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

[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 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

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]

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.

 

[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?

 

[morphism]

Not by parts.

 

[cristo]

well how do we do the integral then?

Try the substitution u=arcsinx..

 

[Gib Z]

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

 

[morphism]

Try the substitution u=arcsinx..

Which is what he did here:

I can tell by inspection, it being of the form f'(x)/f(x), that the answers ln (arcsin x)

 

[d_leet]

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]

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.

 

[d_leet]

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.