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Let $f(x)=\arcsin (x)$ To calculate the interval $f([-1,1])$ we have to check if $f$ is descresing or incresing so that we know if $$f([-1,1])=\left (\lim_{x\rightarrow -1}f(x), \lim_{x\rightarrow 1}f(x)\right )$$ or $$f([-1,1])=\left (\lim_{x\rightarrow 1}f(x), \lim_{x\rightarrow -1}f(x)\right )$$ or not?

Or can we say that since the $\sin(x) : \mathbb{R}\rightarrow [-1,1]$ then for the inverse function it holds that $f([-1,1])=\mathbb{R}$ ?

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The way $\sin(x)$ works, $\sin(-x)$ = -\sin(x)$.

Because of that, $\arcsin(-x) = -\arcsin(x)$.

Therefore, $\arcsin$ from $[-1,0]$ is a flipped image of $[0,1]$.

$\int_{-1}^1 \arcsin(x) \,dx = 0$.

There is no need to check for increasing or decreasing.

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    Let $g=\sin (x) : \mathbb{R}\rightarrow [-1,1]$. We have that $g^{-1}([-1,1])=\{x\in \mathbb{R} \mid g(x)\in [-1,1]\}$. Since $[-1,1]$ is the range of $g$ we have that $g(x)\in [-1,1]$ for all $x\in \mathbb{R}$. Therefore $g^{-1}([-1,1])=\mathbb{R}$. Is this justifiction correct? – 2017-01-04