Prove the definition of the arcsin(s).

Question

I am given $\arcsin: S \rightarrow (-\pi/2,\pi/2) $ is the inverse function of sin(t) (restricted to [$-\pi/2,\pi/2$]). I'm trying to prove that $\arcsin(s)$= $\int_{0}^{s}1/\sqrt{1-x^2}$ .

My initial thoughts for an attempt of a proof:

By definition, we know:

$\sin(\arcsin(x)) = x$

Thus by the chain rule:

$(\sin(\arcsin(x)) = x)' $

$ \cos(\arcsin(x)) \times d (\arcsin(x))/dx = 1$

Also by definition, we know: $\sin(x)^2 + \cos(x)^2 = 1$

It has been given that $\cos(x) > 0 $ over the interval $(-\pi/2,\pi/2)$, which implies

$\sin(\arcsin(x))^2 + \cos(\arcsin(x))^2= 1$.

Therefore, $\cos(\arcsin(x)) = \sqrt{(1-x^2)}$

So, $\sqrt{(1-x^2)}\times d (\arcsin(x)) /dx = 1$ and

$d (\arcsin(x)) /dx = 1/\sqrt{(1-x^2)}$

From there I think we will use parts of the MVT or the fundamental theorem of calculus to prove the result $\arcsin(s)=\int_{0}^{s}1/\sqrt{1-x^2}$, but I'm having trouble with the formalities.

Answer 1

Problem: $$\int_0^s\frac 1{\sqrt{1-x^2}}dx$$

Let $$x=\sin \theta$$ then we get: $$\frac{dx}{d\theta}=\cos \theta$$ By derivating both side by $x$.

$$\therefore \int_0^s\frac 1{\sqrt{1-x^2}}dx$$ $$=\int_0^{\arcsin s}\frac {\cos\theta}{\sqrt{1-{\sin^2 \theta}}}d\theta$$ $$=\int_0^{\arcsin s} 1 d\theta$$ $$=\left[ \theta \right]^{\arcsin s}_{\arcsin 0} d\theta$$ $$=\arcsin s$$.

Answer 2

You can use the fact that$$\arcsin'(x)=\frac1{\sin'(\arcsin x)}=\frac1{\cos(\arcsin x)}=\frac1{\sqrt{1-x^2}}.$$So, and since $\arcsin(0)=0$, you get that$$\arcsin s=\int_0^s\frac{\mathrm dx}{\sqrt{1-x^2}}.$$