Crazy Integral!

  • Thread starter Gib Z
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  • #1
Gib Z
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Main Question or Discussion Point

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 heres 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?
 

Answers and Replies

  • #2
morphism
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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.
 
  • #3
Gib Z
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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
morphism
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Not by parts.
 
  • #5
cristo
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well how do we do the integral then?
Try the substitution u=arcsinx..
 
  • #6
Gib Z
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>.<" Ok..Im feeling really stupid right now...Someone please shoot me.
 
  • #7
morphism
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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
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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),
I don't think that that is correct as stated. It looks more like 1/[f'(x)*f(x)] to me.
 
  • #9
dextercioby
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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.
 
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  • #10
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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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