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I have found this question, here:

$$I=\int\frac{\sin(x)}{\arcsin(-x)}dx$$

I know it is going to have a non-elementary but I have ran out of ideas for how to do this: my first try was: $$I=-\int\frac{\sin(x)}{\arcsin(x)}dx=-\int\frac{\sin(\sin(u))\cos(u)}{u}du$$ Now from here there isn't much we can do other than use series or expand one of the trig functions using $\Re$ or $\Im$ of $e^{jx}$ like this: $$I(a)=-\int\frac{\sin(\sin(u))\cos(au)}{u}du$$ $$I'(a)=\int\sin(\sin(u))\sin(au)du$$ This doesn't seem to lead anywhere though, so I tried: $$I(b)=-\int\frac{\sin(\sin(bu))\cos(u)}{u}du$$ $$I'(b)=-b\int\cos(bu)\cos(\sin(bu))\cos(u)du$$ but this also looks worse. It can also be expressed as: $$I=-\Im\int\frac{\cos(u)}{u}\sum_{r=0}^\infty\frac{j^r\sin^r(u)}{r!}du$$ Without the $u$ on the bottom this would be trivial, but with it the problem becomes difficult. I have also got it in the following forms: $$I=-\Im\int\left(\sum_{r=0}^\infty\frac{j^r\sin^r(u)}{r!}\right)\left(\sum_{k=0}^\infty\frac{(-1)^kx^{2k}}{(2k)!}\right)du$$ and $$I=-\Im\int\sum_{k=0}^\infty \frac{(-1)^ke^{j\sin(u)}u^{2k}}{(2k)!}du$$ If anyone knows any easier ways that would be much appreciated. Thanks

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    $\begingroup$ I have not been able to do anything. The problem seems to be very difficult; I did not find any way even with special functions. For the fun of it, I worked some "amazing" approximations. $\endgroup$ – Claude Leibovici Mar 20 at 10:08
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This is not an answer.

Concerning the initial integral $$I=\int\frac{\sin (x)}{\sin ^{-1}(x)}\,dx$$ it is interesting to notice that the integrand can be quite well approximated using a $[2n,2]$ Padé approximant built at $x=0$. Even the simplest one $$\frac{1-\frac{11 }{30}x^2}{1-\frac{1}{30}x^2}$$ is not so bad. For sure, increasing $n$, the approximation becomes better and better and the integrand reduces to something looking like $$\frac{\sin (x)}{\sin ^{-1}(x)}=P_{2n-2}(x)+ \frac{a}{1-b x^2}$$ that is to say that $$I\approx \frac{a }{\sqrt{b}}\tanh ^{-1}\left(x\sqrt{b} \right)+\int P_{2n-2}(x)\,dx $$

For example, using $n=4$, the Padé approximant would be $$\frac{\sin (x)}{\sin ^{-1}(x)}=\frac{ 1-\frac{15398 }{14553}x^2+\frac{100619 }{436590}x^4-\frac{2603 }{261954}x^6+\frac{754087 }{550103400}x^8} {1-\frac{10547 }{14553}x^2 }$$ For illustration purposes, considering $$J=\int_0^t\frac{\sin (x)}{\sin ^{-1}(x)}\,dx$$ we would obtain the following results ("exact" standing for "numerical integration") $$\left( \begin{array}{ccc} t & \text{approximation} & \text{exact} \\ 0.05 & 0.049986 & 0.049986 \\ 0.10 & 0.099889 & 0.099889 \\ 0.15 & 0.149625 & 0.149625 \\ 0.20 & 0.199110 & 0.199110 \\ 0.25 & 0.248262 & 0.248262 \\ 0.30 & 0.296994 & 0.296994 \\ 0.35 & 0.345223 & 0.345223 \\ 0.40 & 0.392862 & 0.392862 \\ 0.45 & 0.439823 & 0.439823 \\ 0.50 & 0.486019 & 0.486019 \\ 0.55 & 0.531356 & 0.531356 \\ 0.60 & 0.575739 & 0.575739 \\ 0.65 & 0.619066 & 0.619066 \\ 0.70 & 0.661230 & 0.661229 \\ 0.75 & 0.702107 & 0.702105 \\ 0.80 & 0.741560 & 0.741553 \\ 0.85 & 0.779420 & 0.779399 \\ 0.90 & 0.815467 & 0.815404 \\ 0.95 & 0.849394 & 0.849183 \\ 1.00 & 0.880708 & 0.879697 \end{array} \right)$$

I tried similar things with $[2n,4]$ Padé approximants; they do not really improve the results and the formulae become quite nasty because of the required partial fraction decomposition.

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