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I'm having trouble with this integral
$$\int\arcsin(\sin x)\,\mathrm dx$$

The problem is with the intervals of definition for each function :/ if someone could dumb it down for me.

  • $\arcsin\colon [-1, 1] \to [-\pi/2, \pi/2]$.
  • $\sin:\mathbb{R}\to [-1, 1]$.

right?

But what about $\arcsin(\sin x)$ ?

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Note that the arcsine has range $[-\frac{\pi}{2},\frac{\pi}{2}]$, $\sin(x+\pi)=-\sin\;x$, and that $\sin\;x$ has period $2\pi$... –  J. M. Dec 12 '10 at 10:19
3  
As a matter of fact, why not try plotting it and see for yourself? –  J. M. Dec 12 '10 at 10:22

2 Answers 2

My previous hint was a bit misleading - hence here is a (the comments are still appropriate)

CORRECTED HINT

On $I= [-\pi/2,\pi/2]$ the sine function grows from $\sin(-\pi/2)=-1$ to $\sin(\pi/2)=1$, hence it is invertible on $I$. The inverse, $\arcsin$, is defined on $[-1,1]$ and has the property $\arcsin(\sin x) = x$ if $-\pi/2\le x\le \pi/2$. Since $\sin$ is $2\pi$-peiodic we must have $\arcsin(\sin (x+2n\pi))=\arcsin(\sin(x))$ for all $x$ and integers $n$. Hence it is sufficient to understand $\arcsin(\sin x)$ for $\pi/2\le x\le 3\pi/2$, to achieve that use $\sin(x+\pi)=\sin (-x)$.

When you see the function there will be not problem to perform any integration.

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isn't it the otherway around ? sin(arcsin(x)) = x ? –  andrei Dec 12 '10 at 10:22
    
@andrei: they are inverse functions... one undoes the other within a suitably restricted domain. –  J. M. Dec 12 '10 at 10:24
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That reminds me... –  J. M. Dec 12 '10 at 10:27

Since $\sin(x)$ can take $ \mathbb R$ and outputs $[-1, 1]$ and $\arcsin(x)$ can take $[-1, 1]$ and output $[-\pi/2, \pi/2]$ , then it stands to reason that the constraints for $\arcsin(\sin(x))$ would be $\mathbb R \rightarrow [-\pi/2, \pi/2]$.

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