Integrate: $\int \frac{\sin x}{\sin ^{-1}x}dx$

I just faced an integral:$$\int \frac{\sin x}{\sin ^{-1}x}dx$$All I can do is, consider the denominator as $z$ and substitute:$$x=\sin z$$$$dx=\cos z$$And finally, I have:$$\int \frac{1}{z}\sin (\sin z)\cos z dz$$Now what can I do? I tried integration by parts but it seemed a bit complex. But I do not think any kind of usual integration procedure shall work here, since there seems to be no elementary antiderivative present for this expression.

Still, the integration by parts may look like:$$\sin(\sin z)\cos z\int\frac {dz}z-\int\left(\frac d{dz}(\sin(\sin z)\cos z) \int \frac {dz}z\right)dz$$$$=\sin(\sin z)\cos z\ln z-\int\left(\frac d{dz}(\sin(\sin z)\cos z) \ln z\right)dz$$Things seem to get even messier. Any help?