This Integral Will Make You Better At Calculus
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- Опубликовано: 6 фев 2022
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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
#math #brithemathguy #integral
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Hey. Do the sum of 1/x^x, from x = 1 to infinity
1/1¹ + 1/2² + 1/3³ + 1/4⁴ + ...
@@LorddualDesigner ~1.2915
Used desmos
@@manioqqqq hey
How did you do it?
Solving integral of tanx dx: 🙂
Solving integral of sqrt. tanx dx: 🙁
Better at calculus? More like "Thank you for schooling us!" I really liked the showcase of these different techniques and how you can break things up with algebra.
Glad you enjoyed it!
@@BriTheMathGuy Yes, it was inte-great!
@@PunmasterSTPouldn't you agree the multiplying and dividing by the same thing or adding and subtracting the same thing is a trick thst comes out or nowhere and I don't see anyone thinking of that unless shown it before.I LOATHEit and wish it would be abolished...with regard to the 1/u^2 here?
@@BriTheMathGuyinstead of all the adding and subtracting can't you just rewrite as (u +1/u)^2 minus 2 and then so the subsitution z equals u plus 1/u and then procees from there and just write the derivative 1/u^2 interms of z?
@@leif1075 Thank you for sharing! After all my life experiences up to this point, it seems like life is just a whole bunch of experimentation. Someone finds a way to do something, so they do it. Maybe it appeals to them, maybe because they’ve seen something similar before. It’s a great big mystery.
I’m by no means an integration master, but I try to follow along and leave a odd comment to show support and see what responses I get. It’s been fun, and if it can lead to someone showing a better way, I’m stoked!
Just to clarify, the inverse of hyperbolic tangent function is artanh, not arctanh, with "ar" meaning area instead of arc; also, the domain of artanh is (-1, 1), which does not include any of (u+1/u)/√2. Therefore, the result should be arcoth instead of artanh.
Anyway thanks for sharing this problem, it's fun to solve.
battle cat
What are y'all guys talking about?😕
@@Ah-nf2vs hyperbolic tangent and its inverse
Or arg (for argument)
@@seroujghazarian6343 Honestly I would rather just do partial fraction bc i dont really see the point of using another function to find integral of 1/x^2-a^2 , it can be done as so and then everything will be in recognisable function like ln and arctan, i dont know if anyone would use arccoth or arctanh
I do this integral with out any help. It took me an hour but Im very proud of it.
I didn't do it the same way the video did. I use partial (bcuz this is on the exercise on partial in Cal1)
@@nontth5355 hey can u elaborate the method, this question was asked in my highschool tests, did partially the same method what he did in the video but got stuck at the end , so tried another way by using by parts , it still got messy
@@AbsoluteArtist ok I’ll start at integral of 2u^2/(u^4+1) du
u can factor u^4+1 like this:
u^4+1 = (u^4+2u^2+1)-2u^2
= (u^2+1)^2 - (sqrt(2)u)^2
= (u^2+sqrt(2)u+1)(u^2-sqrt(2)u+1)
solve for coefficient A,B,C and D of
(Au+B)/(u^2+sqrt(2)u+1) + (Cu+D)/(u^2-sqrt(2)u+1) = 2u^2/(u^4+1)
then u get 2 integral slightly easier to work with I’ll show u how to do the first one
int (Au+B)/(u^2+sqrt(2)u+1) du
split into
(Au+B)/(u^2+sqrt(2)u+1) = (Au+A/sqrt(2))/(u^2+sqrt(2)u+1) + (B-A/sqrt(2))/(u^2+sqrt(2)u+1)
first one just let the denominator be a new variable and use chain rule u’ll got something like (A/2)ln(u^2+sqrt(2)u+1)+C
second one is quite tricky u can write the denominator as (u+1/sqrt(2))^2+(1/sqrt(2))^2)
and use the fact that int 1/(x^2+a^2) dx = (1/a)arctan(x/a)+C. u can find the integral of the second one.
solve for the last one in the similar way then u get the solution.
(sorry for bad english. i suc)
genius
I couldn't remember my trigonometric derivatives, outside sin and cos (though honestly I'm not sure why I didn't just do quotient rule), so I tried to make it something more manageable using Euler's formula and after like 10 minutes of work, I managed to circle my way back around to sqrt(tan(x)) = sqrt(tan(x)).
atleast u didnt do anything wrong
Thank you!
1:42, right here, you can also use a partial fraction decomposition. Write the denominator as follows:
u⁴+1 = u⁴ + 2u² + 1 - 2u² (adding zero)
= (u²+1)² - (√(2) u)²
= [ u² - √(2) u + 1] [ u² + √(2) u + 1]
And now we have written the denominator as the product of two quadratic factors, which we can split using partial fractions. Then we are just integrating a linear term over a quadratic term, which has a fairly standard type of solution involving logarithms and inverse tangents.
this one would have been my approach!
Excellent video!
Great intro (or lack thereof)! I like the just jumping right into it and not wasting any time to tackle this monster!
Much appreciated!
You could use the natural log version of the formula instead of the hyperbolic arctangent one, as more people are familiar with that one. That's the one they taught me in high school.
You my man, have earned ALL your subscriptions
My head is super warm
I don’t know if I’m better at Calculus now, I know I was very informed.
Thank you for this channel.
Happy to help!
I did it by just factoring u^4+1 as (u^2-usqrt(2)+1)(u^2+usqrt(2)+1). You get the same answer although it does take a lot more work.
I recently learned how to use power series to expand functions like this and get an approximation and honestly it feels liberating to not have to focus on getting an exact function, especially since applied math makes me worry about my future career
Quite interesting...
I'm just a high school student and I've started learning calculus these days so this type of question are quite challenging for me. But I love challenges 😁
Thanks for sharing such type of question.
Great to hear!
The RUclips algorithm and you, Sir, just made me a better person.
Кстати, после замены на u, можно было дальше представить дробь в виде суммы элементарных дробей методом неопределенных коэффициентов
my favorite integral
Glad you like it!
God I remember doing this integral a few years ago 🤦♂️
Cool!💯💯
This question was asked in high school tests, left it after solving it a bit
Great technique! I only know it with partial fractions, which is a lot more work although it does work for every nth root of tan.
In itegrals usually is better to write derivative of tangent as 1+tan^2(x)
If you calculate derivative of tangent by the limit you will get 1+tan^2(x)
Why it is better because in most cases you will rave to replace function depending on old variable with function depending on new variable
What happens when 0 has a value in a function but is undefined in its integral like this one? How can we compute the definite integral from 0 to 1 for example if 0 is undefined in the formula, but 0 to 1 definitely has an area under the curve?
Approximations (although not always as pleasing) can work very well. You could also try a different form integration (Lebesgue for example).
Take the limit as x->0+
tan(0) = 0 is well-defined, though.
@@angelmendez-rivera351 I said it in my comment. Defined in the function but undefined in its integral.
@@jamirimaj6880 Your comment explains nothing. What does it mean for 0 to be undefined in the integral? The integral is not a function you can plug numbers in.
On my exam last year I had to do integral
ln(x²+1)e^sqrt(tan(x)) and it split up basically on ln(x²+1) which was not that hard.But for sqrt(tan(x)) it was little bit complicated, I did everything the same till 1:42 when I used partial fractions.I remember it took me very long time to calculate coefficients 'cause I got somehow all of them zero so I tried three times till I finally resolved them, then it took some of work to finish it but it was very challenging, exam took two hours and I was doing this for about 45 minutes.
ln(x²+1)exp(√(tan(x))) certainly does not have an elementary antiderivative
@@violintegral I just forgot a parenthesis so it should be ln((x²+1)*e^sqrt(tan(x))).So you use rule of logaritms ln(a*b)=lna+lnb si ln(x²+1)e^(sqrt(tan(x)))=ln(x²+1)+lne^(sqrt(tan(x))) and than by rule of logaritms
lne^sqrt(tan(x))=sqrt(tan(x))*lne and we know that lne=1 so it leaves just sqrt(tan(x)).
@@aleksandardashich oh ok that makes a lot more sense lol
at last the formula for INTGRAL 1/x2-2 =is actually in form of log [ ]
Now differentiate it to prove that it indeed equals √tanx
mixing times new roman with computer modern (the latex font) in the last slide is a bit erm
Nice, now I can solve this particular equation... probably
i just had a brain aneurysm
I vaguely remember about this one! 😀
Truly a neat and important for all its aspects integral! Nice video as always! 😀
My brain is dizzy .
amazing
love you from INDIA 🇮🇳
Me too
Crazy
Hey! I am wondering if you could do a video on
how to solve for each variable in the compound
interest formula. Isolate it on the left hand side. I
keep getting stuck solving for n. A=P(1+r/n)^nt. I
can't get it off my mind.
There is no nice solution to this. Just use approximations with newton method or any other
why use the artanh definition of the integral of 1/(x^2-a^2) instead of partial fraction decomposition to terms that integrate to natural logs? are there pros and cons to one method or the other?
It doesn’t matter. The answers you get in the end are equivalent, assuming you’ve done it correctly. It just looks sexier to have artanh and tan and arctan in the same expression
Wow...
At 3:11 why dont we integrate
(1+(1/u^2)/(u^2+(1/u^2) ? Is it because the denominator will be u^4+1, numerator will be u^2+1?
Very impressive, but can you do the integral of (tan(x))^(1/5)dx?
Will fenymans trick work
Is it (tan^2(x) + 4)/400tan^2/5(x) + C?
Damn I would never be able to do this by myself lmao. Cool integral though
I'm sure you could! Have a great day!
I'm at the last year of high school and I don't know what are hyperbolic functions are (actually we didn't take it at school)
But really this is so hard for me to do all of those steps!
Also I'm quite fascinated that the derivative of this long function is only √tan(x).
integral of 1/(x^2 - a^2) = 1/(2a) • ln[(x-a)/(x+a)] + C if i remember correctly
@@cucginel1941yep thats correct
Bro integration by parts is good
This integral got asked in our board exams which is considered as one of the easy exams.....
I got notification
6 days after video is out
this is just regular calculus for us indian students preparing for iIT-JEE
Last part incorrect formula that isnt going to be arc tan x it is going to be 1/2a log (x-a/x+a).Obviously log cant take negative value so take modulus inside.
Integration of 1/x²-a² = (1/2a)ln|(x-a)/(x+a)| +c,ez
yes, the arctanh(x) function can actually be represented as (1/2)(ln((1+x)/(1-x))
@@david_varela_pt But this is a better integral compared to the arctanh and arccoth that has the same integral but different domain
Ngl, it's pretty cool, but why are we expected to be able to think of such methods when facing that kind of integrals or any other similar problem during a test, especially when running out of time 😭
There is another easy way of doing this
This can be easily solved by beta function (trigonometric form)
Pls. How about integral of square root sin x ? Help pls
such a small and innocent question..
Id just multiply by one and treat tanx as u and one as dv
Hey. Do the sum of 1/x^x, from x = 1 to infinity
Okay well if I do it I will need it for the first semester and I can take a look at the next level and I can make a more sense to get a more dramatic sense and I can understand how to make a more dramatic change to my goal is that we can make it a better day for us and I can make it a lot more for me to do it.
This integral was in my frist exam of calculus 2 and nobody can do it
no way in hell i can think of these kind of random algebraic splits and substitutions. i'd do partial fractions instead which takes much longer and gives a messy answer. anyway good video!
This is a general question given in every class 12 maths book
Not mine lmao. Gotta stop assuming your experience is the universal one 😬
11th grade actually
@@ron-jr5qw they taught this to me the day I learnt my first letter.
@@ron-jr5qw nah, the book which taught me alphabets, precisely
@@SuperShadowify Probably this guy's an Indian or Chinese high school student, this kind of problem is very elementary if so, hence why he exaggerated with the "every text book" part.
There is a much simpler way to solve this. Write
√tanx = 0.5[(√tanx+√cotx)+(√tanx-√cotx)]
That makes things simpler
Wow
When you solve the problem but forget to add c
2:21 Famous first step or why does this work?
And how tf am i supposed to know all that during my exam
4:59 the blue is hard to see
Can anybody tell me if such problems are in high school or in college ?
Highschool
Now try with sqrt(cot(x))
🤔
@@BriTheMathGuy Just differentiate x w.r.t to the solution to the integrand mentioned in this video and then integrate the result w.r.t to dx and that's your ans. This will work ryt?
This made me worse at calculus
tanx≥0?
April Fool's isn't for another couple months ...
After watching your other videos, this one feels like you made this from your bed. xD. But a good one nonetheless.
Any suggestions for improvements? Thanks for watching!
@@BriTheMathGuy Well, you are the youtuber here, so I'd be stupid to think I know better than you. Its just my personal opinion that the videos where you write on the glass are more engaging. But for a math-enthusiast, I don't mind this one either. Looking forward to more calculus.
4:53 who was an "Einstein" that thought it's a good idea to use very DARK blue font on the BLACK background, to make it almost unreadable? 🤔🤔🤔
Just write "do it yourself" in a more visible colors or something :P
NCERT HAI BISI
I came to a different solution:
(4√tanx)(1-tan^2(x)) / (1 + tanx)^2
Edit: wait, that was wrong. My bad
the cursive d is giving me eye cancer
🧐
@@BriTheMathGuy cursive is only for mathematical objects, d is not an object
Why
I challenge you to derive the final answer 🤣 good luck
Why just why
because its fun
You’re ans is WRONG
A M A Z I N G new sus!
∫
1000th like!
I hate math
That's a 10th grade calculus problem
I made a comment about a completely different way to evaluate this integral, but for some reason it got deleted :( Anyway, you can integrate sqrt(tan(x)) by considering the integral of sqrt(tan(x) - sqrt(cot(x)) and the integral of sqrt(tan(x)) + sqrt(cot(x)) separately, then taking their sum and dividing by two. As it turns out, both of these integrals can be evaluated with easy substitutions after some intense algebraic manipulation. A Math Stack Exchange post describing this method can be found here: math.stackexchange.com/questions/828640/evaluating-the-indefinite-integral-int-sqrt-tan-x-mathrmdx
Eww
😬
@@BriTheMathGuy Great stuff, man.
wow, last comment
Boring af... all this shit to end not use AT ALL!!! 😡
You lost me after 1 minute in😂
WHO WOULD THINK OF COMING UP WITH ALL THE ALGEBRA TRICKS BRUUUUUUUUUH..... maybe calc is not for me lol