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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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  • $\begingroup$ Note that the arcsine has range $[-\frac{\pi}{2},\frac{\pi}{2}]$, $\sin(x+\pi)=-\sin\;x$, and that $\sin\;x$ has period $2\pi$... $\endgroup$ Dec 12, 2010 at 10:19
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    $\begingroup$ As a matter of fact, why not try plotting it and see for yourself? $\endgroup$ Dec 12, 2010 at 10:22

3 Answers 3

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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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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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  • $\begingroup$ isn't it the otherway around ? sin(arcsin(x)) = x ? $\endgroup$
    – andrei
    Dec 12, 2010 at 10:22
  • $\begingroup$ @andrei: they are inverse functions... one undoes the other within a suitably restricted domain. $\endgroup$ Dec 12, 2010 at 10:24
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    $\begingroup$ That reminds me... $\endgroup$ Dec 12, 2010 at 10:27
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Let's have $f(x)=\arcsin(\sin(x))$ and calculate $\displaystyle F(x)=\int_0^x f(t)dt$

Since $f$ is $2\pi-$periodic and $F(2\pi)=0$ by symetry, we have that $F$ is $2\pi-$periodic too.

enter image description here

The idea to introduce $s(x)=\operatorname{sgn}(\cos(x))$ will be useful to find a closed form.

$F(x)=x\arcsin(\sin(x))-\frac{s(x)}2x^2+C(x)$

In fact his formula is correct on $1$ period interval under the condition that we assign a constant value for $C(x)$ in each of the $3$ intervals described below.


But I'll propose a slightly different form : $F(x)=s(x)(\frac 12f(x)^2)+B(x)$ where $B(x)$ is piecewise constant too, but zero on two of the intervals.

Under this form the periodicity of $F$ is more obvious knowing that $f$ is periodic.


$\displaystyle x\in[0,\frac{\pi}{2}];\quad F(x)=\int_0^x tdt=\frac {x^2}2$

$\displaystyle x\in[\frac{\pi}{2},\frac{3\pi}{2}];\quad F(x)=F(\frac{\pi}2)+\int_{\frac\pi2}^x (\pi-t)dt=\frac {\pi^2}8+\bigg[\pi t-\frac{t^2}2\bigg]_{\frac{\pi}2}^x=\pi x-\frac{x^2}2-\frac{\pi^2}4$

$\displaystyle x\in[\frac{3\pi}{2},2\pi];\quad F(x)=F(\frac{3\pi}2)+\int_{\frac{3\pi}2}^x (t-2\pi)dt=\frac {\pi^2}8+\bigg[\frac{t^2}2-2\pi t\bigg]_{\frac{3\pi}2}^x=\frac{x^2}2-2\pi x+2\pi^2$


$\begin{cases} x\in[0,\frac{\pi}{2}] & f(x)=x & s(x)=+1 & \frac 12s(x)f(x)^2=\frac {x^2}2 & B(x)=0\\ x\in[\frac{\pi}{2},\frac{3\pi}{2}] & f(x)=\pi-x & s(x)=-1 & \frac 12s(x)f(x)^2=-\frac{\pi^2}2+\pi x-\frac {x^2}2 & B(x)=\frac{\pi^2}4\\ x\in[\frac{3\pi}{2},2\pi] & f(t)=x-2\pi & s(x)=+1 & \frac 12s(x)f(x)^2=\frac{x^2}2-2\pi x+2\pi^2 & B(x)=0\\ \end{cases}$

Now if we consider $\displaystyle \frac{1-s(x)}2=\chi_{[\frac{\pi}{2},\frac{3\pi}{2}]+2k\pi}$ then $B(x)=\big(\frac{1-s(x)}2\big)\frac{\pi^2}4$

Finally a simplified closed form for $F(x)$ can be given by :

$\displaystyle \int_0^x \arcsin(\sin(t))dt=\frac{\pi^2}8+\operatorname{sgn}(\cos(x))\bigg(\frac{\arcsin(\sin(x))^2}2-\frac{\pi^2}8\bigg)$

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