How to solve this math question ??

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July 27th, 2022 at 2:40:49 AM permalink

∫ (1+1/x^4)^0.5 dx = ?

unJon

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July 27th, 2022 at 3:32:34 AM permalink

I would copy it into Wolfram Alpha.

https://www.wolframalpha.com/input?i=∫+%281%2B1%2Fx%5E4%29%5E0.5+dx+

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ThatDonGuy

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July 27th, 2022 at 4:31:07 PM permalink

First thought: rewrite it as (sqrt(x^4 + 1) / x^2) dx, then see if you can integrate it by parts

ssho88

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July 27th, 2022 at 6:12:55 PM permalink

Quote: ThatDonGuy
First thought: rewrite it as (sqrt(x^4 + 1) / x^2) dx, then see if you can integrate it by parts
link to original post

Thanks for your reply. I agree that we can rewrite it to (1+x^4)^0.5 / x^2

The hard part is (1+x^4)^0.5, which I can't figure out.

I would appreciate if someone could show me the solution step by step.

gordonm888
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July 28th, 2022 at 2:40:54 PM permalink

Quote: ssho88
Quote: ThatDonGuy
First thought: rewrite it as (sqrt(x^4 + 1) / x^2) dx, then see if you can integrate it by parts
link to original post

Thanks for your reply. I agree that we can rewrite it to (1+x^4)^0.5 / x^2

The hard part is (1+x^4)^0.5, which I can't figure out.

I would appreciate if someone could show me the solution step by step.
link to original post

I don't have time to work this out,, but let me suggest the following: When using integration by parts you need to find the integral of (1+x^4)^0.5. Try setting x^2 = tan y with (2x dx ) =sec^2(y)dy. Thus, (1+x^4)^0.5 dx becomes sqrt(1 + tan^2(y))* (1/2*tan²(y))/sec^2(y)dy.

However, (1+tan^2(y)) = sec^2 (y) so sqrt(1 + tan^2(y))*(1/2*tan²(y))/sec^2(y)dy =
1/2*tan^-2 (y)dy. That should be integrable, I imagine.

I hope I didn't make a mistake . . . .

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ssho88

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July 28th, 2022 at 7:44:48 PM permalink

Thanks gordonm88,

You mentioned : "Thus, (1+x^4)^0.5 dx becomes sqrt(1 + tan^2(y))* (1/2*tan2(y))/sec^2(y)dy."

With substitution x^2 = tan(y), I get :-

(2x dx ) =sec^2(y)dy, and x = sqrt(tan(y))

dx = (1/2) * (1/x) * sec^2(y)dy

(1+x^4)^0.5 dx = sqrt(1 + tan^2(y)) * (1/2) * (1/x) * sec^2(y)dy
= sqrt(1 + tan^2(y)) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy
= sec(y) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy
= sec^3(y) * (1/2) * (1/sqrt(tan(y))) dy
= ????

It does not have an elementary anti-derivative.

Last edited by: ssho88 on Jul 28, 2022

gordonm888
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July 28th, 2022 at 9:59:31 PM permalink

Quote: ssho88
Thanks gordonm88,

You mentioned : "Thus, (1+x^4)^0.5 dx becomes sqrt(1 + tan^2(y))* (1/2*tan2(y))/sec^2(y)dy."

With substitution x^2 = tan(y), I get :-

(2x dx ) =sec^2(y)dy, and x = sqrt(tan(y))

dx = (1/2) * (1/x) * sec^2(y)dy

(1+x^4)^0.5 dx = sqrt(1 + tan^2(y)) * (1/2) * (1/x) * sec^2(y)dy
= sqrt(1 + tan^2(y)) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy
= sec(y) * (1/2) * (1/sqrt(tan(y))) * sec^2(y)dy
= sec^3(y) * (1/2) * (1/sqrt(tan(y))) dy
= ????

It does not have an elementary anti-derivative.
link to original post

yes, sorry, I typed that out quick before running out the door. I've been gone all day, until this moment. The basic idea with the substitution x^2 = tan y was to reduce sqrt(1 + x^4) to a simple trigonometric expression, while accepting whatever you get for dx. I guess it didn't work out but I don't know any other way of attacking an integral with a term sqrt(1 + x²ⁿ).

Can we pursue an antiderivative of X = sec^3(y) * (1/2) * (1/sqrt(tan(y))) by using sec^2(y) =tan^2(y) + 1 ?

X = 0.5*sec(y) *(tan^2(y) +1)/tan^(-0.5)(y) = 0.5* sec(y)*tan^(1.5)(y) +0.5*sec(y)*tan^(-0.5)(y)

This form of X might possibly have an anti-derivative with an analytic form. It is a sum of two terms that are both of the form sec x * tan^a.

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teliot

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July 29th, 2022 at 10:04:32 AM permalink

Quote: ssho88
∫ (1+1/x^4)^0.5 dx = ?
link to original post

Only a very few functions have a closed form for their anti-derivative expressible in elementary functions. I have no idea if what you've posted is such a function, but it smells like one to me. Wolfram Alpha says the anti-derivative needs a "hypergeometric function."

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ssho88

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July 29th, 2022 at 10:52:24 AM permalink

My actual intention is to find the length of curve, f(x) = 1/x, the expression ∫ (1+1/x^4)^0.5 dx was derived as below :-

y=1/x, dy/dx= -1/x^2

(ds)^2 = (dx)^2 + (dy)^2

ds = dx * (1+(dy/dx)^2)^0.5 = (1 + 1/x^4)^0.5 dx

s = ∫ (1+1/x^4)^0.5 dx .

Then I got stuck here.

camapl

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July 29th, 2022 at 11:49:40 AM permalink

I don�t know if you�re expected to provide an accurate answer, but aren�t there tables with approximate answers for certain �difficult� functions?

ETA: Wait a minute� The �length of curve, f(x) = 1/x�. Without citing boundaries (boundless), wouldn�t that be infinite?

Last edited by: camapl on Jul 29, 2022

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unJon

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July 29th, 2022 at 12:53:34 PM permalink

If you have wolfram alpha pro you can just type in arc length 1/X and click show step by step solution.

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gordonm888
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July 29th, 2022 at 2:38:04 PM permalink

Recognizing that this was either an unsolvable integral (i.e., no analytic form for the anti-derivative) or at least one for which the solution is not widely known, I was just typing in notes in my posts as I studied it. I realized that my posts were not directly helpful, sho88, I was just trying to come up with a good idea on how to start the analysis.

I did notice that the integral of (1/sqrt(a²+u²) = arsinh (u/a) + C This is the closest form that I have found for a simple integral relevant to your problem.

Also, the integral of(1/(u*sqrt(a²+u²)) = arcsch |(u/a)| + C where I think arcsch is an arc-hyperbolic cosecant.

All of which seems to be in sync with teliot's comment about Wolfram's advice to look at hyperbolic functions. However, hyperbolas have different mathematical properties than the curve you are studying and I'm not sure if they will reduce to your curve by, say, setting the distance between the two foci to 0.

Certainly not an "easy math problem."

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ssho88

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July 29th, 2022 at 6:19:44 PM permalink

Quote: gordonm888
Recognizing that this was either an unsolvable integral (i.e., no analytic form for the anti-derivative) or at least one for which the solution is not widely known, I was just typing in notes in my posts as I studied it. I realized that my posts were not directly helpful, sho88, I was just trying to come up with a good idea on how to start the analysis.

I did notice that the integral of (1/sqrt(a²+u²) = arsinh (u/a) + C This is the closest form that I have found for a simple integral relevant to your problem.

Also, the integral of(1/(u*sqrt(a²+u²)) = arcsch |(u/a)| + C where I think arcsch is an arc-hyperbolic cosecant.

All of which seems to be in sync with teliot's comment about Wolfram's advice to look at hyperbolic functions. However, hyperbolas have different mathematical properties than the curve you are studying and I'm not sure if they will reduce to your curve by, say, setting the distance between the two foci to 0.

Certainly not an "easy math problem."
link to original post

Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

Any comments ?

Last edited by: ssho88 on Jul 29, 2022

gordonm888
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July 30th, 2022 at 1:20:57 PM permalink

Nice. But with y = x^4, is there any convergence issue when x<1?

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teliot

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July 30th, 2022 at 5:38:26 PM permalink

Quote: ssho88

Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

Any comments ?
link to original post

For a numerical approximation, this is exactly what I would do.

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ssho88

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July 30th, 2022 at 7:00:38 PM permalink

Quote: gordonm888
Nice. But with y = x^4, is there any convergence issue when x<1?
link to original post

Convergence issue is my concern too. Thanks for your comment.

ssho88

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July 30th, 2022 at 7:01:58 PM permalink

Quote: teliot
Quote: ssho88

Possible to solve this problem with Taylor series ?

Taylor series, (1+y)^0.5 = 1 +y/2 - y^2/8 + y^3/16 - (5*y^4)/128 + (14*y^5)/256 +. . . . . .

For my case, just replace y with 1/x^4 . . . .then the integration will be very easy !

Any comments ?
link to original post
For a numerical approximation, this is exactly what I would do.
link to original post

Thanks for your comment.

gordonm888
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August 6th, 2022 at 5:14:38 AM permalink

Did the Taylor series converge? Were you able to solve this integral?

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ssho88

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August 6th, 2022 at 6:53:50 AM permalink

Here is the integral solution:-

However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?

unJon

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August 6th, 2022 at 7:09:14 AM permalink

Quote: ssho88
Here is the integral solution:-

However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?
link to original post

Just the same comment I made twice before.

https://www.wolframalpha.com/input?i=arc+length+1%2FX&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22a%22%7D+-%3E%221%22&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22b%22%7D+-%3E%223%22

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ssho88

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August 6th, 2022 at 7:33:55 AM permalink

Quote: unJon
Quote: ssho88
Here is the integral solution:-

However, I think that series only converge if -1 < x < 1. So I think still not able to find length of arc 1/x from x = 1 to x = 3. Any comment?
link to original post

Just the same comment I made twice before.

https://www.wolframalpha.com/input?i=arc+length+1%2FX&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22a%22%7D+-%3E%221%22&assumption=%7B%22F%22%2C+%22ArcLength%22%2C+%22b%22%7D+-%3E%223%22
link to original post

Thanks for your reply.

So Taylor series is not a viable way to find the arc length of 1/x ?

unJon