# How to integrate $\int \dfrac{1}{\sqrt{1-x^2-y^2}}\; dy$?

by D. Qa   Last Updated July 12, 2018 01:20 AM

I have this integral:

$$\int \dfrac{1}{\sqrt{1-x^2-y^2}}\; dy$$

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:

$$-\mathrm{i}\operatorname{arsinh}\left(\dfrac{y}{\sqrt{x^2-1}}\right)$$

And now I am confused as to why the answer is different? Which method of solving this integral would be correct?

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