Another arctan integral 
$$\int_{-1}^{1}\arctan(zx){\mathrm dx\over x\sqrt{1-x^2}}=\pi\sinh^{-1}(z)\tag1$$

$x=\sin{y}$ then $dx=\cos{y}dy$
$$\int\arctan(z\sin{y}){\mathrm dy\over \sin{y}}\tag2$$
$$\sum_{k=0}^{\infty}{(-1)^k\over z^{2k+1}}\int{\mathrm dy\over (\sin{y})^{2k+2}}\tag3$$

How do we show that $(1)=\pi\sinh^{-1}(z)?$

 A: Setting
$$I = \int^1_{-1} \tan^{-1} (zx) \frac{dx}{x \sqrt{1 - x^2}}.$$
If we let $x = \sin \theta$, then
$$I = \int^{\pi/2}_{-\pi/2} \frac{\tan^{-1} (z \sin \theta)}{\sin \theta} \, d\theta. \tag1$$
Now from the Maclaurin series expansion for the inverse tangent function, namely
$$\tan^{-1} t = \sum^\infty_{n = 0} \frac{(-1)^n t^{2n + 1}}{2n + 1},$$
replacing $t$ with $z \sin \theta$ one has
$$\tan^{-1} (z \sin \theta) = \sum^\infty_{n = 0} \frac{(-1)^n z^{2n + 1} \sin^{2n + 1} \theta}{2n + 1}.$$
So (1) becomes
$$I = \sum^\infty_{n = 0} \frac{(-1)^n z^{2n + 1}}{2n + 1} \int^{\pi/2}_{-\pi/2} \sin^{2n} \theta \, d\theta = 2 \sum^\infty_{n = 0} \frac{(-1)^n z^{2n + 1}}{2n + 1} \int^{\pi/2}_0 \sin^{2n} \theta d\theta,$$
where the summation and the integration sign have been interchanged and we note the integrand is an even function between symmetric limits.  
Since (see here)
$$\int^{\pi/2}_0 \sin^{2n} \theta \, d\theta = \frac{\pi (2n)!}{2^{2n + 1} (n!)^2},$$
we have
$$I = \pi \sum^\infty_{n = 0} \frac{(-1)^n (2n)!}{2^{2n} (n!)^2 (2n + 1)} z^{2n + 1} = \pi \sinh^{-1} z,$$
where we recognise the infinite sum as corresponding to the Maclaurin series expansion for the inverse hyperbolic sine function (see here).   
