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".

$\begingroup$ probably some contour integration magic... $\endgroup$– Dima PasechnikCommented Feb 19, 2015 at 8:31

$\begingroup$ Are you sure the LHS is an antiderivative, or is the bound $z$ mistaken for some other value? Otherwise you might as well differentiate both sides... $\endgroup$– Loïc TeyssierCommented Feb 19, 2015 at 8:48

$\begingroup$ @LoïcTeyssier: there is a $z$ in the integrand, so differentiating the LHS is not as easily done, I think. $\endgroup$– Willie WongCommented Feb 19, 2015 at 10:23

$\begingroup$ Yep, sorry, my mistake. $\endgroup$– Loïc TeyssierCommented Feb 19, 2015 at 11:04

2$\begingroup$ @Zurab Silagadze expand the integrand in powers of $\sin x/\sin z$ and then integrate term by term. $\endgroup$– juanCommented Feb 19, 2015 at 12:06
2 Answers
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 $$ \int_0^{\pi/2} \frac{\arcsin(k\sin\theta)}{k\sin\theta \,\,(1k^2\sin^2\theta)^{(1/2)}}\ \mathrm{d}\theta = \int_0^{\pi/2} \sum_{n=0}^{\infty} c_n\,k^{2n}\sin^{2n}\theta\,\mathrm{d}\theta, $$ 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 wellknown formula $$ \int_0^{\pi/2} \sin^{2n}\theta\,\mathrm{d}\theta = \frac{\pi}{2^{2n+1}}\ {{2n}\choose{n}}, $$ one obtains $$ \int_0^{\pi/2} \frac{\arcsin(k\sin\theta)}{k\sin\theta (1k^2\sin^2\theta)^{(1/2)}}\ \mathrm{d}\theta = \pi \sum_{n=0}^{\infty} \frac{k^{2n}}{(4n{+}2)} = \frac{\pi}2 \sum_{n=0}^{\infty} \frac{\sin^{2n}z}{(2n{+}1)}\ . $$ The rest should be clear.
This integral is due to Lobachevskii. He gave it in more general form as follows which can be found in his work "Application of imaginary geometry to certain integrals" (1836). Also see equation 3.842.7 in Gradsteyn and Ryzhik. The integral in OP's post follows from this formula in the limit $~\alpha\to i\infty$.