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?
 A: 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.
A: $$\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}$$
$$\int\frac{dx}{\sqrt{2-x^2}}=\int\frac{dx}{\sqrt{{(\sqrt2})^2-x^2}} = \sin^{-1}\frac{x}{\sqrt2}$$
or, doing it other way
$$\int\frac{1}{\sqrt{2-x^2}}dx=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{1-(\frac{x}{\sqrt{2}})^2}}dx$$ 
now put , t=$\frac{x}{\sqrt2}$
$$\frac{1}{\sqrt{2}}\int \frac{\sqrt2 \, dt}{\sqrt{1-t^2}}= \int\frac{dt}{\sqrt{1-t^2}}=\sin^{-1}t+c = \sin^{-1}\frac{x}{\sqrt2}+c$$
A: I'm going to use a $u$-substitution to make it clearer why this is the case. Let $x = \sqrt{2}\sin \theta$ so $dx = \sqrt{2} \cos \theta d\theta $. We have that
\begin{eqnarray}
\int \dfrac{dx}{\sqrt{2-x^2}} &=& \int\dfrac{\sqrt{2}\cos \theta d\theta}{\sqrt{2 - (\sqrt{2}\sin \theta)^2}} \\
&=& \int\dfrac{\sqrt{2}\cos \theta d\theta}{\sqrt{2 - (\sqrt{2}\sin \theta)^2}} \\
&=& \int\dfrac{\sqrt{2}\cos \theta d\theta}{\sqrt{2 - 2\sin^2 \theta}}\\
&=& \int\dfrac{\cos \theta d\theta}{\sqrt{1 - \sin^2 \theta}} \\
&=& \int\dfrac{\cos \theta d\theta}{\sqrt{\cos^2\theta}} \\
&=& \int\dfrac{\cos \theta d\theta}{cos\theta} \\
&=& \int d\theta = \theta+C.
\end{eqnarray}
Since $x = \sqrt{2}\sin \theta$, we have $\sin \theta = \dfrac{x}{\sqrt{2}}$ and so $\theta = \sin^{-1}\left(\dfrac{x}{\sqrt{2}}\right)$. Therefore 
$$
\int \dfrac{dx}{\sqrt{2-x^2}} = \theta+C = \sin^{-1}\left(\dfrac{x}{\sqrt{2}}\right) + C.
$$
