Simplify $\int \frac{1}{\sqrt{2-x^2}}\, dx$

Question

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?

Answer 1

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.

Answer 2

$$\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a}$$

$$\int\frac{dx}{\sqrt{2-x^2}}=\int\frac{dx}{\sqrt{{(\sqrt2})^2-x^2}} = \sin^{-1}\frac{x}{\sqrt2}$$

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=$\frac{x}{\sqrt2}$

$$\frac{1}{\sqrt{2}}\int \frac{\sqrt2 \, dt}{\sqrt{1-t^2}}= \int\frac{dt}{\sqrt{1-t^2}}=\sin^{-1}t+c = \sin^{-1}\frac{x}{\sqrt2}+c$$

Answer 3

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 $$ \int \dfrac{dx}{\sqrt{2-x^2}} = \theta+C = \sin^{-1}\left(\dfrac{x}{\sqrt{2}}\right) + C. $$