Integral of arctan(x^2) from 0 to 1 using power series, calculus 2 tutorial
ฝัง
- เผยแพร่เมื่อ 30 ก.ย. 2024
- Integral of arctan(x^2) from 0 to 1 using power series,
Please subscribe for more math content!
Check out my T-shirts & Hoodies: teespring.com/...
support bprp on Patreon (there's a discount code to bprp Teespring store for everyone) / blackpenredpen
Check out my site & social media
😃 blackpenredpen...
😃 / blackpenredpen
😃 / blackpenredpen
Thank you for your support!
blackpenredpen | 曹老師
![ผัวเถื่อน - ยูริ โตเกียวมิวสิค [ OFFICIAL MUSIC VIDEO ]](http://i.ytimg.com/vi/P08ORRVDGl8/mqdefault.jpg)
![ผัวเถื่อน - ยูริ โตเกียวมิวสิค [ OFFICIAL MUSIC VIDEO ]](/img/tr.png)
Plzz integrate
ln(ln(ln(lnx)))
LOVE IT!!!!
Math 1052 ¿?¿?¿?
Ah finally I relate more to these .. now studying engineering and actually my final in calc 2 is in 4 days 💙 keep it up LOOOVE your channel
Could anyone tell me a god book for calculus.... By the way I am your huge fan sir..
I used Stewart for calc2 and I like it!
Why not use sigma notation, problems like these are much easier to look at if you just write it like that
How curious, you can do integration by parts to integrate that :D. And you don't need any top notch functions like erf, Si and those.
Perfect!!! This is what I needed to see. Cleared it up well. Thanks.
You could've used u= arctan (x^2) and integrate
Oh dear God, don't try this. The integration gets VERY nasty with all that trig. Secant and cosecants to fractional powers. Absolutely horrible
@Mathias ICARTE MANCILLA this method also ends in a VERY nasty state. Do no try
GhostyOcean That is not that terrible, actually, because you can always convert functions to fractional powers into related functions with integer powers. I would take it over integrating arctan of a function.
I think x^2 = u is wiser, since du = 2x dx => du/2SqRt(u) = dx gives arctan x^2 dx = arctan(u)/2SqRt(u) du. Now we can integrate by parts, choosing 1/2SqRt(u) to integrate and arctan(u) to differentiate. As for the boundaries, u = 0 and u = 1, they remain the same. Now the first part of the answer is SqRt(u)•arctan(u) from 0 to 1, which is π/4. The second part of the answer requires integrating SqRt(u)/(1 + u^2). Now if one must let SqRt(u) = v, then dv = du/2SqRt(u) = du/2v, so du = 2v dv. Then our integrand is 2v^2/(1 + v^4) dv, the bounds still remain the same. Now we are getting somewhere. Notice that 1 + v^4 + 2v^2 = 1 + 2v^2 + v^4 = (1 + v^2)^2. Hence 1 + v^4 - 2v^2 + 2v^2 = 1 + v^4 = (1 + v^2)^2 - 2v^2 = (1 + v^2)^2 - (SqRt(2)•v)^2 = [v^2 + SqRt(2)•v + 1]•[v^2 - SqRt(2)•v + 1]. Let me call this quantity D for the sake of typing convenience. Now we get to do the strangest partial fraction decomposition exercise I have ever done in my life. 2v^2/D = (Av + B)/(v^2 - SqRt(2)•v + 1) + (Cv + D)/(v^2 + SqRt(2)•v + 1) = [(Av + B)(v^2 + SqRt(2)•v + 1) + (Cv + D)(v^2 - SqRt(2)•v + 1)]/D => (A + C)v^3 + (ASqRt(2) + B - CSqRt(2) + D)v^2 + (A + BSqRt(2) + C - DSqRt(2))v + (B + D) = 2v^2, such that A + C = 0, B + D = 0, A + BSqRt(2) + C - DSqRt(2) = 0 & ASqRt + B - CSqRt(2) + D = 2. From the first 2 equations, C = -A and D = -B, so A + B•SqRt(2) - A + B•SqRt(2) = 2B•SqRt(2) = 0 and A•SqRt(2) + B + A•SqRt(2) - B = 2 = 2A•SqRt(2) = 2. Now we can conclude that B = D = 0 & A = 1/SqRt(2) & C = -1/SqRt(2).
Recollecting everything, our original integral, arctan(x^2), from x = 0 to x = 1, is now equal to π/4 minus the [integral of {(1/SqRt(2))•v/(v^2 - SqRt(2)•v + 1) - (1/SqRt(2))•v/(v^2 + SqRt(2)•v + 1)} from v = 0 to v = 1]. This can be rewritten as π/4 + 2^(-/2)•Integral{v/(v^2 + SqRt(2)•v + 1); v = [0, 1]} - 2^(-/2)•Integral{v/(v^2 - SqRt(2)•v + 1); v = [0, 1]}. Whoa! We came a long way, did we not?
Now, we need to solve the remaining two integrals, which is far easier than the original integral. For the first one, we should use the fact that v^2 + SqRt(2)•v + 1/2 = (v + 1/SqRt(2))^2, so v^2 + SqRt(2)•v + 1 = [v + 2^(-/2)] + /2. Let w = v + 2^(-/2), so that the boundaries become w = 2^(-/2) and w = 1 + 2^(-/2). Then the denominator is w^2 + [2^(-/2)]^2, and the numerator is w - 2^(-/2). This means the first integral I1 can be decomposed into two integrals, the first one being w/(w^2 + [2^(-/2)]^2) with the given bounds. Let y = w^2 + 1/2 so that 2w dw = dy, the integrand becoming 1/2•dy/y. The boundaries for this are y = 1 and y = 1 + SqRt(2) + 1/2 + 1/2 = 2 + SqRt(2). This integral simplifies to Ln(SqRt(2 + SqRt(2))), where Ln is the natural logarithm. The second part of the integral is -2^(-/2)/(w^2 + [2^(-/2)]^2). The antiderivative of this is -2^(-/2)•1/2^(-/2)•arctan(w/2^(-/2) = - arctan(SqRt(2)•w). Then evaluating at the boundaries and subtracting we get - arctan(1) + arctan(SqRt(2) + 1) = arctan(SqRt(2) + 1) - π/4. Then I1 = arctan(SqRt(2) + 1) + Ln(2 + SqRt(2))/2 - π/4. In the expression π/4 + 2^(-/2)•Integral{v/(v^2 + SqRt(2)•v + 1); v = [0, 1]} - 2^(-/2)•Integral{v/(v^2 - SqRt(2)•v + 1); v = [0, 1]}, we have that I1 is multiplied by 1/SqRt(2), so this amounts to arctan(SqRt(2) + 1)/SqRt(2) + Ln(2 + SqRt(2))/(2•SqRt(2)) - (π/4)/SqRt(2). Combining this with the π/4, we have π/4•δ/SqRt(2) + arctan(SqRt(2) + 1)/SqRt(2) + Ln(2 + SqRt(2))/(2•SqRt(2)), where δ is the silver ratio SqRt(2) - 1.
Now all that remains to do is calculate the second integral, I2, multiply it by 1/SqRt(2), and subtract it from the beast constant we already obtained in the previous step. This will gives us the final result for the definite integral of arctan x^2 from x = 0 to x = 1.
The second integral has the integrand v/(v^2 - SqRt(2)•v + 1). Note that v^2 - SqRt(2)•v + /2 = (v - /SqRt(2))^2, so the denominator is (v -/SqRt(2))^2 + 1/2. Let w = v - /SqRt(2). This has dw = dv and for the bounds, w = -/SqRt(2) and w = δ/SqRt(2). The integrand becomes (w + /SqRt(2))/(w^2 + (1/SqRt(2))^2). This can be decomposed into two parts. The first part has integrand w/(w^2 + (/SqRt(2))^2) with antiderivative Ln(w^2 + 1/2)/2. Evaluated at the boundaries, we have Ln(1)/2 - Ln((δ^2 + 1)/2)/2 = Ln(2)/2 - Ln(2 - SqRt(2))/2. The second part is /SqRt(2)•/(w^2 + (/SqRt(2))^2). This has antiderivative arctan(SqRt(2)•w). Evaluated at the boundaries, we have arctan(-1) - arctan(SqRt(2) - 1) = -π/4 - arctan(SqRt(2) - 1). Multiplying by -1/SqRt(2), we get π/4•1/SqRt(2) + arctan(SqRt(2) - 1)/SqRt(2) + Ln(2 - SqRt(2))/(2•SqRt(2)) - Ln(2)/(2•SqRt(2)).
Combining the beastly constants, the result is π/4 + (arctan[SqRt(2) + 1] + arctan[SqRt(2) - 1])/SqRt(2). Notice how calculating this definite integral, despite difficult, required only algebraic manipulations. This means one can actually calculate the antiderivative of arctan(x^2). I will not do this right now because it will take much time.
Okay, I learned the integral wrong. It seems I made some algebraic mistakes along the way. I was not too far off from the real answer, in any case.
Cool, keep up the good work!
N! = N^N solve n I just thought it and it isn't possible right
N=1
@@rot6015 lol other from 1 I know 1 is a solution
For real values theres only 1 and if you take the limit also 0, i dont know about complex solutions tho
STEEEEEEEVE please, it's time to make a (complete) thorough playlist on VECTOR CALCULUS
IT'S TIME TO STOP with mortal calculus and... transcend
vector is also meh.
@@98danielray is it though?
Check Out Dr. Peyam! He has many good videos on them already.
daniel Yes, it is very meh. Tensor calculus, though... now that is truly the calculus of the gods. I would like to see a series on that.
@@angelmendez-rivera351 me too man, but as someone who is just starting to check out that subject, I would say that vector calculus let's you ease into tensor calculus.
I like this comment!
like me
First comment
DAB