How to integrate this function?

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Homework Help Overview

The problem involves integrating the function $$ \int dx =\int \frac {d\phi} {\phi \sqrt {1 - \phi²}} $$, with the known outcome being $$ x = \operatorname {arcsech}{\phi} $$.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods for integration, including the use of trigonometric substitutions and algebraic manipulations. Some express uncertainty about the steps taken and the signs involved in the process.

Discussion Status

There is ongoing exploration of different approaches to the integration problem. Participants have suggested substitutions and noted potential issues with signs in the calculations. No consensus has been reached, but guidance has been provided regarding possible methods.

Contextual Notes

Participants acknowledge that they are aware of the final answer but are struggling with the integration process itself. There is a mention of using apps for assistance, indicating a reliance on external tools for verification.

Mutatis
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Homework Statement


I've got to integrate the following $$ \int dx =\int \frac {d\phi} {\phi \sqrt {1 - \phi²}}. $$

Homework Equations

[/B]
I already know the answer but not how to get it. The answer that I got from solution is ## x = \operatorname {arcsech}{\phi} ##.

The Attempt at a Solution


I've used two apps to help me get this answer. First I did it using Mathway and I got:
73jgcy.jpg
.

Then, WolframAlpha told me that ## x = \operatorname {arcsech}{\phi} ## is equal to:
90536r.jpg
.

After that I took that first equation in the first image and with some algebraic manipulations I got:
2v1n4hw.jpg
.

So I don't know how to procedure beyond here. This was the closest I achieved from the final answer.
 

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Mutatis said:

Homework Statement


I've got to integrate the following $$ \int dx =\int \frac {d\phi} {\phi \sqrt {1 - \phi²}}. $$

Homework Equations

[/B]
I already know the answer but not how to get it. The answer that I got from solution is ## x = \operatorname {arcsech}{\phi} ##.

The Attempt at a Solution


I've used two apps to help me get this answer. First I did it using Mathway and I got:
View attachment 227957 .

Then, WolframAlpha told me that ## x = \operatorname {arcsech}{\phi} ## is equal to:
View attachment 227958 .

After that I took that first equation in the first image and with some algebraic manipulations I got:
View attachment 227959 .

So I don't know how to procedure beyond here. This was the closest I achieved from the final answer.
One approach would be to use a trig substitution, with ##\sin(u) = \sqrt{1 - \phi^2}##, and ##\cos(u)du = \frac{-2\phi d\phi}{\sqrt{1 - \phi^2}}##. The result you get probably won't look like the solution you show, but it should be equal to the solution above within a constant.
 
Mutatis said:
I already know the answer but not how to get it. The answer that I got from solution is ## x = \operatorname {arcsech}{\phi} ##.
You dropped a sign somewhere. The derivative of arcsech is the negative of the integrand.

I'd use the substitution ##\phi = \cos u##. It's pretty much the same as Mark's suggestion, but it makes the algebra a little simpler.
 
Thank you guys! I'm going to try to do it again watching the signs and doing what you've said!
 

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