<s>[INST] Integral of $\frac{1}{\sqrt{2-x^2}}$ I know that the integral of $\frac{1}{\sqrt{1-x^2}} = arcsin(x)$, but what is the the integral of $\frac{1}{\sqrt{2-x^2}}$? Is this how you do it?
$$\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2}(1-\frac{x^2}{\sqrt{2}})} = \frac{1}{\sqrt{2}} \arcsin\left(\frac x{\sqrt{2}}\right)$$
 [/INST] By considering the general integral $$\int \frac{dx}{\sqrt{a^2-x^2}}$$ and  applying the substitution $$x=a\sin u$$ 
You can calculate that $$ \int \frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right) $$
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