Integrating the Complex Expression: 1/(√(1-x²) · arcsin(x))

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Discussion Overview

The discussion revolves around the integration of the complex expression \(\int \frac{1}{\sqrt{1-x^2} \cdot \arcsin(x)} \, dx\). Participants explore various methods for solving the integral, including integration by parts and substitution techniques, while addressing misunderstandings and corrections related to the approach.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant suggests that the integral resembles the form \(f'(x)/f(x)\) and proposes that the answer is \(\ln(\arcsin(x))\).
  • Another participant points out the importance of including the constant of integration in the evaluation of the integral.
  • There is a suggestion to use substitution \(u = \arcsin(x)\) as an alternative method for solving the integral.
  • Some participants express uncertainty about the correctness of the initial assumption regarding the form of the integral, with one stating it appears to be \(1/[f'(x) \cdot f(x)]\).
  • Disagreement arises over the method of integration, with one participant asserting that integration by parts is not suitable for this integral.
  • Another participant acknowledges a mistake in their reasoning and agrees with the correction made by others regarding the derivative's placement in the integral.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct method for solving the integral, with multiple competing views and approaches being discussed throughout the thread.

Contextual Notes

Participants express uncertainty about the application of integration techniques and the handling of constants of integration, indicating potential limitations in their understanding of the problem.

Gib Z
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Ok well The integral is :

[tex]\int \frac{1}{\sqrt{1-x^2}} \frac{1}{\arcsin x} dx[/tex].

I can tell by inspection, it being of the form f'(x)/f(x), that the answers ln (arcsin x), but I was hopping Integration by parts could do it for me as well. But here's my Problem:

u=1/(arcsin x)
du=(-1)/ [sqrt(1-x^2) (arcsin x)^2] dx

dv= 1/(sqrt(1-x^2) dx
v= arcsin x

uv- integral:v du

[tex]1+\int\frac{1}{\sqrt{1-x^2}\arcsin x}[/tex].

The 2nd part is the Original integral. letting it equal I,

I=1+I.

What did i do wrong?
 
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This is what happens when you disregard the constant of integration. Your expression should actually be I + C_1 = 1 + I + C_2, where C_1 and C_2 are some constants.
 
Ahh Thats what came to mind after I posted, but then i thought the constants would cancel >.<", well how do we do the integral then?
 
Not by parts.
 
Gib Z said:
well how do we do the integral then?

Try the substitution u=arcsinx..
 
>.<" Ok..Im feeling really stupid right now...Someone please shoot me.
 
cristo said:
Try the substitution u=arcsinx..
Which is what he did here:
Gib Z said:
I can tell by inspection, it being of the form f'(x)/f(x), that the answers ln (arcsin x)
 
Gib Z said:
Ok well The integral is :

[tex]\int \frac{1}{\sqrt{1-x^2}} \frac{1}{\arcsin x} dx[/tex].

I can tell by inspection, it being of the form f'(x)/f(x), that the answers ln (arcsin x),

I don't think that that is correct as stated. It looks more like 1/[f'(x)*f(x)] to me.
 
d_leet said:
I don't think that that is correct as stated. It looks more like 1/[f'(x)*f(x)] to me.

No, no, he's right. The derivative is in the numerator.

This integral is one good example of the situation when some integrals have a unique method of solving.
 
Last edited:
  • #10
dextercioby said:
No, no, he's right. The derivative is in the numerator.

This integral is one good example of the situation when some integrals have a unique method of solving.

Yes, of course, you're right, I don't know what I was thinking.
 

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