Can the Integral of 1/sqrt(1-x^2) be Solved without Using Trig Functions?

[Sparky_]

Greetings,

I know $$\int \frac {1} {{\sqrt{1-x^2}} dx$$ is arcsine(x)

My question is can $$\int \frac {1} {\sqrt{1-x^2}} dx$$
be solved by some technique (parts, substitution...) and the answer be in terms of x and NOT an expression of arcsine?

Meaning, I would like the solution to the integral in terms of x and possibly other functions (ln for example) but not in terms of trig functions.

If so, can you show me?

Thanks
Sparky_

 

[d_leet]

It can be done by an appropriate trig substitution ie. x=sin(y), but the answer will end up identically. I'm fairly certain that arcsin(x) is the only answer you will ever get though.

 

[ZioX]

Antiderivatives are unique up to a constant. Ie, all antiderivatives of that will be arcsine(x)+C.

 

[Hurkyl]

The only two other (sane) ways I can think of to write that antiderivative is as a power series, or in terms of a complex logarithm. But any of those ways will simply be a different way of writing arcsin x.

 

[DeadWolfe]

One could certainly express arcsin in terms of logs using Euler's formula...

 

[Gib Z]

One could certainly express arcsin in terms of logs using Euler's formula...

Yea, Hurkyl already mentioned that.

...or in terms of a complex logarithm

So yes I think the OP is happy,

possibly other functions (ln for example)[\QUOTE]

 

[Sparky_]

What I am wanting is an expression for arcsine that uses other functions (hopefully LN and powers of x) but I am hoping to work in the real domain.

Gib Z help me quite a bit with a derivation regarding arcsine.

I was hoping some clever integration could come up with another expression equal to arcsin by integrating the derivative of arcsine.

Any thoughts?

Thanks
Sparky_

 

[HallsofIvy]

If trig substitution, which just about everyone mentioned, is not a "clever integration", I don't know what is!

 

[Sparky_]

I agree trig substitution is clever, I was hoping for a solution that doesn't include trig functions.

-Sparky_

 

[Gib Z]

I remember what sparky wanted from his previous thread that i tried to help him in. He had $$\arcsin x = -i \ln ( ix \pm \sqrt{1-x^2})$$ as he should, but could not prove why the expressions were equal.

 

[Sparky_]

Correct Gib_Z and you did help me prove it - you helped me get the true solution and in that I ended up with some other solutions that I assumed involved the other quadrants.

Using that expression (arcsin = -iln(...) has become very involved due to working in the complex domain.

I'm now searching for a similar type of solution in the real domain only. (I suppose this rules out Euler's)

-Sparky_

 

[Gib Z]

Well then the best answer I can give you is the Taylor Series for Arcsin x, which If i remember correctly is this:

$$\arcsin x = \sum^{\infty}_{n=0} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1}$$

 

[Sparky_]

Thank you all for your help.

Sparky_