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

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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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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?
 
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:
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.