integral of 1/(x^4+1) by brutal-force partial fraction, • Integral of 1/(1+x^4) ... Derivative of tanh^-1(x) and coth^-1(x), • THE CONFUSING DERIVATIVES Integral of sqrt(tan(x)), • integral of sqrt(tan(x... blackpenredpen
The mind boggles. Thing is, the reason why I love your videos mate, is that I can follow each step easily (I have some university mathematics to rely on) ... but it's just the creativity of coming up with the next step that impresses me all the time! I would take one of those steps and while understanding it, would not even think of it in the first place. Really good stuff my friend, inspiring to an old bloke like me!
11:38 You could also have used the formula (integral)dx/(x^2-a^2) = (1/2a)ln|(x-a)/(x+a)| + C That's a much quicker method, but the good part about your method is that....I finally got to know the derivatives for inverse hyperbolic tangent and cotangent :)
14:00 I've been wondering how often that happens. Usually markers are alcohol based, and alcohol is volatile, and the two color switch technique requires to keep the markers uncapped. I read there are water based erasable markers, perhaps those last longer, I haven't tried.
it could be done in a short way. The integral = 1/2 integral of (1-x^2)/(x^4 +1) + (1+ x^2)/(x^4 + 1), the first integral could end up with a log function after letting t = 1/x + x, and the second integral could end up with an arctan function after letting t = 1/x - x, then consider the integral limits,
Awesome video....😘👌 But I have another question about differential equation... That 🤔In 1715,20 may ,Leibniz revealed the solution of the differential equation x^2.y"=2y and I don't know how to find the solution of this differential equations can you please make videos on it ...🙋
If we want to calculate numerically it is good idea to use change of variable u=1/x then we will get Int(u^2/(1+u^4),u=0..1) Answer is close to one quarter If we need to have initial guess one quarter is good option
To them who find this problem difficult , you might get disappointed but this problem is fuckin easy, its just 4-5 min problem. Well that my opinion. -Jee aspirant,Indian.
Can we just put x² = u and just calculate the primitive of 1/1+u². Then we can write the primitive of this function as a limit [arctan(u) ] between 1 and infinite. Because the function is continue on I=[1,inf] And the function 1/1+u², in +Inf, is equivalent to a Riemann function with t^-2 which converge α=2 >1. So, by comparison theorem 1/1+u² CV. We can write the integral is equal to the Primitive's limit in +Inf minus the value of the primitive in 1. u = x² does not change the result: π/2 - π/4 = π/4
here's a deceivingly simple integral, you could go down the rabbit hole of trig identities, or you could do it the "easy way": integral of (7sin(x)+9cos(x))/(4cos(x)+5sin(x)) dx The easy way would be to say "wouldn't it be nice" and express 7sin(x)+9cos(x) as A(4cos(x)+5sin(x))+B(4cos(x)-5sin(x)) for some constants A and B: 7sin(x)+9cos(x) = A(4cos(x)+5sin(x))+B(4cos(x)-5sin(x)) 7sin(x)+9cos(x) = (5A-5B)sin(x)+(4A+4B)cos(x) 7=5A-5B 9=4A+4B solving the linear system of equations with gives A=73/40 B=17/40 Then the integral becomes integral of [(73/40)(4cos(x)+5sin(x))/(4cos(x)+5sin(x))+(17/40)(4cos(x)-5sin(x))/(4cos(x)+5sin(x))] dx = integral of [(73/40) + (17/40)(4cos(x)-5sin(x))/(4cos(x)+5sin(x))] dx which can be solved by simple u-substitution.
This is like a weirder best friend, who wants part of him to have a fourth power and part not, part negative of the original and part not, and his x wants to go from infinity to 1 instead of being trapped from -1 to 1. This best friend wants to explore his limits and see if he can go to infinity lmao
You did not finish to the end of simplification of final answer. arc cot h (x) = 1/2 ln ((x+1)/(x-1)). So, arc cot h (sqrt(2)) = 1/2 ln [(sqrt(2)+1) / (sqrt(2)-1)] = ln(sqrt(2)+1). So, final answer = [ pi/sqrt(32) + ln(sqrt(2)-1) / sqrt(8) ] = same as your answer but much simpler without "inverse cot h". Use cosh(x) + sinh(x) = exp(x) to compute inverse-cot-h.
Assuming that each of the 2nd through 4th terms are entirely under their square roots, the solution is as follows: Given x + √x(x+1) + √x(x+2) + √(x+1)(x+2) = 2, x > 0 , then x + √(x^2+x) + √(x^2+2x) + √(x^2+3x+2) = 2, x > 0. Plugging in x = 0 we get: 0 + √(0^2+0) + √(0^2+2*0) + √(0^2+3*0+2) = √(0+0) + √(0+0) + √(0+0+2) = √0 + √0 + √2 = 0 + 0 + √2 = √2, which is < 2. Plugging in x = 1 we get: 1 + √(1^2+1) + √(1^2+2*1) + √(1^2+3*1+2) = 1 + √(1+1) + √(1+2) + √(1+3+2) = 1 + √2 + √3 + √6, which is > 2. Notice that all 4 terms always increase with x when x is > 0. Therefore, the solution must be x = some number between 0 and 1. Going further, x + √x(x+1) + √x(x+2) + √(x+1)(x+2) = 2 turns into √x * [√x + √(x+1)] + [√x + √(x+1)] * √(x+2) = 2 [√x + √(x+1)] * [√x + √(x+2)] = 2, and that's all I got. However, Wolfram Alpha says that the answer is 1/24. (1/√24 + 5/√24) * (1/√24 + 7/√24) = (6/√24) * (8/√24) = 48/24 = 2.
I may be wrong, but did he make a small error at 13:30 . 1/1-x^2 is not actually (1/w^2+1)-1 . So he could nit change thr sign of the 1/2 outside the integral to positive.
The graphs of arctanh(x) and arccoth(x) don't really look similar, although they start and end at the same points so I guess you can say they togther extend the domain of the integral of 1/(1-x²) with x = ±1 being a singularity. That being said, their sister functions arctan(x) and -arccot(x) (notice the minus) *are* off by a constant (although the constants depends on the value of x because arctan(x) + arccot(x) is π/2 if x ≥ 0 and -π/2 if x < 0).
@@MarioFanGamer659 is the singularity at just the value where re=1 or -1 and the imaginary value is 0, or for all imaginary values for real value 1 or -1?
@@alextaunton3099 As a rule, if something isn't mention, it's implied to be zero, not any value. In fact, this is quite consistent, I don't know how you came to the idea I implied _all_ imaginary numbers (i.e. a line of singularities) and not just two values with no imaginary component (i.e. two points).
Watching normally his videos(with blue and black pen) but you see blue pen entering(something serious is going to happen ) and at last green pen enters 🤯🤯🤯🤯🤯🤯🤯🤯🤯🤯🤯🤯🤯🤯
Hey can you please link me to one of your videos explaining the 11:02 step / or someone can please explain to me ? I don't understand this step with tan-1. Thank you
L.G.F.C Raise both sides to the power of x: x = i^x Then take ln of both sides: ln(x) = x ln(i) x = e^ln(x) so we can write ln(x) = e^ln(x) * ln(i) Divide both sides by -e^ln(x): -ln(x) * e^-ln(x) = -ln(i) Now we can use the W function because we have something of the form z * e^z: -ln(x) = W(-ln(i)) so x = e^-W(-ln(i)) ln(i) = i*pi/2 so x = e^-W(-i * pi/2) Hope this helped
@@blackpenredpen 1. This is a definite integral, not an antiderivative, using the same bounds as used in the video. 2. Wolfram says the value of the integral is 2 Gamma(5/4)^2/sqrt(pi) ~ 0.92704, using the gamma function, which has featured on this channel before. It would be interesting to see how that is actually obtained by hand.
[EDIT: I thought the integral was on R⁺ LOL ,anyway, using u-sub u=1/x, we can show that the integral on R⁺ is 2 times the integral on [1, infinity) @@blackpenredpen Hi (from france btw xD) blackpenredpen and mike4ty4, for the defined integral, here’s a BEGINNING of proof, using Ramanujan’s Master theorem: en.wikipedia.org/wiki/Ramanujan%27s_master_theorem Let’s call F(x)=sqrt(1/(1+x⁴)) First notice that the mellin transform of F is nice : www.wolframalpha.com/input/?i=mellin+transform+of+1%2Fsqrt(1%2Bx%5E4) We can define (R) as : (R) : mellin transform of F = gamma(s/4)*[gamma(1/2-s/4)/(4*sqrt(pi))] and M(s)=mellin transform of F Considering s’=s/4, using u-sub u=t⁴ such that we can have a term in u^(s-1) we have : M(s’)=1/4*mellin transform of 1/sqrt(1+u) = M(s) www.wolframalpha.com/input/?i=mellin+transform+of+1%2F4*sqrt(1%2F(1%2Bx)) We want to prove the relation : (R’) : M(s’)= gamma(s’)*[gamma(1/2-s’)/(4*sqrt(pi))] And then we want to compute the value : M(s’=1/4) which will give us the answer. (R)(R’) for some values of s (convergence for the values we are interested in, can be shown with Riemann, for x==> infinity ; 1/x^alpha converges alpha >1) It turns out : sum of (-x)^k / k !*[gamma(1/2+k)/(4*sqrt(pi))] from 0 to inf= 1/sqrt(1+x) for |x|
One can compute z:=coth^{-1}(√2) by writing √2=coth(z)=(e^{2z}+1)/(e^{2z}-1), multiplying both sides by e^{2z}-1, solving for e^{2z}, and then taking the log to solve for z. He should have done that instead of leaving his final result as he has, with coth{-1} still flying around.
Dániel Mácsai Yeah, just square the dare root. I mean, if you want only the square root being positive, fine, but then you just like to nitpick. Squaring cancels with the square root, and I used the polynomial of (a+b)^2=a^2+2ab+b^2 to solve (x^2-5)^2.
Very tricky and not direct. Very non intuitive. The majority of us do not think like this. Most of us would not see to do this slight of hand approach. Unfair problem. Only someone like you could be expected to see to do this.
