# Crazy Integral!

Homework Helper
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?

morphism
Homework Helper
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.

Homework Helper
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?

morphism
Homework Helper
Not by parts.

cristo
Staff Emeritus
well how do we do the integral then?

Try the substitution u=arcsinx..

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

morphism
Homework Helper
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)

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),

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

dextercioby