How to integrate $\int \frac{1}{\sqrt{1-x^2-y^2}}\,dy$? I have this integral: 
$$\int \frac{1}{\sqrt{1-x^2-y^2}}\; dy$$
The way I would integrate it is: 
$$\int \dfrac{1}{\sqrt{(\sqrt{1-x^2})^2-y^2}}\;dy=\sin^{-1} \dfrac{y}{\sqrt{1-x^2}}$$
$\int \dfrac{du}{\sqrt{a^2-u^2}}=\sin ^{-1} \dfrac{u}{a}, \; \text{where} \; a=\sqrt{1-x^2}$. 
However, using the integral calculator, I get: 
$$-i\operatorname{arcsinh}\left(\dfrac{y}{\sqrt{x^2-1}}\right)$$
And now I am confused as to why the answer is different? Which method of solving this integral would be correct?

The two expressions are equal, as $\sinh^{-1}(z)=\frac1i\sin^{-1}(iz)$.  Look here.
