how about this calculus with this formula (in degrees) limit for x approach for x pi=x\2*sin (360\(x+2)) \ sin((x\4)*(360\(x+2))) I want to know how approach you can get even with a calculator (I tried this formula and works as much as you dont round the numbers (less the calculator to do)) and approach as much as bigger as x is (I sugggest to began aproach with x= 100 to get first digit 3 and for second you need x= apraxch 1000 so almost 10 to the power of 3 plus number of digit you want (for smaller values of diggits ) gets the most approach you can get by hand and calculator
What a show man! And also big congrats to Lars for his winning the cancer finally! You two are awesome integral-fighter and cancer-fighter respectively 😎
I got A+ in calculus 21 years ago Integration was my favorite game but after these years I totally lost my skill I love math more than any other science but unfortunately I left studying it to have better job in engineering to gain more money Math is the science that all inventions based on it Love math
The 379th Hero you can split into products as tan^2(x)*tan(x)=(sec^2(x)-1)tan(x)=tan(x)sec^2(x)-tan(x). The integral will come out to be (tan(x))^2/2+ln(abs(cos(x)))+C. You might end up the answer in terms of secant as (sec(x))^2/2-ln(abs(sec(x)))+C since it end up a different constant and using properties of logarithms.
Those who are womdering why he used hyperbolic cot instead of hyperbolic tan, it is because with tan there is a limitation, namely |x| 1 so that limitation works here
The other method is partial fraction decomposition where you divide u^4-2u^2+2 by u^2-au+b. After working this out you'll get: a=sqrt(2+2sqrt(2)) and b=sqrt(2).
Way easier to start off with the substitution 1 + tan(x) = cos^2(u) and then do some simple trig until you can integrate by partial fraction. Gives a MUCH nicer answers too: 1/sqrt(2) ln[sqrt(2) + sqrt(1+tan(x)) / sqrt(2) - sqrt(1+tan(x))] + c
I agree, I thought the choice between tanh^(-1) and coth^(-1) was based on the domain and since it's an indefinite integreal it's arbitrary. And maybe (1/2)ln|(1+x)/(1-x)| would be best since its domain includes all real numbers except +/- 1. It's not as fun as the inverse hyperbolic functions though
The polynomial P = u^4 + 2 u^2 + 2 factors over the reals (the only irreducible polynomials over R are linear or quadratic with negative discriminant). Here is a factorization P = (x^2 - a x + b) ( x^2 +a x + b) where a = sqrt(2sqrt 2 + 2) and b = sqrt 2 Once you compute this, the integral is straightforward.
It’s hard to tell without the use of parentheses. If you meant the integral of sin(x^2), then it’s non-elementary but will come out as the answer of the sine Fresnel integral as S(x)+C. If you meant the other way as the integral of (sin(x))^2, then you have to change the identity as (1-cos(2x))/2. Doing so will get the answer to be x/2-sin(2x)/4+C.
Hi BlackPenRedPen! Didn't know if you noticed that in the inverse tan of your answer you switched plus for minus the answer should have been: 1/sqrt(2sqrt(2)-2)*arctan([1/sqrt(2sqrt(2)-2)](sqrt(1+tanx)-2/sqrt(1+tanx)))-1/sqrt(2sqrt(2)-2)*arctanh([1/sqrt(2sqrt(2)-2)](sqrt(1+tanx)+2/sqrt(1+tanx)))+C. But that was an awesome job. Integrals can be tricky, but you do an amazing job.
let 1+tanx = u^2 sec^2xdx = 2udu so the original integral becomes (u.2udu)/(1+u^2) then simplifying it we get 2du - 2du/(1+u^2) so we get 2u-2tan^-1(u) 2*(sqrt(1+tanx)) - 2*tan^-1(sqrt(1+tanx)) Won't this be easier? Or am I wrong somewhere?
I did tried using integración by parts multiple times and some regular substitution in the middle, i got to a point were i had de integral of sqrt(tanθ), that's when i stopped because i know that isn't pretty (or is way too pretty, depending on how messed up you are).
the chessmate tan(x)^3 = tan(x)·tan(x)^2 = tan(x)·[sec(x)^2 - 1]. Use linearity, and now you can consider the integrals of tan(x)sec(x)^2 and the integral of tan(x). The integral of tan(x) is given by ln|sec(x)| + C(x), where C(x) is an arbitrary piecewise step function with discontinuities whenever x = πn + π/2 for some integer n. The integral of tan(x)sec(x)^2 can be calculated by letting u = tan(x) => du = sec(x)^2 dx, which gives the integral of u with respect to u, which is equal to u^2/2 + C, or tan(x)^2/2 + C. Adding the integrals results in tan(x)^2/2 - ln|sec(x)| + C(x). Is there a better way to write the antiderivative? Other than switching tan(x)^2 for sec(x)^2, which is permitted because they differ by a constant, no, there is not a better way, as far as I am concerned.
@@angelmendez-rivera351 alright thank you i wrote it with sec instead of tan but i know some people said they wrote it with ln(cos) instead of ln(sec) wanted To know if it was any better or worse lol
have to say that i felt a little emotional at the beginning of the video, I'm also an engineer student and fighted cancer exactly 3 months ago, everything is fine now but anyways I'm in quarantine watching Big Integrals playlist AGAIN hah it's just sooo good. Cheers to Lars
Hey, Blackpenredpen! I was trying to find the area under the graph of y=|[x^3]| from x=-2 to x=3. [.]- Greatest Integer Function and |.|-Modulus. Result involves a series of summation of the cube roots of first n integers. How do I go about solving this? Because the answer in my text book is given in the form of a numerical value. I really enjoy all of your integral videos, btw!
Btw it is ln not log. I didnt learn hyperbolic functions yet but based on what i know about them, they have the same shape as ln. However, they have different domains. U just use them because they are more convenient instead of spending time doing more trig sub or partial fractions
@@nvapisces7011 You can say log if you want. Please read this first paragraph here on Wikipedia en.m.wikipedia.org/wiki/Natural_logarithm#:~:text=The%20natural%20logarithm%20of%20a,is%20implicit%2C%20simply%20log%20x.
on the 4th line do we get inverse cot instead of natural logarithm (the 2nd integral)? maybe i am confused but the derivative of inverse cot is -1/(1+x^2) not 1/(1-x^2)
That is a hairy answer... Does it ever make sense to convert the integrand to a Taylor series and just integrate the resulting (infinite degree) polynomial?
Why we can't made this integral by part and then we can call 1+tanx=u with du=dx/cos^2(x) I don't see why it doesn't work, i mean I find it much easier to do, so it must be wrong😅...(in the integration by part i considered 1 equals g'(x), and I derived sqrt(1+tanx)...). Any help?
Integration by parts does not work because you are forgetting the square root of the integrand. In this case, the function being differentiated is sqrt(1 + tan(x)), not 1 + tan(x). This yields a much different integrand than what you expected.
It was great and understandable although I would recommend not using the inverse hyperbolic functions. Their usage is severely limited and they can be rewritten using natural logs which seems to be a better idea, especially considering that calc 2 students dont really use those things. But apart from that, it was a pleasant experience as always :)
Logarithmic term as solution represents only the real part of the solution, whereas inverse hyperbolic function gives complete solution without neglecting imaginary terms. For the sake of completeness, I think it's better to write ans in inverse hyperbolic function.
If f(x) = [f^(-)](x), then [f^2](x) = x. In other words, your question is equivalent to the question of what is the solution to sin[sin(x)] = x. Once this is acknowledged, realizing that the only solution is x = 0 is trivial.
The integral of f(x) from a to b is the same as the integral of f(a+b-x) from a to b. Could someone please correct me on this if it isn't always the case. But in this case it is so we'll let: I=integral from 0 to π/2 of sin³x/(sin³x+cos³x) and we'll also plugin 0+π/2-x into it so we get I=integral from 0 to π/2 of sin³(π/2-x) /(sin³(π/2-x) +cos³(π/2-x)). We notice that sin(π/2-x) =cos(x) and that cos(π/2-x) =sin(x) So we have sum the two I from which we get: 2I=integral from 0 to π/2 of (sin³x+cos³x) /(sin³x+cos³x) or the integral of 1dx So we get 2I = x [0 to π/2] or that 2I=π/2 in the end getting that the integral is π/4
Fede Fubbi In general, for any arbitrary function f that is well defined on the real interval (0, 1), the integral from 0 to π/2 of f[sin(x)]/(f[sin(x)] + f[cos(x)]) with respect to x is equal to π/4. This uses the similar reasoning as the comments above. You can exploit the fact that sin(x) = cos(π/2 - x) and cos(x) = sin(π/2 - x) to prove this. In the special case of your integral, f(x) = x^3, but this could have been done with any other function with domain (0, 1). Even a discontinuous function would have done. There is also somewhat of similar trick with tan(x) and cot(x). For example, consider any function f : R -> R with the properties f(x)f(y) = f(xy), f(x^n) = f(x)^n, and f(1) = 1. Then the integral from 0 to π/2 of 1/(1 + f[tan(x)]) is the same as the integral from 0 to π/2 of 1/(1 + f[cot(x)]). Since cot(x), you can multiply numerator and denominator by f[tan(x)] to get f[tan(x)]/(f(1) + f[tan(x)]), and since f(1) = 1, f[tan(x)]/(1 + f[tan(x)]). Notice that this is identical to our original integral I, so if we add it to our original integral, then on the one hand, we have 2I, and on the other hand, we have the integral from 0 to π/2 of (1 + f[tan(x)])/(1 + f[tan(x)]) with respect to x, which is trivially equal to π/2. Thus, I = π/4.
I feel so honored :) ... Hello to everyone and warm greetings from Belgium 😀😀😀
Hi Lars 👋😃
@@chirayu_jain Hey... ✌😄
Here’s the man!
@@blackpenredpen 😀
hey lars hope you are doing well
U, w, v
And x 😉
Bro why are you so cool? Pls ans.
Ur Amazing bro
From KASHMIR
how about this calculus with this formula (in degrees) limit for x approach for x pi=x\2*sin (360\(x+2)) \ sin((x\4)*(360\(x+2)))
I want to know how approach you can get even with a calculator (I tried this formula and works as much as you dont round the numbers (less the calculator to do)) and approach as much as bigger as x is (I sugggest to began aproach with x= 100 to get first digit 3 and for second you need x= apraxch 1000 so almost 10 to the power of 3 plus number of digit you want (for smaller values of diggits ) gets the most approach you can get by hand and calculator
And if I'm the one doing this, I will not use w and v, instead writing down the algebraic twin
Something like: d(u-sqrt(2)/u), d(u+sqrt(2)/u)
Did you pause & try?
Also, check out my 100 integrals where I first mentioned Lars. th-cam.com/video/dgm4-3-Iv3s/w-d-xo.html
Math tricks for any competition plz....
I don't understand how you keep so motivated doing math.
when you changed the W to the X you wrote + instead of -
btw check the answer via derivative :D
Came here just to say that hahahah
th-cam.com/video/_RwQLGYu5yk/w-d-xo.html
16 minute int just to fuk up at the end lol
Nearly as painful as if he’d have forgotten the + C
What a show man! And also big congrats to Lars for his winning the cancer finally! You two are awesome integral-fighter and cancer-fighter respectively 😎
Thank you!!! Yay!! We are all happy!
@@blackpenredpen & you are so awesome!!!!!!!
Dabloo?
I got A+ in calculus 21 years ago
Integration was my favorite game but after these years I totally lost my skill
I love math more than any other science but unfortunately I left studying it to have better job in engineering to gain more money
Math is the science that all inventions based on it
Love math
cool story :)
i found them back in my job :))
@@annevanderbijl3510 hello
@@aashsyed1277 hahahh hi ive seen you many times before
@@annevanderbijl3510 :0
now prove this by differentiating
The 379th Hero you can split into products as tan^2(x)*tan(x)=(sec^2(x)-1)tan(x)=tan(x)sec^2(x)-tan(x). The integral will come out to be (tan(x))^2/2+ln(abs(cos(x)))+C. You might end up the answer in terms of secant as (sec(x))^2/2-ln(abs(sec(x)))+C since it end up a different constant and using properties of logarithms.
th-cam.com/video/_RwQLGYu5yk/w-d-xo.html
Justin Lee lol
Please check the answer by differentiation.
there is a misstake sir !!!
when you were substituting u in the w expression
you puted a (+) insted of a (-)
I love your work
Do the proof that sqrt(2) is irrational in under sqrt(2) minutes!
th-cam.com/video/_RwQLGYu5yk/w-d-xo.html
Can be done in 2 mins
@@bharatipatel5076 He means "square root of 2" minutes, shorter than 2 minutes.
@@bharatipatel5076 it's about 84,85 seconds
Keep making more of these amazing videos! The world needs more of this!
I love the hidden symmetry in this integral!
Great ... I just enjoyed.
Thank you so much ❣️
when you went from u to x when integrating, you did + instead of - for u! all good though, nice job!!:)
Those who are womdering why he used hyperbolic cot instead of hyperbolic tan, it is because with tan there is a limitation, namely |x| 1 so that limitation works here
The other method is partial fraction decomposition where you divide u^4-2u^2+2 by
u^2-au+b. After working this out you'll get: a=sqrt(2+2sqrt(2)) and b=sqrt(2).
Very nice video :-) But you did a mistake in the last line: when resubstituting the w you wrote a plus instead of a minus :D
Your change of face expression at 2:41
😂😂
youtube must encourage these typeof educative channels
Wow, this one does not even involve special function in the answer. But the steps were really long. Nice video!
Noticed that too
Do it for Lars!
That would be a cool harshtag!
#doitforLars
Way easier to start off with the substitution 1 + tan(x) = cos^2(u) and then do some simple trig until you can integrate by partial fraction. Gives a MUCH nicer answers too: 1/sqrt(2) ln[sqrt(2) + sqrt(1+tan(x)) / sqrt(2) - sqrt(1+tan(x))] + c
13:20 can't we do the second integral using partial fractions?
This is such a big brain math play to make two integrals this way
I did not find it.
This was a very difficult one !
I prefer your aswer to that of wolframalpha, which goes needlessly to the complex world.
Thanks very much. I was stuck in integration of similar kind, and your videos did provide me with a solution.
U r a cool teacher🤟👍
Greetings from India
You could also write the answer with tanh^(-1). The differentiation for this is MONSTROUS (and fairly tedious), but it's doable.
I agree, I thought the choice between tanh^(-1) and coth^(-1) was based on the domain and since it's an indefinite integreal it's arbitrary. And maybe (1/2)ln|(1+x)/(1-x)| would be best since its domain includes all real numbers except +/- 1. It's not as fun as the inverse hyperbolic functions though
Please check the answer by diferentietion :)
A sign error happened at 16:34! Note that w = u - sqrt(2)/u.
Why did you choose coth^-1 rather than tanh^-1? Both can be differentiated to the form you need, right?
The polynomial P = u^4 + 2 u^2 + 2 factors over the reals (the only irreducible polynomials over R are linear or quadratic with negative discriminant).
Here is a factorization
P = (x^2 - a x + b) ( x^2 +a x + b)
where a = sqrt(2sqrt 2 + 2) and b = sqrt 2
Once you compute this, the integral is straightforward.
Yeah i did it this way, doing a long division by using u^2-au+b as the division factor.
7:37 Who else remembered Arthur??
Plz solve integral of sin theta^2
It’s hard to tell without the use of parentheses. If you meant the integral of sin(x^2), then it’s non-elementary but will come out as the answer of the sine Fresnel integral as S(x)+C. If you meant the other way as the integral of (sin(x))^2, then you have to change the identity as (1-cos(2x))/2. Doing so will get the answer to be x/2-sin(2x)/4+C.
@@justabunga1 th-cam.com/video/_RwQLGYu5yk/w-d-xo.html
Niraj Roy :Motivational and Teacher you’re doing the infinite nested square root derivative. He already did that in the video to show work.
@@RoyEduworks isko Hindi kaise smjega
@@justabunga1 yeah,my doubt was first case sin(x^2),got it
from South Africa! Hugs fan of yours! hope to be as profiecient as you are someday! much love.
what an insane integral!!!!
Proce by differentiation def of derivative and epsilon delta
Hi BlackPenRedPen! Didn't know if you noticed that in the inverse tan of your answer you switched plus for minus the answer should have been:
1/sqrt(2sqrt(2)-2)*arctan([1/sqrt(2sqrt(2)-2)](sqrt(1+tanx)-2/sqrt(1+tanx)))-1/sqrt(2sqrt(2)-2)*arctanh([1/sqrt(2sqrt(2)-2)](sqrt(1+tanx)+2/sqrt(1+tanx)))+C.
But that was an awesome job. Integrals can be tricky, but you do an amazing job.
let 1+tanx = u^2
sec^2xdx = 2udu
so the original integral becomes
(u.2udu)/(1+u^2)
then simplifying it we get
2du - 2du/(1+u^2)
so we get
2u-2tan^-1(u)
2*(sqrt(1+tanx)) - 2*tan^-1(sqrt(1+tanx))
Won't this be easier? Or am I wrong somewhere?
Tricky one without knowing with substitution to use.
I did tried using integración by parts multiple times and some regular substitution in the middle, i got to a point were i had de integral of sqrt(tanθ), that's when i stopped because i know that isn't pretty (or is way too pretty, depending on how messed up you are).
14:03 should be minus.
Much respect for you!
Imagine fail just for that.
easy peasy lemon sqeezy for that one. holy moly
Whats the best way to write the integral of tan^3(x) or are they no better writings of the answer?
the chessmate tan(x)^3 = tan(x)·tan(x)^2 = tan(x)·[sec(x)^2 - 1]. Use linearity, and now you can consider the integrals of tan(x)sec(x)^2 and the integral of tan(x). The integral of tan(x) is given by ln|sec(x)| + C(x), where C(x) is an arbitrary piecewise step function with discontinuities whenever x = πn + π/2 for some integer n. The integral of tan(x)sec(x)^2 can be calculated by letting u = tan(x) => du = sec(x)^2 dx, which gives the integral of u with respect to u, which is equal to u^2/2 + C, or tan(x)^2/2 + C. Adding the integrals results in tan(x)^2/2 - ln|sec(x)| + C(x). Is there a better way to write the antiderivative? Other than switching tan(x)^2 for sec(x)^2, which is permitted because they differ by a constant, no, there is not a better way, as far as I am concerned.
@@angelmendez-rivera351 alright thank you i wrote it with sec instead of tan but i know some people said they wrote it with ln(cos) instead of ln(sec) wanted To know if it was any better or worse lol
@@Simplement724 Well lnIcos(x)I = -lnIsec(x)I
Hello blackpenredpen,
May I ask, what is the limit of the expression "(W(x)/ln(x))^x" as x approaches a sideways 8 (infinity)?
Eso si es de gánster, muy buen video siempre es genial ver el nivel hard de estos ejercicios
Check the answer via differentiation
Hey bprp, I checked your apparel, why doesn't your "for every ϵ > 0" t shirt have the rest of the limit definition on the back side? thanks :)
It was requested by someone who just wanted that to be in the front. And the good thing is the cost is cheaper if it’s just one side print, too.
I love how you go and just say "pause the video and try this first :)"
have to say that i felt a little emotional at the beginning of the video, I'm also an engineer student and fighted cancer exactly 3 months ago, everything is fine now but anyways I'm in quarantine watching Big Integrals playlist AGAIN hah it's just sooo good. Cheers to Lars
How long should I try to solve an integral before giving up?
Hey! you should check the answer by differentiating it! just to be sure you got it right
Mistake 14.04 -- minus ave
You should do the diferenciation
Thank you for your videos. But did you see my question on integration of x^(x^2)? Is it integrable? Thank you
I don’t think it has a nice answer
very helpful! thx a lot!
Hi bprp, w=(u-(sqrt2/u)), the good thing is, the arctg() argument is not a typo because you wrote a + sign there?
Sir, have you uploaded the video of riemann sum proof??
what about its derivative ?
Let's see... so, we have f(x) = sqrt(1 + tan(x)), therefore f'(x) = sec(x)^2 / (2 * sqrt(1 + tan(x))). :P
Sir where you put the main function of 'w' there will be a minus sign
Hey, Blackpenredpen! I was trying to find the area under the graph of y=|[x^3]| from x=-2 to x=3. [.]- Greatest Integer Function and |.|-Modulus. Result involves a series of summation of the cube roots of first n integers. How do I go about solving this? Because the answer in my text book is given in the form of a numerical value. I really enjoy all of your integral videos, btw!
I am confused on how hyperbolic integrals work. I instead use (1/2a)log((x + a)/(x - a)) on 1/(x^2 - a^2)
Btw it is ln not log. I didnt learn hyperbolic functions yet but based on what i know about them, they have the same shape as ln. However, they have different domains. U just use them because they are more convenient instead of spending time doing more trig sub or partial fractions
@@nvapisces7011 You can say log if you want. Please read this first paragraph here on Wikipedia en.m.wikipedia.org/wiki/Natural_logarithm#:~:text=The%20natural%20logarithm%20of%20a,is%20implicit%2C%20simply%20log%20x.
Please make a video checking this via derivative. Do it for Lars
more substitutions than the pokemon move
on the 4th line do we get inverse cot instead of natural logarithm (the 2nd integral)? maybe i am confused but the derivative of inverse cot is -1/(1+x^2) not 1/(1-x^2)
What could dx/dy look like or mean in general?
I like Scott Joplin. Who is the musician ?
W=U - note + sqrt2/u
What's the integral of (x^(1/(1+x)) - x^(1/(1-x)))dx?
My bro I see a mistake for w substitution . Put - not +. Or maybe I can't see properly but I think you forgot to subtract not addition
Really similar to integrating sqrt(tan(x)), making algebraic twin again!
the correct factor in the denorminator is {u-(1/u)}²-4
please kindly disregard my above comment i wasn't thinking very well thanks so much for sharing this video you're highly appreciated sir
🗣️DIVIDEE BY YOU SQUARE🔥
That is a hairy answer... Does it ever make sense to convert the integrand to a Taylor series and just integrate the resulting (infinite degree) polynomial?
Love you ❤️❤️
Why are the graphs of sqrt(1+tan(x)) and 2x^2/(x^4 - 2x^2 + 2) so different ?They have the same integral.
Joluju 😂 he did a u sub, thats why they are different
Check answer via derivative.
imagine in a test, this question comes and you forgot to write the C at the end.🙂
Check the derivative!!!
So if you d/dx all that mess, you get sqrt(1/tan-1 x) ?
Yes because in the answer he made a mistake with the arctan function, instead of a minus he wrote a plus, that's why your answer came out different.
Why we can't made this integral by part and then we can call 1+tanx=u with du=dx/cos^2(x)
I don't see why it doesn't work, i mean I find it much easier to do, so it must be wrong😅...(in the integration by part i considered 1 equals g'(x), and I derived sqrt(1+tanx)...).
Any help?
Integration by parts does not work because you are forgetting the square root of the integrand. In this case, the function being differentiated is sqrt(1 + tan(x)), not 1 + tan(x). This yields a much different integrand than what you expected.
@@angelmendez-rivera351 I took that in mind, you get xsqrt(1+tanx) - 1/2 integral of (sec^2/sqrt(1+tanx))dx wich you can do by sustitution, ain't it ?
@@angelmendez-rivera351 Sorry, just found the error, i forgot an x in the integration by part, my mistake 🙌🏻
Really I had done EXACTLY same
Check the answer
In 13:14 you make a little mistake . You should write tanh-1(v) not coth-1(v) !!! I'm right ???
It was great and understandable although I would recommend not using the inverse hyperbolic functions. Their usage is severely limited and they can be rewritten using natural logs which seems to be a better idea, especially considering that calc 2 students dont really use those things. But apart from that, it was a pleasant experience as always :)
Logarithmic term as solution represents only the real part of the solution, whereas inverse hyperbolic function gives complete solution without neglecting imaginary terms. For the sake of completeness, I think it's better to write ans in inverse hyperbolic function.
@@itsviv1 dont absolute value signs make it better
I think that you have a mistake, in the first part of the answer we have sqrt(1+tan(x))-sqrt(2÷(1+tan(x))) not a positive sign
Nice Lars ! Gg for having fought against that shitty cancer !!! This will turn into an old bad memory now ;)
Amazing
What value of sinx = sin inverse x pls tell me and How ???????
sin(x)=sin^-1(x) when x=0
Mirunal meenakshi Zero
If f(x) = [f^(-)](x), then [f^2](x) = x. In other words, your question is equivalent to the question of what is the solution to sin[sin(x)] = x. Once this is acknowledged, realizing that the only solution is x = 0 is trivial.
Check by Differentiation😎😎😎
What is the integral of (-1)^2
X
M if it’s with respected to M
@@blackpenredpen (steve) if it's in respect to (steve)
@@durgeshnandinijha6054 + C :)
Quick question couldn’t you just substitute tan(x) for tan^2(θ). This would make this equation way more simple.
you get sectheta as the the main thing but the dx is the problem
Ohhmmgggg please then differentiate the answer back to the question 😭
That would be lovely 😅😅😅
'It wasnt painful at all...' - BPRP
I'd like to be sure of something: when we integrate at the end, we can replace inverse coth by inverse tanh, it doesn't change anything?
What happened with all those copied videos yesterday?
He deleted them because they were accidental, check the community post.
# Teacher of Chirayu Jain #!
Brutal!!!!!!!!
lost all hope bro used the formula for the derivate of the inverse hyperbolic cotangent
Error in minute 12 and 5 sec ! Taking out 1/sqrt(2 sqrt(2)-2) should be 1/(2 sqrt(2)-2) same for 1/(2 sqrt(2)+2)
can you do integral of sin^3x/(cos^3x+sin^3x) from 0 to pi/2? thanks sir in advance
The integral of f(x) from a to b is the same as the integral of f(a+b-x) from a to b. Could someone please correct me on this if it isn't always the case. But in this case it is so we'll let:
I=integral from 0 to π/2 of sin³x/(sin³x+cos³x) and we'll also plugin 0+π/2-x into it so we get
I=integral from 0 to π/2 of sin³(π/2-x) /(sin³(π/2-x) +cos³(π/2-x)).
We notice that sin(π/2-x) =cos(x) and that cos(π/2-x) =sin(x)
So we have sum the two I from which we get: 2I=integral from 0 to π/2 of (sin³x+cos³x) /(sin³x+cos³x) or the integral of 1dx
So we get 2I = x [0 to π/2] or that 2I=π/2 in the end getting that the integral is π/4
@@mislavplavac6641 thank you!!
Goran Plavac Yes.
Fede Fubbi In general, for any arbitrary function f that is well defined on the real interval (0, 1), the integral from 0 to π/2 of f[sin(x)]/(f[sin(x)] + f[cos(x)]) with respect to x is equal to π/4. This uses the similar reasoning as the comments above. You can exploit the fact that sin(x) = cos(π/2 - x) and cos(x) = sin(π/2 - x) to prove this. In the special case of your integral, f(x) = x^3, but this could have been done with any other function with domain (0, 1). Even a discontinuous function would have done.
There is also somewhat of similar trick with tan(x) and cot(x). For example, consider any function f : R -> R with the properties f(x)f(y) = f(xy), f(x^n) = f(x)^n, and f(1) = 1. Then the integral from 0 to π/2 of 1/(1 + f[tan(x)]) is the same as the integral from 0 to π/2 of 1/(1 + f[cot(x)]). Since cot(x), you can multiply numerator and denominator by f[tan(x)] to get f[tan(x)]/(f(1) + f[tan(x)]), and since f(1) = 1, f[tan(x)]/(1 + f[tan(x)]). Notice that this is identical to our original integral I, so if we add it to our original integral, then on the one hand, we have 2I, and on the other hand, we have the integral from 0 to π/2 of (1 + f[tan(x)])/(1 + f[tan(x)]) with respect to x, which is trivially equal to π/2. Thus, I = π/4.
@@angelmendez-rivera351 👌👌👌
Integration of sqrt of 1+xtanx ????
This is an non-elementary integral.