integral of sqrt(tan(x)) by brute force

[@blackpenredpen]

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!

[@zackthotho7459]

Np. it is fine

[@lxathu]

This one's more worth the storage area of google than a dozen 9 percent of the content of YT.

[@KeyMan137]

blackpenredpen Thanks

[@AhnafAbdullah]

Did you have to redo the video? or just edit the definition that showed on screen

[@subinmdr]

How do we factorize (u^4 + 1) ? plz

[@barqueros2001]

you know something is hard when blackpenredpen uses five colours

[@seshnarayan7972]

I only noticed 4 colours

[@spooky2526]

@@seshnarayan7972 blue black red green purple!

[@createyourownfuture5410]

@@spooky2526 Where is purple?

[@esajpsasipes2822]

@@createyourownfuture5410 he mentioned he uses purple for the second part of observe section (10:39) although it doesn't look that diffirent from blue

[@eganrabiee627]

"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..."

[@megauser8512]

Lol

[@TheLucidDreamer12]

"I just integrated 1/(x²+tan(x)) dx"

[@jakepfeiffer6577]

∫√t̅a̅n̅x̅ dx

[@createyourownfuture5410]

@@jakepfeiffer6577 WHERE. DID. YOU. GET. THAT. INTEGRAL. SIGN. FROM?

[@jakepfeiffer6577]

@@createyourownfuture5410 on Mac it’s just option + b Idk about windows but you can copy/paste it

[@whiz8569]

"1+1 is 2, right?" Calculus, everybody.

[@blackpenredpen]

Yes. But in Z2, 1+1=0.

[@blazep5881]

well 1+1 is 1 if you account for boolean algebra :)

[@ganaraminukshuk0]

You know you're high on math(s) when you forget what 1+1 is.

[@aronquemarr7434]

1+1=10 in binary

[@guisilva9815]

1+1=3

[@rileywells3045]

Mr Math Man. Math Me a Man. Make him integrate the square root of tan.

[@kastormorgan1536]

stealing my joke, still love you though babe

[@Chrisuan]

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"

[@tesfayebabore6862]

this is pure game. There are very very few maths teachers at level of you. Thank you.

[@프로틴요플레]

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!

[@Awai_quotes]

And what you doing? In uni?

[@프로틴요플레]

@@Awai_quotes Electrical and Computer Engineering, but I think I failed the exam. I might just quit and make indie game

[@alfredomulleretxeberria4239]

@@프로틴요플레 So...how did it go for you?

[@GustavoMerchan79]

Evil integral to place on an exam ... :/

[@rawn4203]

Would have to be like the only question or 1 of 2 questions.

[@kishorekumarsathishkumar1562]

I'm from India, and i am practising for this exam, I can assure you, there are more brutal questions.

[@rahimeozsoy4244]

@@kishorekumarsathishkumar1562 yeah for Indiana this is ez or normal

[@sgsnake2x]

@@kishorekumarsathishkumar1562 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

[@Wu-Li]

This one is an easy problem

[@RyanLucroy]

14:41 "This is the two, so what should I do?" What a rhyme :O

[@DGCubes]

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. :)

[@Someone-cr8cj]

DGCubes what are YOU doing here??¿

[@DGCubes]

What can I say, I like calculus. :P

[@blackpenredpen]

thank you DGCubes!!

[@metalmathprofessor1467]

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!

[@thephysicistcuber175]

BOI!

[@weerman44]

15:02 WIZARD! Where can I buy this magic blackpen??

[@agfd5659]

WTF! I didn't even notice it before!! How did he do that?!

[@morganmitchell4017]

He edited it because he forgot the (-) sign

[@weerman44]

Haha I understand that Just a silly comment ;)

[@vvsutar6179]

Morgan Mitchell Johnson

[@habiburrehman7108]

hahaha i also noticed after you comment.....

[@Johan-st4rv]

god damn I love math

[@blackpenredpen]

Me too!!!

[@FingertipsOfTheNight]

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

[@morganthem]

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.

[@blackpenredpen]

This is great! It's good to work out the problem on your own or along the way.

[@tanmayagarwal4981]

Got to see this question first time on my exam today... carrying weightage of 6marks.. was totally fucked up😑

[@kava2214]

Which exam?

[@rakshithgowda1606]

And pretend nothing happened!?? Great lines

[@CasualGraph]

Man, so many colors! If you wrote blackpenredpen in that empty space and taken a picture, you'd have a pretty good channel banner.

[@jilagamnagendrakumar5522]

10:43 that's your brilliancy sir

[@anthonyjaas]

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!

[@cruciflux8634]

Did you guys notice how he changed his pens at 05:23? That was awesome!

[@mahj1]

I LITERALLY LOVE YOU SO MUCH YOU DESERVE THE WHOLE WORLD

[@DriverMate]

6:13 out of context is beautiful

[@aashishkarki7867]

You made me fall in love with mathematics!❤️ Thank you!

[@pankajkumarpandey6658]

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

[@TheYoshi463]

sqrt(-1) just love your videos! I'm gonna start studying math very soon and your videos really hype me for it^^

[@blackpenredpen]

THANK YOU!!! I AM VERY HAPPY TO HEAR THIS!!

[@agfd5659]

Your pun made me cringe. I hope you're happy! :D

[@mrocto329]

@@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!

[@Vaibhav-ye6to]

prolly the first time im actually being happy after a maths answer

[@ashotdjrbashian9606]

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.

[@williamwen7190]

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]

undefined or complex, but either way I think you are right -- see my comments

[@williamwen7190]

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.

[@CrypticalGaming]

You could've used another formula/ method for integrating the 1/(t^2 -2) just by using : integral of 1/(x^2-a^2) = (1/2a)*( ln( |x-a|/|x+a| ) . that would have been more easy and intiutive than hyperbolic inverse function.(just saying) BTW very nice explanation sir. Great content. You're a wonderful teacher. PEACE

[@pasodirect]

Ez annyira bonyolult, hogy nincs értelme ennyit számolni, de blackpenredpen igazi zseni !

[@gregoriousmaths266]

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

[@ernestschoenmakers8181]

This integral can also be done by partial fractions decomposition. 1+u^4 can be factorized into (1+sqrt(2)u+u^2)(1-sqrt(2)u+u^2).

[@ngonotseg719]

You are a brilliant Math teacher

[@sharpnova2]

i'm surprised you don't have a video up on how to integrate 1/(x^2 - a^2) it's very different from the arctan stuff for 1 / (x^2 + a^2) it's a nice problem that requires a clever rewriting of a numerator or (equivalently) p.f.d.

[@meraldlag4336]

Am I completely missing the question or can you use partial fractions to integrate it Yes I haven’t fully studied integration lol

[@KING-ni4ze]

Put u=tan x and then differentiate it, sec^2 x= 1+tan^2x

[@div_07]

The hardest part about this is to remember the +C.

[@TheMauror22]

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!)

[@blackpenredpen]

Mauro Castañeda I would need another white board for that tho. Lol

[@AlejandroRodriguez-lq9mz]

Watching for second time, now i got it! :D Great video!!!

[@blackpenredpen]

yay!! Thank you!

[@bbxrhythm1127]

3:50 take t = u^2

[@imme3024]

If we use fractional decomposition to integrate 1/(x^2-a^2) , we'll get 1/(x^2-a^2) = [1/(2*a)] * [ 1/(x-a) - 1/(x+a) ] and therefore its integral is equal to [1/(2*sqrt(2))] * ln( |(t-sqrt(2)) /(t+sqrt(2))| ), by replacing "a" with sqrt(2). That's not the same as (1/a) * tanh-1 (x/a)

[@PunmasterSTP]

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...

[@tomatrix7525]

Your reslly entertaining and it is very interesting. I love your out of the box thinking to manipulate things to make them work!,,

[@2008abhik]

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

[@jamescollier3]

yeah. I had some sort of flashback

[@pankajkumarpandey6658]

I solved this problem long back. ie in 1982 when I was in class 12th.

[@zhongyuanchen8424]

This is fantastic. I am better at understanding the calculus than I am at manupulating algebra. Every time I see this kind of twsting or algebra tecniques, I wonder how the people could come up with it in a test..... By the way, have you ever used it in the final exame of your class? If so, has anyone got it right? 只是好奇啦

[@gohilpratapsinh3510]

I think that when you divided with u^2 that is wrong But you have to do that [(u^2+1)+(u^2-1)] After that give them saprets tum

[@brucefrizzell4221]

As my Cal 3 Professor used to say " What could be simpler ."

[@ethanbartiromo2888]

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?

[@alejrandom6592]

No, cuz then you wouldn't have the +1/u² term nor the -1/u². You need them both to cancel each other.

[@ny6u]

Every integral tackled for the first time is a journey into the unknown...

[@aditya3984]

was trying to do on my own but was stuck after first substitution this video helped.

[@disgruntledtoons]

Brute force is taking Taylor's expansion of the square root of the tangent, and integrating *that*.

[@jaishkhan7442]

50% math 50% flexing different colored pens

[@Quantris]

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).

[@amritanshubajpai5297]

We have studied another formula: integral of 1/x²-a² = 1/2a [ln (x+a/x-a)]

[@stapler942]

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.

[@sauravthegreat]

Fantastic video. Very well explained. Thx

[@RB_Universe_TV]

Ahh yes BlackpenRedpenBluepenGreenpenPurplepen

[@dhanushsk1549]

Hey bprd I have suggestion in 3:31 , we can take u^2 as t and then do partial fraction , it will be easy , after doing partial fraction substitute t as u^2 .

[@alvaroarizacaro3451]

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

[@RitaSahay-g9p]

3:34 split it in the form of u^2+1 and u^2-1 to save the trouble..

[@vinod9764]

Amazing explained

[@niyogijoydeep3351]

Light insufficient ; kindly use contrast /dark colour of pens for better visibility

[@hassanbasil1350]

i remember i gave this integral to my professors in 2011 and they couldn't solve it so I gave them the solution they just bring it in the 2nd try in last exam

[@haninyabroud7810]

Damn Tan(x) is easy Only 1/2 changes everythink

[@americansoul5563]

That's log(sec(x))

[@pi17]

Hold my ⅓,¼,⅛

[@maximusmeridius2060]

Can you break up the integral of 1/(t^2-2) as a partial fraction A/t-sqrt(2) + B/sqrt(2) and integrate that way?

[@mohammadshahabeezurrahman6646]

If we put tanx=sec^2(@) and then we multiply numerator and denominator by cos^4(@) and then we substitute t=cos(@) we end up getting I=integral of(-2dt/(1+t^4)). This last equation is quite easy to evolve do you think the same.

[@art3m1s9]

Just use the horseshoe integration

[@packofskittles1232]

Meanwhile 12th Grade Indian students casually having such questions in their textbooks 💀

[@chrysophylaxs7208]

Makes video on why we rationalize denominators, doesn't rationalize the denominators 😋. Still loved the video :D

[@blackpenredpen]

Chrysophylax Dives lol I know!!!

[@aryanbansal624]

Do the integral of sqrt( secx ) thats a challenge ...... no one seems to be able to do it

[@josebarona271]

Best birthday gift , thanks

[@AkashChaudhary-ux9pg]

Int of 1 upon log x ka kya hoga sir

[@ameyadeshpande6508]

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.

[@dekz0911]

Mind blown after watching this after I have forgotten mathematics...gotta study again

[@shubhx_23]

The formula for 1/ x² - a² = (1)/(2a) log (x-a)/(x+a) + C In my NCERT Book....

[@sudhanvab]

Again Why dont you use the identity integral 1/(x^2-a^2). = (1/2a) log((x-a)/(x+a))

[@cnekanba]

tanh-1(x) domain is |x|

[@danieljulian4676]

This isn't calculus. It's algebra and becoming adept at using the relations; definitions, useful substitutions, and simplifications of the combinations of functions that turn up. Still, this is what we have to learn how to do. These operations have been fashioned into algorithms. Still beautiful, but it's now like seeing whether a human can beat a computer at chess. We can do this integral with Sagemath (fast!) and get a big slug of functions added together that may or may not look like the solution we saw for the original integral. The integrals that a computer algorithm cannot do, we may or may not be able to do, either, given how much work this one is. The insight is figuring out the first substitution, completing squares, and so on. Know when a tactic is taking you farther away instead of closer. Do these videos try to teach that? Can we obtain any intuition about whether we have a useful result from using CAS? That's what we should really be thinking about, here. Math olympics? Meh. Perform like a trained seal.

[@anuraagrapaka2385]

Can you please use logarithmic functions , i am in class 12 And there isnt calculus related to hyperbolic functions in our syllabus Thank you very much

[@irongosse8366]

He brought out the blue, green and purple pen 💀

[@averageasiankid3105]

What 3 year old me sees when my brother teaches me fractions.

[@davisnganga6266]

The trick is to complete the square. Got it.👏👏👏👏

[@leek8656]

Interesting… when I did this I didn’t divide by the u^2, and my answer had two natural logs and two inverse tangents

[@nasaxd1862]

Purple pen = turbo

[@xandersafrunek2151]

I will always laugh when you say, "isn't it?"

[@blackpenredpen]

Isn’t it?!

[@benjaminmoller3992]

you are awesome men this appear on my exam from yesterday i lov u 💕

[@darthnihilius6757]

Why not substitute the u^2 by t for example and then have \int{\frac{1}{t^2+1}}dt? then the answer is tan^-1(tan^2x)

[@HaitaoWang268]

Yesss! This question was on my calc 2 final exam and I got it right!

[@juanandrespenaloza7068]

Por qué no sacaste el 2 de la integral, y sumaste 1 arriba y restaste 1, luego queda cancelar terminos, y super fácil separando las fracciones

[@SamudrarajOfficial]

1/t²-2 can also be integrated as 1/2a log(t+a/t-a) right? (a=√2)

[@ilanpi]

You should have used: u^4 + 1 = -(-1 + Sqrt[2] u - u^2) (1 + Sqrt[2] u + u^2) then used partial fractions.

[@unoriginalusernameno999]

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!

[@unoriginalusernameno999]

well...I'm wrong :'D lol

[@anybodyk3524]

Broo I was going through this math for quite a long time... Then I found this video... I thought this math couldn't be bigger Just some square related terms would be needed to solve this which I couldn't catch it... Turns Out It's a 19 mins long math 🙂🙂🙂🙂

[@I_am_FRANCO]

9:46 But how would I think about it like this? Like... how would I think!!?

[@SV42165]

Lets be honest math professors once sat down and created these methods and you just have to memorize the steps and forms and see how questions are dealt with.

[@shivmohansingh4941]

Value of Integration of 'e'to the power tanx dx

[@subramanyakarthik5843]

I got 1/ sqrt(2) * Ln((sqrt(tan(x)) + 1/sqrt(tan(x)) - sqrt(2))/((sqrt(tan(x)) + 1/sqrt(tan(x)) + sqrt(2)) + c as the answer but your answer is correct way

[@yourcreed104]

Why didn't you use the other formula in step 4?

[@walter9029]

Funny to see, how the answer of a short question turns out to be monstrous..

[@joyboy7679]

Can you please integrate y=xtanx?