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?

$$\int \frac{1}{\sqrt{1-x^2-y^2}}\, dy$$
Set
$$u = \frac{iy}{\sqrt{1-x^2}} \implies dy = -i\sqrt{1-x^2}\,du$$
$$-\int \frac{i\sqrt{1-x^2}}{\sqrt{(1-x^2)u-x^2+1}}\, du$$
$$-i\int \frac{1}{\sqrt{u^2+1}}\, du$$
Set
$$u = \sinh(v)\implies v= \operatorname{arsinh}(u),\,\,dv=\cosh(v)dv$$
$$-i\int\frac{\cosh(v)}{\sqrt{\operatorname{arsinh}^2(v)+1}}dv$$
Let $\operatorname{arsinh}^2(u)+1 = \cosh^2(v)$
$$-i\int1\,dv = v = -i \operatorname{arsinh}(u)$$
Undo subtitution $u$

$$-i \operatorname{arsinh}\left(\frac{iy}{\sqrt{1-x^2}}\right)$$

