The easiest way to solve $\int\arcsin(\sin x)\;\mathrm{d}x$?

Question

Here's the integral:

$$\int\arcsin(\sin x)\;\mathrm{d}x$$

What is the best way to solve it? I was thinking of integration by parts, but it seems to be the most hard way to deal with it. Are there easier ways to proceed?

Answer 1

$$f(x)=\arcsin(\sin x)$$ is a periodic function with $T=2\pi$ $$f(x)=\arcsin(\sin x)=\begin{cases}x & -\dfrac{\pi}{2}\leq x \leq\dfrac{\pi}{2} \\ \pi-x & \dfrac{\pi}{2} \leq x\leq \dfrac{3\pi}{2}\end{cases}$$

Answer 2

hint

In the interval $[-\frac {\pi}{2}+2k\pi,\frac {\pi}{2}+2k\pi] $

$$\arcsin (\sin (x))=x-2k\pi $$

and in $[\frac {\pi}{2}+2k\pi,\frac {3\pi}{2}+2k\pi ] $

$$\arcsin (\sin (x))=x-(2k+1)\pi $$

for example

$$\int_0^\pi\arcsin (\sin (x))dx=$$ $$\Bigl [\frac {x^2}{2}\Bigr]_0^\frac {\pi}{2}+\Bigl [\frac {x^2}{2}-\pi x\Bigr]_\frac {\pi}{2}^\pi $$

Answer 3

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.

The idea by kingW3 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 :

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