Why is the integral of ##\arcsin(\sin(x))## so divisive?

[Saracen Rue]

TL;DR Summary

Various sources give different answers to $\arcsin(\sin(x))$, and none of them seem to actually be fully correct. Why is this the case, and is there an actual answer?

Initially, I was attempting to find the function which expresses the area under enclosed between the function $\arcsin(\sin(x))$ and the $x$-axis (so technically I am looking for $\int_{0}^{x} \arcsin(\sin(t)) dt$ specifically, but got caught up on finding the general antiderivative).

This lead me to attempting to integrate $\arcsin(\sin(x))$ using WolframAlpha, which gave me $x\arcsin\left(\sin\left(x\right)\right)-\frac{x^{2}}{2}\sec\left(x\right)\sqrt{\cos^{2}\left(x\right)}$. After a quick inspection of the function, it became clear that something just wasn't quite right. So, I graphed it against $\arcsin(\sin(x))$ and quickly saw that, which being accurate for the domain $(-\frac{\pi}{2}, \frac{\pi}{2})$, it wasn't at all accurate outside of that domain. I next used Symbolab to attempt to evaluate $\int \arcsin(\sin(x)) dx$ and it gave me the result $\frac{\arcsin\left(\sin\left(x\right)\right)^{2}}{2}$. Again, this seems to be correct for the domain $(-\frac{\pi}{2}, \frac{\pi}{2})$, but not for $(\frac{\pi}{2}, \frac{3\pi}{2})$... but then it is correct for $(\frac{3\pi}{2}, \frac{5\pi}{2})$ but not for $(\frac{5\pi}{2}, \frac{7\pi}{2})$ and so on.

Honestly, this has kind of left me stumped. I tried watching several Youtube videos but they all either have one of the answers above or are just completely wrong in their entirety. Why is it that no one can seem to agree on what this indefinite integral should be? I can understand if there's no way to fully express it in terms of elementary functions, but why not say that? WolframAlpha and Symbolab will both say there's no standard solution for integrating something like $x^x$, so why do they both confidently give incorrect or at very least only partially correct answers to this integral? At the very least they should tell you that the antiderivative is only valid over a certain domain.

Finally, if anyone can find a way to evaluate $\int_{0}^{x} \arcsin(\sin(t)) dt$ it would be greatly appreciated! Thank you for your time.

 

[renormalize]

Finally, if anyone can find a way to evaluate $\int_{0}^{x} \arcsin(\sin(t)) dt$ it would be greatly appreciated! Thank you for your time.

I don't understand your issue. Isn't $\arcsin\left(\sin\left(t\right)\right)\equiv\sin^{-1}\left(\sin(t)\right)=t+2n\pi$ (where $n=0,\pm1,\pm2,\ldots$)? So doesn't your integral simply evaluate to $\frac{1}{2}x^{2}+2n\pi x$? (Note that for $n=0$, that's precisely the answer returned by both Wolfram Alpha and Symbolab.)

 

[PeroK]

I don't understand your issue. Isn't $\arcsin\left(\sin\left(t\right)\right)\equiv\sin^{-1}\left(\sin(t)\right)=t+2n\pi$ (where $n=0,\pm1,\pm2,\ldots$)? So doesn't your integral simply evaluate to $\frac{1}{2}x^{2}+2n\pi x$? (Note that for $n=0$, that's precisely the answer returned by both Wolfram Alpha and Symbolab.)

The $\arcsin$ function gives a value in the range $[-\frac \pi 2, \frac \pi 2]$. So:

For $0 \le t \le \frac \pi 2$, we have $\arcsin(\sin t) = t$

For $\frac \pi 2 \le t \le \frac {3\pi} 2$, we have $\arcsin(\sin t) = \pi - t$

For $\frac {3\pi} 2 \le t \le \frac {5\pi} 2$, we have $\arcsin(\sin t) = t - 2\pi$

This is a sawtooth function, with period $2\pi$.

It's not clear how you would describe the integral in terms of a single expression. The Wolfram answer works in each separate interval, but it's not going to work over the whole domain.

 

[PeroK]

Finally, if anyone can find a way to evaluate $\int_{0}^{x} \arcsin(\sin(t)) dt$ it would be greatly appreciated! Thank you for your time.

The simplest approach is to reduce $x$ to the interval $[0, 2\pi]$. The integral is the same for any $x + 2n\pi$. Then $$\int_{0}^{x} \arcsin(\sin(t)) dt = $$$$\frac{x^2}{2} \ (0 \le x \le \frac \pi 2)$$$$\pi x - \frac{x^2}{2} - \frac{\pi^2}{4} \ (\frac \pi 2 \le x \le \frac {3\pi} 2)$$$$\frac{x^2}{2} - 2\pi x +2\pi^2 \ (\frac {3\pi} 2 \le x \le 2\pi)$$