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

In summary, the conversation discusses the integral \int \frac{1}{\sqrt{1-x^2}} \frac{1}{\arcsin x} dx and the attempt to solve it using integration by parts. However, it is discovered that disregarding the constant of integration leads to incorrect results and the integral cannot be solved by parts. The conversation then suggests using the substitution u=arcsinx as a potential method of solving the integral.
  • #1
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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  • #2
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.
 
  • #3
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?
 
  • #4
Not by parts.
 
  • #5
Gib Z said:
well how do we do the integral then?

Try the substitution u=arcsinx..
 
  • #6
>.<" Ok..Im feeling really stupid right now...Someone please shoot me.
 
  • #7
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)
 
  • #8
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.
 
  • #9
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.
 

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

What does "integrating the complex expression" mean?

Integrating a complex expression means finding the antiderivative or the reverse operation of differentiation. It involves finding a function whose derivative is equal to the given expression.

Why is the expression "1/(√(1-x²) · arcsin(x))" considered complex?

This expression is considered complex because it involves both a square root and an inverse trigonometric function, which can be challenging to work with and manipulate in mathematical calculations.

What are the steps for integrating this complex expression?

The steps for integrating this complex expression involve using substitution to rewrite the expression in a simpler form, applying trigonometric identities to simplify further, and then using integration techniques such as integration by parts or u-substitution to find the antiderivative.

Can this expression be integrated using basic calculus techniques?

Yes, this expression can be integrated using basic calculus techniques such as substitution, trigonometric identities, and integration by parts. However, the process may be more complicated and involve multiple steps compared to integrating simpler expressions.

What applications does integrating this expression have in science?

Integrating this expression can be useful in various scientific fields, such as physics, engineering, and statistics. It can help in solving problems related to motion, electrical circuits, and probability distributions, among others.

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