My goal is to prove that the function $\arcsin:\left [ -1,1 \right ]\rightarrow \mathbb{R}$ can be defined as $$x \mapsto \arcsin x \equiv \int_{0}^{x}\frac{1}{\sqrt{1-t^2}} \ dt,$$ which is odd and continuous. Assume nothing about the sine function is known.
Showing the function is continuous on $(-1,1)$ just follows from the definition.
Showing the function is odd should be as simple as showing that $$f(-x)=-f(x).$$ Plugging this in gives $$f(-x)=\int_{0}^{-x}\frac{1}{\sqrt{1-t^2}} \ dt=-\int_{-x}^{0}\frac{1}{\sqrt{1-t^2}} \ dt.$$ I am not too sure what to do next.
Lastly, I think the bulk of the question is to prove that the function we are integrating is Riemann integrable. That is, it is a valid definition. I am having trouble on this part as well because I am not sure how to construct my partition $P$ such that $$U(f;P)-L(f;P)<\epsilon.$$ I know this a lot- I am pretty confused! Can anyone provide me some feedback and help?
Thanks so much in advance!
Powers and sums are continuous, and compositions of continuous functions are continuous, so the integrand is continuous on $(-1,1)$ hence RI, by the FTC, $\sin^{-1}(x)$ is continuous. For oddness, use the change of variables $u=-x$.
Assume $x\ge 1/2$. Because the integrand is positive, the integral increases in $x$, so the change of variables $u=1-t^2$ expresses value of $|\arcsin 1-\arcsin x| $ as $$ \int_x^1 \frac {\text{d}u}{2\sqrt u \sqrt{1-u}}\le \frac 1{\sqrt 2}\int_{x}^1\frac {\text{d}u}{\sqrt{1-u}}=\sqrt 2\sqrt{1-x}\to 0 \quad \text{as $x\to 1$}$$ Thus, $\arcsin x$ is continuous at $1$. The same procedure can be applied at $-1$.