- #1

Gib Z

Homework Helper

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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?

[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?