Integrating a Filthy boi - Papa's Improvised Session #1 [ integral sqrt(tan(x)) ]
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Improvised Session Playlist: • Improvised Sessions
Arctan integral: • Integral 1/(x^2+1) und...
ln integral: • The integral 1/(x+1) a...
Complete square: • Completing the square:...
Partial Fractions outtake: • Behind the Meme: Parti...
That boi has been a lot of work, but it was fun giving it a shot! =) If you like me improvising sht then pls say so in the comments =)
#FreshToadwalker
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![Integrating a Quadratic boi - Papa's Improvised Session #2 [ integral sqrt(ax^2+bx+c) ]](http://i.ytimg.com/vi/cfE5ZPnp-Qg/mqdefault.jpg)
![Integrating a Quadratic boi - Papa's Improvised Session #2 [ integral sqrt(ax^2+bx+c) ]](/img/tr.png)
16:03 it should be -1 not +1
Fun. Let t= tan x ? A little faster.
@GreenMeansGO I’m a little late to respond obv but I noticed that too
Super epic integration tho!
@@janepianotutorials how does that help make it faster?
@@cooldawg2009 the integrand becomes [(t)^1/2 ] / [t^2 + 1]. Still complicated to solve after that. Has anyone differentiated his answer to confirm the solution? I just tried. No go, but I probably goofed. Curious, I checked WolframAlpha, they did what I did! But came up with a different answer. Hmm…. Headache
be sure to check by differentiation
Lol
agree!
It requires more work than integrating lol
Omar Farouk th-cam.com/video/e23vv1nQ0EQ/w-d-xo.html
Omg blackpenredpen
U r a superhero
Honestly if i got this on exam
I prefer a 0
Everybody gangsta until flammable math pulls out the lambda
*Giorno's theme plays*
Flammable: Intergrates the sqrt(tan(x))
Desmos: Impossible
Maxima:
integrate(sqrt(tan(x)),x);
(-log(tan(x)+sqrt(2)*sqrt(tan(x))+1)/2^(5/2)+log(tan(x)-sqrt(2)*sqrt(tan(x))+1)/2^(5/2)+atan((2*sqrt(tan(x))+sqrt(2))/sqrt(2))/2^(3/2)+atan((2*sqrt(tan(x))-sqrt(2))/sqrt(2))/2^(3/2))
Whenever an integral involves sqrt you know that things are gonna go sqrrrrrt 🔥🔥
i breathed really hard at this
Me every tuesday: If that answer isn't right, then I'm going to hang myself in this very room.
Everyday*
*me writing 1=2 on purpose*
oh no…
anyway
Please use me as your next substitution variable
You're hard to draw!
Márcio Amaral Thank you.
Abhishek Minz Die Deutschen sind halt einfach die Besten!
Hashtag proud to be German
Ahh that's hawt
What does zeta means
I really think these improvised sessions are better at teaching integration than planned-out sessions
Because we can really see how you think approaching these problems, and what can we try in similar situations.
Thanks a lot Papa Flammy!
Pro tip: instead of t^2 = tan x in the first step, use t^2 = 2 tan x. Also, use v = t - 2 and w = t + 2. This eliminates all the surdy bois
I did this one in front of my class
Took about 15 minutes
So relieved to find that my answer was correct
Dat's one filthy boi
You Dope man my book says sqrt tanx cannot be integrated.
konsi book use karta hai?
Maxima says otherwise: -(sqrt(2)*log(tan(x)+sqrt(2)*sqrt(tan(x))+1)-sqrt(2)*log(tan(x)-sqrt(2)*sqrt(tan(x))+1)-2^(3/2)*atan((2*sqrt(tan(x))+sqrt(2))/sqrt(2))-2^(3/2)*atan((2*sqrt(tan(x))-sqrt(2))/sqrt(2)))/4
ATHARV BHAGYA Flammy, defying books since 1547
it can't be integrated by parts, maybe that's what your book meant ?
Another way to solve this, flammy bois: let I = integral sqrt(tan(x)) dx and J = integral sqrt(cot(x)) dx. Evaluate I + J and I - J. Note that sin(2x) = 1 - (sin(x) - cos(x))^2 = (sin(x) + cos(x))^2 - 1. You should get to an equivalent answer in terms of sines and cosines.
You are a brave man, my boi, doing such monsters for all the internet to see
Thank you Flammable boi, very cool.
No thats our PAPA come on man...
great video, u integrated this quite nicely using simple techniques but still solving a difficult integral. one of your best vids imo
It's time for you to differentiate this guy and see is that correct or not
Early on papa writes + instead of - when completing the square but papa effortlessly sticks the landing 🙌🙌🙌
I suggest this channel to be renamed to “Integral on Steroids”
I first saw that integral in 1999, when I was taking my calculus class.
It took me around 10 pages of my homework excercise book and around three times to solve it. Then I realized I missed some substitutions then corrected then .
Voila everything fine!
After that when I was going to buy a graphic calculator I used to used this integral as a bench. At that time only the hp49 was the only capable to solve this integral.
Now I only remember few methods to solve integral but this video brought many remembrances about that moment.
There’s an easier way, express 2t^2/(1+t^4) as (t^2+1+t^2-1)/(t^4+1) which becomes 2 integrals. (t^2+1)/(t^4+1) and the other one, this can be computed by dividing both top and bottom by t^2 and using the fact that t^2+1/t^2=(t-1/t)^2+2. A similar thing can be done with the second one. You end up integrating 1/u^2-2 and 1/u^2+2
If you improvise so well, I don't know when you prepare the session.
Your channel is amazing and I love each and every video. Its like a dopamine rush every time you get to the end especially when I try to solve in parallel and reach the same result. Keep going and wish you the best.
but there's an easier way :'( Just sub t for e^u and divide top and bottom of the fraction through by e^2u. It becomes the sum of hyperbolic functions which are easy to integrate.
Gamma Knife very, very nice. And with the right hyperbolic identities, the worst thing I had to integrate was a polynomial with 2 distinct linear factors in the denominator. 👏🏾👏🏾👏🏾👏🏾👏🏾👏🏾👏🏾👏🏾👏🏾👏🏾👏🏾👏🏾👏🏾👏🏾
Can you explain a little more the e^u substitution. The term u being what? It's just that I'm integrating 1/sqrt(tanx) and I think your method might help. Thanks.
@@TavoLL1511 bro just do the same substitution that pappa did you will get a t on the denominator and it will cancel with the t on the numerator, then do partial fractions
how does it simplify?
i mean with tan(e^u) you get e^e^u stuff which doesn't simplify
15:02 you can write ln(...) instead of ln|...| because you're adding a square (always positive) and a positive number. So it's always positive, thus no absolute values needed
I would like to suggest an easier method. After substituting t=√tanx, we write numerator as 2t^2 = 1+t^2 + t^2 -1
Now we split into two integrals and divide by t^2 on both num and dem. Then we substitute t-1/t=u and t+1/t=k respectively. We get the solution easily. Without partial fractions or any lengthy calculations
was this a freestyle boi?
Well, you can write 2t^2 =(t^2+1)+(t^2-1).
Then divide Numerator and denominator by t^2,
Then adding +2 and-2 in denominator will give a very easy method to solve it.
this is great! Improvising this showed me how you can think through a problem if you haven't seen it before... don't get that much at the university, especially me as an undergrad, but it's so important :) thanks!
Imagine spending all that time differentiating that fucking monster and at the end all you get is tan^1/2
At 6:20 you can add and subtract the missing derivative of the denominator so then you can just split this into integrals that give you arctan and ln directly.
PS: In Canada we have been having a heat wave for the last month and a half :(
"Add this bitch boi on both sides" I legit spat out my drink when you said that
Another way:
I = integral (sqrt(tanx) + sqrt(cotx))
J = integral (sqrt(tanx) - sqrt(cotx))
required integral = (I+J)/2
I and J can be evaluated by decomposing into sin(x) and cos(x) and then a little manipulation.
Could you elaborate on the manipulation further? I cant seem to make this method work
You made me spit my food out laughing. "Let's be fucking bad ass. ".
This in my country is considered a very basic problem for integration. Just divide the numerator and denominator by t^2 and in N^r add and subtract 1/t^2 and accordingly make D^r a perfect square.
I suppose if you simply memorize unintuitive keys to common problems rather than solve it organically for the first time, it's probably quite easy.
India I would guess?
Thank you Flammy! But I dare you to write the arctan in terms of the LN function and see if the whole thing can be symplified by the end! Congrats from a brazilian boi that love your videos!
Not sure arctan has a ln form.
Artanh does though
today i learned i could use "w" as a substitution variable instead of just reusing "u" and "v" and confusing myself
thank you
You can use literally any symbol. You can use a small drawing of a turtle if you want.
Nice video dude as always keep it up !
You might simplify the formula a bit,
Since arctan(a) + arctan(b) = arctan( (a+b)/(1-ab))
And also theres a way to make the logarithm look better
You’ll have something like that :
I = 2^(-1/2)*( Arctan(2a/2-a^2) + 1/2*ln( 1 + 2^(3/2)*t/(t^2 + 2^(1/2)*t +1)))
3:31 Did Euleroid just smack you on the head? xD
Thanks sir for stepping ahead towards my question
Only flammable maths can make me integrate at 4:24 AM
Thank you for teaching!
that natural log factors in an interesting way: [(t-(-1/sqrt(2)+i/sqrt(2)))(t-(-1/sqrt(2)-i/sqrt(2)))]/[(t-(1/sqrt(2)-i/sqrt(2)))(t-(1/sqrt(2)+i/sqrt(2))] , t -> sqrt(2)*t , [(t+1-i)(t+1+i)]/[(t-1+i)(t-1-i)] now we can use the identity arctan(x)=1/(2i)*ln((x−i)/(x+i)) somehow to rewrite it in terms of arctangent.
Papa Flammy confirmed to be a hot boi.
classic! glad to see you integrate this boi.
I love this board
me too :p
Looks like something Himmler would dream up.
except the error you are a genius man
I’m so happy to finally find someone who did it the same way as me
lol. What is the other way of doing it?
Everyone always forgets that fact sqrt(x) equeals to x^(1/2 )
You can write the numerator 2t^2 as (t^2+1) +(t^2-1) then break it into 2 integrals and then in both divide by t^2 and put t+1/t =u in one of them and put t-1/t=z in second integral ...Then you don't have to apply partial fractions.
A VERY LONG JOURNEY
ok ill sub even tho the only integral i know is x where its (1/2)x^2
I love your opening sounds
Don't do partial fraction here @3:13 .split 2t^2 into (t^2-1)+(t^2+1) and then divide numerator and denominator by t^2.and substitute for t+1/t for term in first term and t-1/t for second term.and using this you can solve this integral in 5 minutes.and then thank me later.
This nigga just fortnite danced on us I love you papa
I think you've done this the hard way. From what I can remember there is a simpler solution, but well done!
Came for the math, stayed for the hips
Oh nice, a spontaneous boi
This integral was calculated many times
Try this integral
Int(\frac{x}{\sqrt{\exp{x}+\left(x+2
ight)^2}}\mbox{d}x)
Wolfram claims that antiderivative cannot be expressed with elementary functions but it is not true
My hints
Add the zero to the integrand to be able to use linearity of integral
Multiply integrand by one and then use substitution
What is this zero and one ?
Try to guess
We can do that problem quiet easier........in the Int[2t^2/(t^4+1)] we can do 2t^2=(t^2+1)+(t^2-1).....and then seperate the terms and divide with 1/t^2 on numerator and denominator...... Then, t^2+1/t^2 = (t+1/t) ^2 -2 = (t-1/t)^2 -2........ Then take t+1/t =z & t-1/t =y ....... Then, find dz&dy...... And then write integral interms of z&y and apply formulae.........
This is much better to eliminate writing plenty of steps........ Time saving method.😇.....
We can do that problem in a different way...... Take √tanx=[(√tanx+√cotx)/2]+[(√tanx-√cotx) /2]......and then 2√tanx= {(sinx+cosx) /√sinxcosx}+{(sinx-cosx)/√sinxcosx}......
Sinx.Cosx={[(Sinx+Cosx)^2-1]/2}
={[1-(sinx-cosx) ^2]/2}
And find sinx+cosx=t .....find dt!
Then sinx-cosx=z......find dz! And then,
Write integral in terms of t&z....
That's it... ✌
May I get your mail ID!!!!!
Atleast name in Facebook Or profile details...... 👨🏻📖
Just to interact...... 😍
This is how we landed a probe on Mars.
17:23 heart started beating again
You could also add these arctans
u = tan(x) ma boi
Commenting at 3:20
Could you not write t² (1 / t⁴ +1)? You can the use arctangent and integrate by parts?
(Will try on paper and update)
Ich habe öffentliche Hänge so gerne und Du hast einen Fehler auf der vorletzten Zeile bei der 16:03 Marke im Video gemacht. Das "+1" im ersten mathematischen Begriff innerhalb der eckigen Klammern sollte auf "-1" korrigiert werden. Wird ein Programm für die Hinrichtung gedruckt, das andere öffentliche Vergnügungen und Ablenkungen für die Kinder anzeigt? Wir wollen die Kleinen nicht schockieren, wenn ein Mathematiker die Strafe erhält.Doch im Ernst, Du hast eine äußerst inspirierende mathematische Leistung vollbracht. Der einzige Fehler, den ich erkannt habe is ganz egal! Habe Dir diese Post auf Deutsch geschickt so dass die Sache nur unter vier Augen bleiben wird.
Ich habe Ihren Kanal abonniert. Die mathematische Fähigkeit ist offensichtlich außergewöhnlich und Deine zufällige Witzchen sind auch gut zum Lachen.
Und es erscheint dass unser Geheimnis nicht mehr geheim sei
The answer I have is [ 1/√2 arcCtg( 1/√2 ( √Ctgx - √tgx ) ) - 1/√2 arCth( 1/√2 ( √Ctgx + √tgx ) ) ]
th-cam.com/video/KF8Wrqc_vjw/w-d-xo.html
It could be easily checked by differentiation.
The Russian notations means:
arcCtg ≡ acot ≡ cot^-1
arCth ≡ acoth ≡ coth^-1
Ctg ≡ cot
tg ≡ tan
You could substitute u=sqrt(cosx) then you differentiate both sides and give an expression of sinx in terms of u. Substitute everything in, then you only need to solve the integral of -2du and substitute back in.
The solution is -sqrt(cosx). Ps: sqrt(tanx) = sqrt(sinx)/sqrt(cosx)
Huh? When you solve for sin you have to take sqrt of dU… is that even possible?
@@cooldawg2009 solve u = sqrt(cosx) for x and insert into sin when considering the differential du. Then you are good to go.
@@shayanzahedi5735 i dont think that works. You have no way to get rid of the sinx that appears after taking derivative of sqrt(cosx)
pls make more, we want 1h video of you struggling to solve disgusting boiii
Hi Flammable Maths,
Thanks for the video and this amazing integral, but there's too much complications that you can avoid them, after the first substitution you have to integrate (2u^2/1+u^4), here you can do a trick look:
(2u^2/1+u^4)=(u^2+1+u^2-1)/(1+u^4) so you split the integral in two parts (u^2+1/u^4+1)+(u^2-1/u^4+1) wich are equal to ((1+1/u^2)/(u^2+1/u^2))+ ((1-1/u^2)/(u^2+1/u^2)) and now for the first term let v=u-1/u and for the second term let w=u+1/u then you have to intergate (dv/v^2+2)+(dw/w^2-2) too much easy to calculate
I'm a big fun of you,
Hope you appreciate my method
Bye
I was going to post the mistake at 16:03, but someone beat me to it. I guess I'll have to watch your videos quicker
Can you solve the integral of square root of sin x using the elliptic integral??
It was amazing man
Instead of partial fraction u can use algebric twins.
What is Algebraic twins method?
U could have used the 2t^2 like t^2+1 and t^2 -1..to have avoided the partial fractions.....(btw it's really good I got someone like you teaching me maths ...)
2:34 You have committed crimes against BOIrim
"...Is nothing but..."
Multiply the numerator and denominator by sec^2x den it comes to a 5 min problem
Now for the fifth root of tan(x)
Beim Ln brauchst du kein Betrag weil alles positiv ist durch das Quadrat
You complain about the heat, well, it was over 96 degrees F with 80% humidity today. However, I remember when I was working for the Arizona Department of Transportation that it got over 127 degrees F with little to no humidity and we were working outside all day long! The surface temperature of the AC was over 165 degrees F. I know because I had a thermometer.
10:30 chalk ASMR
Jesus christ, someone told me to try and integrate this clusterfuck. I never got to integrate it except the first substitution part after that I got stuck.
Like i did the intégration it was pretty fun ❤
This question came in cbse class 12th exam!! Eazy question 😂😂
:D
No way in hell is this an easy question for cbse 12th 🤣
it's one of the most famous ones
15:52 he went from tan to ta
In Indian students are solving this question in high school/12th standard
beware of -1/2 and +1/2, after changing the sign inside the binomial squared: the result changes... @7:56
Pretty impressive!
Couldn't you skip a bunch of work @ 2:56 by using integration by parts? Int(t*2t/(1+(t^2)^2)dt)=t*arctan(t^2)+int(arctan(t^2)dt) then do another substitution?
INTEGRATE SIN(X)^X !!!!
Jack Wagner Its probably going to be similar to his x^-x, just using a mclaurin series of sinx or some spicy stuff like that, but it does look like a very challenging problem.
Jack Wagner Most probably non elementary
Did I catch a "bitch boi" at 2:05?
a salute to this guy
Amazing
It's a good subject, very nice. I am looking for a different solution method for this integral. If I can fix it, I will post it on my youtube channel soon.
5:12 Electro Shuffle
5:13 Floss
5:16 Fresh
Ohh Papa Flammy. I see what you did there.
BTW, I got lost into your substitution int the part where you squared t and tan(x) twice. Can you repeat or explain the process again?
Are you talking about the part about 2 minutes in?
1=secx^2 - tanx^2, so secx^2=1+tanx^2. And from the very first line of work, t=sqrt(tanx). So secx^2=1+t^4.
If not, give a timestamp.
Lemur Potatoes i'm talking about that one. Did he squared it in order to utilize the trig identities?
That was badass
papa euler would get pissed off >:v
Just a very very very small mistake that in the first term there should be - 1 in place of +1. But please don't mind. It is still very good work of you.
The most impressively ridiculous way to factor a quartic is the Ferrari solution. This one has no real roots, so the resolvent cubic should have only one root....