Numerical evidence shows the validity of the following identity
$$\int\limits\_0^z\frac{xdx}{\sin{x}\sqrt{\sin^2{z}-\sin^2{x}}}=\frac{\pi}{4\sin{z}}\ln{\frac{1+\sin{z}}{1-\sin{z}}},\tag{1}$$
if $0< z< \pi/2$.
How can it be proved? An indirect proof can be found in the paper <http://link.springer.com/article/10.1134%2FS1547477113010044> (Potential of multiphoton exchange in the scattering of light charged particles of a heavy target, by Yu.M. Bystritskiy, E.A. Kuraev and M.G. Shatnev). Formula (1) is equivalent to (12) from the paper where it is called "the marvelous identity".



Interesting integral
Actually, I now think that the easiest method is to do this: Write $k=\sin z$, so that $|k|<1$, and make the substitution $x = \arcsin(k\sin\theta)$, where $0\le \theta\le \frac\pi2$. The integral becomes
$$
\int\_0^{\pi/2} \frac{\arcsin(k\sin\theta)}{k\sin\theta \,\,(1-k^2\sin^2\theta)^{(1/2)}}\ \mathrm{d}\theta
 = \int\_0^{\pi/2} \sum\_{n=0}^{\infty} c\_n\,k^{2n}\sin^{2n}\theta\,\mathrm{d}\theta,
$$
where the numbers $c\_n$ are the coefficients in the even power series
$$
\frac{\arcsin(t)}{t \,(1{-}t^2)^{(1/2)}} = \sum\_{n=0}^{\infty} c\_n t^{2n},
$$
which are easily calculated to be 
$$
c\_n = \frac{2^{2n} (n!)^2}{(2n{+}1)!}.
$$
Combining this with the well-known formula
$$
\int\_0^{\pi/2} \sin^{2n}\theta\,\mathrm{d}\theta = \frac{\pi}{2^{2n+1}}\ {{2n}\choose{n}},
$$
one obtains
$$
\int\_0^{\pi/2} \frac{\arcsin(k\sin\theta)}{k\sin\theta (1-k^2\sin^2\theta)^{(1/2)}}\ \mathrm{d}\theta
 = \pi \sum\_{n=0}^{\infty} \frac{k^{2n}}{(4n{+}2)}
= \frac{\pi}2 \sum\_{n=0}^{\infty} \frac{\sin^{2n}z}{(2n{+}1)}\ .
$$
The rest should be clear.


