# Crazy Integral!

1. Feb 19, 2007

### Gib Z

Ok well The integral is :

$$\int \frac{1}{\sqrt{1-x^2}} \frac{1}{\arcsin x} dx$$.

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

$$1+\int\frac{1}{\sqrt{1-x^2}\arcsin x}$$.

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

I=1+I.

What did i do wrong?

2. Feb 19, 2007

### morphism

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. Feb 19, 2007

### Gib Z

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. Feb 19, 2007

### morphism

Not by parts.

5. Feb 19, 2007

### cristo

Staff Emeritus
Try the substitution u=arcsinx..

6. Feb 19, 2007

### Gib Z

>.<" Ok..Im feeling really stupid right now...Someone please shoot me.

7. Feb 19, 2007

### morphism

Which is what he did here:

8. Feb 19, 2007

### d_leet

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

9. Feb 20, 2007

### dextercioby

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: Feb 20, 2007
10. Feb 20, 2007

### d_leet

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