL, or you could use horseshoe integration, Indeed one of the most powerful mathematical tools out there.
Still forgot the +c dude 😂
Sólo los pendejos dicen que está facil resolver un problema.
@@neilshah754 *depression intensifies*
@@cesarturanzasfarill2976 calmate viejo, era un chiste. No lo entendiste??
14:41
"This is the two, so what should I do?" What a rhyme :O
15:02 WIZARD! Where can I buy this magic blackpen??
WTF! I didn't even notice it before!! How did he do that?!
He edited it because he forgot the (-) sign
Haha I understand that
Just a silly comment ;)
Morgan Mitchell Johnson
hahaha i also noticed after you comment.....
When you do all of this in the exam and you realize the integral was sqrt(tanx + 1)
god damn I love math
Me too!!!
You, my good sir, are turning calculus into art! Awesome video!
I LITERALLY LOVE YOU SO MUCH YOU DESERVE THE WHOLE WORLD
The most difficulty integral that i ever seen in my entire life! But it was really good xD
Thank you : )
And..... there's the integral of cbrt(tan(x))
@@blackpenredpen imo that’s easier than this
Very well work dude, the resolution was easier than I thought, I had problems when using trinomio. You got a like and a new subscriber!
I like this integral and its anti-derivative. I am impressed with your technique of using trigonometric and u-substitution along with algebraic manipulation to arrive at the answer.
Ez annyira bonyolult, hogy nincs értelme ennyit számolni, de blackpenredpen igazi zseni !
This turned to blackpenredpengreenpenbluepenpurplepen real quick
😆
Simple presentation of the difficult integral in a nice manner . Thanks .
Ok, so in the beginning we have a pretty simple mathematical expression while the result in the end is something horrible. Things tend to evolve naturally from more complex to simpler. I therefore consider it normal to leave this formula unintegrated. Thank you for all your thumbs up ! :D
10/10 what a trip
Yea, I know!
I just differentiated this myself, it starts out ridiculously complex, but it slowly starts to fit in with everything, good luck on doing it! You might need 6 boards to do it, I managed to do it in 1 board, but I had to rub out a lot of the work out I did
Differentiating sqrt(tanx) is not hard at all😂
@@ramking7869 he is about differentiating the antiderivative of sqrt(tanx)
Brilliant. So many techniques in one shot.
This is lengthy problem. Very few can solve in first time. We can only solve this problem if we practice at home. Your explanation is very nice
10:54 defenitely most important part
I majored in physics in school and always preferred the more pure abstract mathematical parts of it. Watching this video is like taking a mental vacation back into the past. I'm happy that I was able to follow it through to the end on my first viewing :)
It feels nice that i solved it all by myself for the most part. I have some amazing teachers. Can't thank them enough
Brute force is taking Taylor's expansion of the square root of the tangent, and integrating *that*.
You made me fall in love with mathematics!❤️ Thank you!
Man, so many colors! If you wrote blackpenredpen in that empty space and taken a picture, you'd have a pretty good channel banner.
Slightly more straightforward (although much longer/messier): Factor x^4 + 1 = (x^2 + sqrt(2)x+1)(x^2 - sqrt(2)x+1), then do partial fractions, complete the square in the denominators and solve. This saves you from having to figure out the trick where you add 1/u^2 and subtract 1/u^2.
And pretend nothing happened!??
Great lines
you are seriously the best maths teacher!
Watching for second time, now i got it! :D Great video!!!
yay!! Thank you!
I think that the most monumental achievement humanity could ever hope to accomplish would be to find an intuitive understanding of how to go directly from the integrand to the antiderivative in one step...
prolly the first time im actually being happy after a maths answer
sqrt(-1) just love your videos!
I'm gonna start studying math very soon and your videos really hype me for it^^
THANK YOU!!! I AM VERY HAPPY TO HEAR THIS!!
Your pun made me cringe. I hope you're happy! :D
@@blackpenredpen 4 years later you are still inspiring people. I'm only 14 right now, so don't have any solid plans for uni etc. (other than studying CS as I like programming) but now you got me interested in maths! I've been spending my afternoons just trying to learn maths for the past few weeks, and it's been really fun so far!
Gratulálok, lenyűgöző végig az átváltások sorozata ! Főleg az (U^2 + (1/U)^2) átalakítása !
19:05 Try to differentiate THIS to give sqrt(tan x)
Ahh yes BlackpenRedpenBluepenGreenpenPurplepen
So...
u substitute for fx
Square u = sqrt (tanx).
See that 2udu = sec^2(x) dx.
Squaring u^2 = tanx gives us tan^2(x) = sec^2 (x) -1 = u^4 which we can see as sec^2 (x) = u^4 + 1.
Thus dx = 2udu/u^4 + 1.
Plug in original equation to have integral of u*(2u/u^4 + 1)du.
Multiply top and bottom by 1/u^2 to get complex fraction with sum of squares in denominator.
Complete the square to get (u + 1/u)^2 - 2, which has derivative of inside equal to 1 - 1/(u)^2.
Now we want two integrals (why? it is not clear unless you see the tanh^-1 and tan^-1 option coming up), one with 1 - 1/(u)^2 in numerator and other with 1 + 1/(u)^2 in our numerator.
Because the completed square can have two forms we can have the appropriate denominators to do two more substitutions, this time with t and say w.
If we do the substitutions correctly we have two integrals, one being of 1/(t^2 - 2), and our other being of 1/(w^2 + 2), both in their respective worlds.
A formula exists for these forms to be integrated neatly into tanh^-1 and tan^-1 forms. Substitute u back in for t, w, and sqrt (tanx) for u.
Do this correctly, and then
Add c and we're done.
I did this mostly for my own understanding, but I'm fairly sure I didn't skip too much for it to act as a quick summary.
This is great! It's good to work out the problem on your own or along the way.
Did you guys notice how he changed his pens at 05:23? That was awesome!
YAY I got this correct
Imma do a video on how I did it and then compare it with your method.
This took me ages btw
Thank youuuu❤, greetings from 🇨🇴
Thank you! :)
Your reslly entertaining and it is very interesting. I love your out of the box thinking to manipulate things to make them work!,,
the sheer joy with which my man just said: "I also have a purple pen :)"
I passed calc 2 this summer and then I looked at this video because I couldn't figure this out on my own. I heard "Hyperbolic tan" and had a mental breakdown and screamed aloud, "SINCE BLOODY WHEN IS THERE A THING CALLED HYPERBOLIC TANGENT!? THEY'RE JUST MAKING NEW THINGS TO MESS WITH ME!" followed by expletives and crying.
10:43 that's your brilliancy sir
You are a brilliant Math teacher
I've seen this video a couple years ago but decided to comment only now. First of all, very nicely done! Without trying to diminish guy's effort and all the excitement of the viewing public, I just wanted to remark that from the view of pure math this is absolutely worthless. Here's why: In all applications (including physics, engineering, and even math) ALL integrals are definite. This integral would be of interest only if integral is from, say 0 to pi/2. In couple of steps now I'll solve the problem of integrability and the value of that integral.
First, simple substitution v=pi/2-x translates this integral to the integral of \sqroot(cotv) over the same integral. Second, cotv is approximately inverse the of v for small v, so the question is \sqroot(1/v) integrable, and the answer is, of course yes. And we are done!
If somebody needs the numeric value, just take integral of \sqroot(1/v) from 0 to 0.01 and then calculate the remaining part from 0.01 to pi/2 by whatever method of approximate integration.
6:13 out of context is beautiful
Every integral tackled for the first time is a journey into the unknown...
Couldn’t you have just done tan inverse with the original u equation after you divided the fraction by u^2, or does it only work with one variable and one constant term?
No, cuz then you wouldn't have the +1/u² term nor the -1/u². You need them both to cancel each other.
I always understand your calculus explanations. Thank you.
Riveting :) Bravo on a brilliant solution! I think I could explain that to someone else now!
Love your videos sir, you make very complex calculus part easy 😃😃
Or you can do integration by parts! (i paused at 2:40 to try myself)
First, use substitution to get integral of sqrt(tan(x)) = integral of u*2udu/u^4+1
Now do integration by parts to get, integral of sqrt(tan(x)) = u*(integral of 2udu/x^4+1) + integral of integral of 2udu/x^4+1
Use substitution again (set z=x^2) to get u*arctan(x^2) + integral of arctan(u)
What's the integral of arctan(u)? it's u*arctan(u) - ln(1+u^2)/2
So we have integral of sqrt(tan(x)) = 2u*arctan(u) - ln(1 + u^2)/2 and substitute u = sqrt(tan(x)) back to get the answer: 2sqrt(tan(x))arctan(sqrt(tan(x))) - ln(1 + tan(x))/2
Now I am going back to the video to see how you did it and see if I'm right or not!
well...I'm wrong :'D lol
We can solve this in some other way too. We can write sqrt tanx as 1/2 {(sqrt tanx + sqrt cotx)+ (sqrt tanx - sqrt cotx)},and then break these into sin cos expressions,and then subtitute sinx + cosx = t and sinx-cosx = k in the first and second integrals respectively,and then apply the standard integral formula. Anyways love you process too!
I couldn't do this integral without u
Fantastic video. Very well explained. Thx
The math is a universal languaje. I'm peruvian, but i can see this video and understand it!🤩
Is this Calc 2?
If you plug in 0.5 for the integral function, then (sqrt(tan 0.5)+sqrt(cot 0.5))/sqrt(2)= 1.4793, but arctanh(1.4793) is undefined. The range of (sqrt(tan x)+sqrt(cot x)/sqrt(2)is alway greater than 1, which makes arctanh(sqrt(tan x)+sqrt(cot x)/sqrt(2)) always undefined for this integral function.
undefined or complex, but either way I think you are right -- see my comments
DougCube But integral from 0.2 to 0.5 of sqrt(tanx) is defined and real from the graph of sqrt(tanx). And all the result of this integral function is complex or undefined.
I did the integral in a slightly different way. I ended up with
(sqrt(2)/4) * log(abs((1/2 + (sqrt(tan(x)) - sqrt(2)/2).^2) / (1/2 + (sqrt(tan(x)) + sqrt(2)/2).^2))) +
(sqrt(2)/2) * atan(sqrt(2*tan(x)) + 1) +
(sqrt(2)/2) * atan(sqrt(2*tan(x)) - 1).
I also let u = sqrt(tan(x)) but in of multiplying through by 1/u^2 I added and subtracted 2u^2 from the denominator and then factoring the result using difference of squares. Once I had it in a factored form, I used partial fractions.
Very good explanation... Thanks a lot
Quite a popular integral in jee exams in the 80s where u had maths 200 marks ,40 questions carried 5 marks each and attempt all questions
yeah. I had some sort of flashback
Finally! Loved the video! Is great! You could do the same integral for the 100k, but now you do it the hard way! (The one with partial fractions and with the factorization of u^4+1, and with the natural log result!)
Mauro Castañeda I would need another white board for that tho. Lol
Got to see this question first time on my exam today... carrying weightage of 6marks.. was totally fucked up😑
Which exam?
At 9:59, I understand why you add the second fraction in, but doesn't the second fraction need the same denominator as the first to make it equal to the fraction on the line above?
Exactly the question i just asked him. That's a violation of fraction rules
And to prove it. When those two fractions are added back they can't be added back... So the figure 2 that was split vanished.
I have a question:
Does the identity x^2 - a^2 = x^2 + (ia)^2 in combination with the integral identities involving arctan(x) and arctanh(x) imply
i•arctan(-ix/a) = -arctanh(x/a) ?
Here is my reasoning:
∫1/(x^2 + a^2)dx = (1/a)arctan(x/a) + C, therefore
∫1/(x^2 + (ia)^2)dx = (1/(ia))arctan(x/(ia)) + C
But, x^2 - a^2 = x^2 + (ia)^2, so ∫1/(x^2 + (ia)^2)dx = ∫1/(x^2 - a^2)dx =
= (1/a)arctanh(x/a) + C
Therefore (1/(ia))arctan(x/(ia)) = (1/a)arctanh(x/a)
Multiplying the argument of arctan by i/i and multiplying both sides of the equation by ai^2
we get i•arctan(-ix/a) = -arctanh(x/a).
Is this right? Am I wrong to cancel out the constants? Thanks for your help and thanks for the great videos!
Bu soru calculus 2 dersinde vize sorusu idi.
I think it's correct.
He didn't reupload this video, we all just have a déjà-vu at the exact same moment.
Sorry for the reupload.
I actually had a mistake on the integral of 1/(x^2-a^2) in the previous video.
I will make up to you guys by checking my answer by differentiation! That video will be done soon!
One rather interesting thing about that expression is that (if we take C = 0) we'll generally get a complex number out of it. The imaginary part is always the same though.
I think you get a nice expression if you introduce complex numbers (starting by factoring (u^4 + 1) into (u^2 +- i). But it seems non-obvious that 1) the manipulations I did after that are totally ok, since I kind of ignore principle value etc... and 2) how to rigouously show this is actually equal to the answer found with real math.
Writing R = (1 + i) / sqrt(2) and u = sqrt(tan(x)) (and Re == real part) then I eventually got that Re[2R*arctan(Ru)] is an antiderivative of sqrt(tan(x)) w.r.t. x
Writing out the def. of (complex) arctan => Re[(1/R) Log [(R - u) / (R + u)]] is an antiderivative.
Numerically checking for some random values I find this does give the same values (up to a constant) as your solution. Though I did assume that tan(x) is non-negative here (that is how I ended up with "real part" in there).
There is simple method (I Feel) put x=(pi/4-t) dx=-dt and integral become sqrt(1-tan(t))/1+tan(t)) after rationalising numerator we get (1-tan(t)/sqrt(1-tan(sq)t) which is (cos(t)-sin(t))/sqrt(cos(sq)t-sin(sq(t)) now split the integrals. Cos(t)/sqrt(1-2sin(sq)t) other integral is sin(t)/sqrt(2cos(sq)t-1) by sqrt(2) manipulation we get u/sqrt(1-u(sq)) other u/sqrt(u(sq)-1) both forms are familiar having formulae
As my Cal 3 Professor used to say
" What could be simpler ."
This is the most chaotic BPRP video I've ever seen
I now know I got this question wrong on my final
you are awesome men this appear on my exam from yesterday i lov u 💕
The trick is to complete the square. Got it.👏👏👏👏
Mind blown after watching this after I have forgotten mathematics...gotta study again
All depens on interval for x. If this integral is for x in (0.5;1) you can use sqrt(tg)=1/(sqrt (cotgx x) And use substitution y=sqrt(cotgx), than you have short solution
Best birthday gift , thanks
The hardest part about this is to remember the +C.
I was wondering if you can use Feynman's method at the start to get a u^3 term at the top then do another sub.
About the integral answer with inverse tanh and so on: is this integral defined anywhere as a function? A trip to Desmos seems to indicate that this function has no real values, and plugging in several of the usual trig values yields "undefined" no matter where I look.
What a gorgeous way to find the answer! Who is the genius first to find u+1/u and u-1/u pairs for this question? It really drive me crazy for such an integral. Thank BPRP.
Yesss! This question was on my calc 2 final exam and I got it right!
Very symmetrical so much so with all those arctans and tan x's you could combine them except for that little h in the inverse hyperbolic tangent it stands for Hell in this integral bc you can't take tanh^-1(n) ||n|| > 1, ~|tan x| + |cot x| w/o screwing up the rest of it unless it's some kind of cubic solution with complex b coefficient ~ 1/3 ln (i)
Point it seems so close to y = f(x); f^-1(y) = x, x could be 0-2π with no trouble you would run into complex numbers but they cancel themselves out ish
Hola, gracias por la explicación. Me parece que hay un problema con la solución: El dominio de la función arcotangentehiperbólica es el intervalo abierto (-1,1); por otro lado los corchetes que suceden esta función contienen una función cuyo recorrido o rango es algo así como el intervalo que va de 1.39... hasta infinito. De esta manera la antiderivada no se puede evaluar para ningún valor de x. ¿Es así?... Gracias.
Hello, thank you for the explanation. I think there is a problem with the solution: the domain of the function hyperbolic arctangent is the open interval (-1,1); on the other hand, the square brackets that follow this function contain a function whose route or range is something like the interval that goes from 1.39... to the infinity. This way the antiderivative cannot be evaluated for any x value. is that so?... Thank you
Tell me if I’m wrong.
When you had the choice beetwin partial fractions and tan-1, tanh-1
Wouldn’t you get 2 different results? One in ln, one in tanh ?
Nvm just remembered tanh^-1 is just made out of ln
I will always laugh when you say, "isn't it?"
Isn’t it?!
Sir I wanted to ask one thing, the integration of type 1/x²+a² should give answer as 1/2a log() in this format then how it got as tan-¹x format ?