Hey Professor, Thanks a lot to you again for a new integral video. I commented on your previous integral video about an indefinite integral which came in my class 12th Maths board exam at the year 2014. The problem was, I myself was unable to recall the solution to that problem. But by watching just a minute of this video, I immediately remember the entire solution and solved it again. If you get the solution of the indefinite integral I mentioned in my previous comment, you will find that a part of the solution which you have shown in this video applies there also. Some complicated integrals become so interesting that its solution remains in the mind even till a decade.
Hello Mr Prime Newtons, hope you're doing well, so I'm a student in a graduate school and the teacher gave us this exercise: (1) Let a,b and s be real numbers such that a < b and 0 < s < b-a . Prove that a number n in the Z group exists such that a < ns < b. Please can you take a look at it and maybe help me solve it please? I'll appreciate it ❤❤❤ , greetings.
Try using the Archimedean property: since s > 0, there exists an integer n such that ns > a. In particular, take the smallest such integer (so that ns > a but (n-1)s a. Now we need to show ns < b. To do this, note that (n-1)s
Mr prime newtons. I am starting a revolution in my school where we are writing -C instead of +C on indefinite integrals. Would you like to join me and spread the word?
@@YohannesGetu-o3x A differential cannot be equal to an algebraic expression by itself. For a given function y=f(x), dy is equal to; (derivative of f(x))*dx. For the second type of mistake I mentioned, he confused the plus sign for a minus sign which made for a basic mistake.
If you turned this into a definite integral on x in [0,1], how do you handle the fact that the solution seems undefined (division by 0). The integral I should have no problem here.
12:50 Isn't it -(1/a)tanh^-1(x/a) formula? Because int{(1/(a^2-x^2)}dx=(1/a)tanh^-1(x/a)+c. Therefore as it has -ve Infront, isn't it +ve value? Or I am getting confused?
cool video! But you forgot to make it negative arctanhx because the formula is 1/(a^2-x^2) not 1/(x^2-a2). So the answer should have a plus in between the arctangent of x and the arctanh of x!
12:05 to early for change the integral from one integral to two integral, must you separate the integral first and put dx in each integral than you can change the numerator with derivatif form of denominator . I guest that my opinion i'm sorry if my English is bad and mispronunciation
I was thinking of splitting the denominator into quadratics but the roots are easy to find X^4+1=0 X^4=-1 X^2=+-i X1= √2/2+i√2/2 or x2=-√2/2-√2/2i for x^2=i and X3=√2/2-i√2/2 and x4=-√2/2+√2/2i for x^2=-i Using vietas formulas and making a polynomial out if x1 and x3: x^2-√2x+1 Making a polynomial from x2 and x4: x^2+√2x+1 So (ax+b)/(x^2-√2x+1)-(cx+d)/(x^2+√2x+1)=1/(x^4+1) Equatig coefficients: x^3: a-c=0 => a=c X^2 b-d+√2a+√2c=0 X: √2d+√2b+a+c=0 b-d=1 => b=d+1 1+2√2a=0 a=-√2/4 2√2d+√2+2(-√2/4)=0 2√2d=-√2/2 d=-1/4 The rest should be easy by grouping up to obtain logarithms and arctangents
DOING BEFORE THE VIDEO Using the residue method when Xi is a simple root, we firstly need to find the roots of xˆ4+1, that are e^iπ/4, e^i3π/4, e^i5π/4 and e^i7π/4, and now, we get that An = 1/4(Xn)ˆ3 by getting the derivative of xˆ4 Therefore: A0= 1/4e^(i3π/4), A1=1/4e^(i9π/4), A2=1/4e^(i15π/4) and A3=1/4e^(i21π/4) So, the integral (1/xˆ4+1) dx =1/4 integral ((∑0->3)1/ (e^(i(3π/4+3πn/2))x - e^(i(π+2πn))) dx which is equal to 1/4 (∑0->3) ln( e^(i(3π/4+3πn/2))x-e^(i(π+2πn)) )/ e^(i(3π/4+3πn/2))
Sir please do a limit question which was came in JEE Advanced 2014 shift-1 question number 57 it's a question of a limit lim as x-->1 [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4 You have to find the greatest value of a It has 2 possible answers 0 and 2 But I want the reason that why should I reject 2 and accept 0 Because final answer is 0 Please help 😢
You made a slight error. The hyperbolic tangent should be negative because integral of 1/(x^2 - a^2) is actually -(1/root(a))tanh^-1(x-root(a)). That tanh^-1 term should be positive.
First thing that comes to my mind would be partial fraction decomposition using the well known identity x⁴ + 1 = (x² + √2 x + 1) (x² - √2 x + 1) so that 1 / (x⁴ + 1) = [ (√2 x + 2) / (x² + √2 x + 1) + (- √2 x + 2) / (x² - √2 x + 1) ] / 4
Ayyyy my favourite youtuber. Thanks for a video on integrals. I love these new videos. Thanks!
Hey Professor,
Thanks a lot to you again for a new integral video.
I commented on your previous integral video about an indefinite integral which came in my class 12th Maths board exam at the year 2014.
The problem was, I myself was unable to recall the solution to that problem. But by watching just a minute of this video, I immediately remember the entire solution and solved it again.
If you get the solution of the indefinite integral I mentioned in my previous comment, you will find that a part of the solution which you have shown in this video applies there also.
Some complicated integrals become so interesting that its solution remains in the mind even till a decade.
Some of the best teachers about integrals, even with english being my 2nd language I understand everything
in the minute 11:15 , it would not be du=(1+x^(-2))dx ?
yes
You’re a happy person I love it!
Thanks!
Thank you
Clean solution. Beautifully explained. ❤
i like the way you say at the end, "beautiful" 💙
Integrate[1/(1+X^4),X]=1/(2Sqrt[2])[Arc Tan[(X^2+1)/(XSqrt[2])]-Arc Tanh[(X^2-1)/(XSqrt[2])]]+C It’s in my head.
bro's traumatized
👍👍 Explanation really clear too !
Great job explaining the solution to a complex math integral.
Hello Mr Prime Newtons, hope you're doing well, so I'm a student in a graduate school and the teacher gave us this exercise:
(1) Let a,b and s be real numbers such that a < b and 0 < s < b-a . Prove that a number n in the Z group exists such that a < ns < b. Please can you take a look at it and maybe help me solve it please? I'll appreciate it ❤❤❤ , greetings.
Try using the Archimedean property: since s > 0, there exists an integer n such that ns > a. In particular, take the smallest such integer (so that ns > a but (n-1)s a. Now we need to show ns < b. To do this, note that (n-1)s
Mr prime newtons. I am starting a revolution in my school where we are writing -C instead of +C on indefinite integrals. Would you like to join me and spread the word?
Leta fcking go
Wonderful 😍 I love integrals very much❤
I think actually du is equal to (x ^ 2 ) + (1 / (x ^2)) not (x ^ 2 )- (1 / (x ^2))
Thanks ❤❤❤
why does he always do the substitutions right, but differentiate them incorrectly? 5:58, 8:08. Another type of mistake: 11:11.
Yeah, I noticed the same thing.
But I don't see any mistake there
@@YohannesGetu-o3xis 11:11 correct? He mixed up + and - there, didn’t he? The rest is correct.
@@Kosekans Okay I saw it. Tnx for the illustration
@@YohannesGetu-o3x A differential cannot be equal to an algebraic expression by itself. For a given function y=f(x), dy is equal to; (derivative of f(x))*dx. For the second type of mistake I mentioned, he confused the plus sign for a minus sign which made for a basic mistake.
We can also use trignometric substitutions
Thank you, I love this VOO clip .
If you turned this into a definite integral on x in [0,1], how do you handle the fact that the solution seems undefined (division by 0). The integral I should have no problem here.
12:50 Isn't it -(1/a)tanh^-1(x/a) formula? Because int{(1/(a^2-x^2)}dx=(1/a)tanh^-1(x/a)+c. Therefore as it has -ve Infront, isn't it +ve value? Or I am getting confused?
cool video! But you forgot to make it negative arctanhx because the formula is 1/(a^2-x^2) not 1/(x^2-a2). So the answer should have a plus in between the arctangent of x and the arctanh of x!
Same thing. I got +ve too. I too commented on it.
Very nice demonstration.
Why not just substitute in sqrt(tan(theta))
12:05 to early for change the integral from one integral to two integral, must you separate the integral first and put dx in each integral than you can change the numerator with derivatif form of denominator .
I guest that my opinion i'm sorry if my English is bad and mispronunciation
Beautiful! (Besides the couple observations that have been made here)
Looks good. Easy to follow.
Wonderful as usual!
Good Explanation
Excelent 😊
∫1/(x²-a²)dx = -1/a arctanh(x/a)
You have a lil wrong but good job i realy learn a lot of you keep going ❤❤
yeah I noticed that but wouldn't that make the final integral have a "+" in between the tan and tanh inverse functions instead of a "-".
@lcex1649 This is exactly what we are aiming for . try the derivative you will understand
This problem can be done by residues as well right?
Nice trick, but I prefer factoring the denominator and then using partial fraction decomposition.
He said he will be doing 3 videos using 3 different methods
@@henrymarkson3758 You are absolutely right
I was thinking of splitting the denominator into quadratics but the roots are easy to find
X^4+1=0
X^4=-1
X^2=+-i
X1= √2/2+i√2/2 or x2=-√2/2-√2/2i for x^2=i and
X3=√2/2-i√2/2 and x4=-√2/2+√2/2i for x^2=-i
Using vietas formulas and making a polynomial out if x1 and x3: x^2-√2x+1
Making a polynomial from x2 and x4: x^2+√2x+1
So (ax+b)/(x^2-√2x+1)-(cx+d)/(x^2+√2x+1)=1/(x^4+1)
Equatig coefficients:
x^3: a-c=0 => a=c
X^2 b-d+√2a+√2c=0
X: √2d+√2b+a+c=0
b-d=1 => b=d+1
1+2√2a=0 a=-√2/4
2√2d+√2+2(-√2/4)=0
2√2d=-√2/2
d=-1/4
The rest should be easy by grouping up to obtain logarithms and arctangents
Yo hi..... try using substitution of x^2=t and then again substitute 1/t^2 +1 it wil be easy
You wrote done the wrong expression for du !!
Substitution of "u" is right but its differentiation "du" is incorrect.
Try watching a little closer, I thought that aswell
It was a typo
The way to arrive at the log expression for the second integral is to expand its integrand in partial fractions.
Why is falcon teaching math?
This is called algebraic twins methodd
DOING BEFORE THE VIDEO
Using the residue method when Xi is a simple root, we firstly need to find the roots of xˆ4+1, that are e^iπ/4, e^i3π/4, e^i5π/4 and e^i7π/4, and now, we get that An = 1/4(Xn)ˆ3 by getting the derivative of xˆ4
Therefore: A0= 1/4e^(i3π/4), A1=1/4e^(i9π/4), A2=1/4e^(i15π/4) and A3=1/4e^(i21π/4)
So, the integral (1/xˆ4+1) dx =1/4 integral ((∑0->3)1/ (e^(i(3π/4+3πn/2))x - e^(i(π+2πn))) dx
which is equal to 1/4 (∑0->3) ln( e^(i(3π/4+3πn/2))x-e^(i(π+2πn)) )/ e^(i(3π/4+3πn/2))
(x4 +1) = ( x2 +1 + rac 2x ) ( x2 +1 - rac2x )
Hey , why is tye channel named prime Newton
Sir please do a limit question which was came in
JEE Advanced 2014 shift-1 question number 57
it's a question of a limit
lim as x-->1
[{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4
You have to find the greatest value of a
It has 2 possible answers 0 and 2
But I want the reason that why should I reject 2 and accept 0
Because final answer is 0
Please help 😢
nice method!
11:17 he has made a mistake. A + symbol was required, instead of - symbol
11:16. The is a mistake:
u=x-1/x
du= 1+1/x^2 dx
du/dx=1+1/x^2
du=(1+1/x^2)dx
as in12:00
11:11 Prof you missed the '+' sign in the differentiation 🫡
Thank you!
wait, antiderivative of (1/( x^2 - a^2) ) = (1/(2a))(ln((x - a)/(x + a)), let try antiderivative of 1/ ( x^2 - 1)
du = (1 + 1/x^2) dx (sign correction)
Radical Racional
thầy giải hộ em nguyên hàm của 1/x^8+1 vs
I tried to integrate 1/(x^5 + 1) but it's so difficult!
Starting from x^3 = t, dx = dt/3x^2 would be much easier
May you please make a video of the Fibonacci sequence
❤
İkinci bölüm hatalı çözüm,dikkat
why cant use (1+x^2)^2 - 2x?
Use logarithm
You made a slight error. The hyperbolic tangent should be negative because integral of 1/(x^2 - a^2) is actually -(1/root(a))tanh^-1(x-root(a)).
That tanh^-1 term should be positive.
Знание сила
First thing that comes to my mind would be partial fraction decomposition using the well known identity
x⁴ + 1 = (x² + √2 x + 1) (x² - √2 x + 1) so that
1 / (x⁴ + 1) = [ (√2 x + 2) / (x² + √2 x + 1) + (- √2 x + 2) / (x² - √2 x + 1) ] / 4
It is not "well-known" identity, but do go that route.
@@robertveith6383 x^4-1=x^4+2x^2+1-2x^2=(x^2+1)^2-2x^2= (x^2+1)^2-(sqrt(2)x)^2
Why can't you change you username? You need to make your own name.
I love my username
Too much talking.
🤡