Indefinite integral $\int \arcsin \left(k\sin x\right) dx$

Solution 1:

Let us denote $$ \mathcal{I}\left(k,x\right)=\int_0^x \eta\left(y\right)dy, \qquad \eta\left(y\right)=\arcsin\left(k\sin y\right).$$ It turned out that this integral can be computed in a rather simple way. Namely, let us make the following observations:

1. We have $$ \eta\left(x\right)=\frac{1}{2i}\left[\ln\left(1+k\, e^{i\left(x+\eta(x)\right)}\right)-\ln\left(1+k\, e^{-i\left(x+\eta(x)\right)}\right)\right]$$
2. Also, \begin{align*} 2\left[1+\frac{k\,\cos x}{\sqrt{1-k^2\sin^2x}}\right]\times\frac{1}{2i}\left[\ln\left(1+k\, e^{i\left(x+\eta(x)\right)}\right)-\ln\left(1+k\, e^{-i\left(x+\eta(x)\right)}\right)\right]=\\ =\frac{d}{dx}\left[\operatorname{Li}_2\left(-k\,e^{i\left(x+\eta(x)\right)}\right)+ \operatorname{Li}_2\left(-k\,e^{-i\left(x+\eta(x)\right)}\right)\right] \end{align*}
3. And finally: \begin{align*} \eta'\left(x\right)=\frac{k\,\cos x}{\sqrt{1-k^2\sin^2x}}. \end{align*}

Combining these three formulas, we get \begin{align*}\boxed{ \;2\mathcal{I}\left(k,x\right)=\operatorname{Li}_2\left(-k\,e^{i\left(x+\eta(x)\right)}\right)+ \operatorname{Li}_2\left(-k\,e^{-i\left(x+\eta(x)\right)}\right)-2\operatorname{Li}_2\left(-k\right)-\eta^2\left(x\right)\;} \end{align*}