Simplify $\int \frac{1}{\sqrt{2-x^2}}\, dx$ Somewhere in the provided answer: 
$$\int \frac{1}{\sqrt{2-x^2}} dx = \sin^{-1}{\frac{x}{\sqrt{2}}}$$
How did they get that? What I have: 
$$\frac{1}{\sqrt{2-x^2}} = \frac{1}{\sqrt{2(1-\frac{x^2}{2})}} = \frac{1}{\sqrt{2} \sqrt{1-\frac{x^2}{2}}}$$
$$\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{1-(\frac{x}{\sqrt{2}})^2}} = \frac{1}{\sqrt{2}} \sin^{-1}{\frac{x}{\sqrt{2}}}$$
So I have an extra $\frac{1}{\sqrt{2}}$ ... I probably had some stupid mistakes?

You made a mistake in the last step. To see why, let $u = \frac{x}{\sqrt{2}}$, $du = \frac{dx}{\sqrt{2}}$.
\begin{align*}
\frac{1}{\sqrt{2}} \int \frac{dx}{\sqrt{1-\left(\frac{x}{\sqrt{2}}\right)^2}} &= \frac{1}{\sqrt{2}} \int \frac{\sqrt{2}du}{\sqrt{1-u^2}} \\
&= \int \frac{du}{\sqrt{1-u^2}} \\
&= \arcsin{u} + c \\
&= \arcsin\left({\frac{x}{\sqrt{2}}}\right) + c
\end{align*}
Basically, you made an implicit variable substitution, but forgot that $dx$ also changes when you change the variable.
