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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?

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2  
HINT: $u=x/\sqrt{2}$ then $dx=?$ (which is missing in your last line (and the title) and is probably the source of the error!) –  draks ... Apr 29 '12 at 12:15
2  
You're making change of the variable in your integral $t=x/\sqrt{2}$. This implies that you need to change your differential: $dx=\sqrt{2}dt$. And this will take care of your extra $\sqrt{2}$ –  Artem Apr 29 '12 at 12:15
3  
Leaving out the $dx$ sounds like a good idea, save some time, save some paper. But people who leave out the $dx$ often make mistakes during the substitution process. –  André Nicolas Apr 29 '12 at 15:49

3 Answers 3

up vote 7 down vote accepted

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.

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

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

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

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

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You forgot the $+c$ –  Stefan Smith Jun 8 '12 at 12:04

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

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