Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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
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
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.

share|cite|improve this answer

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

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

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

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

share|cite|improve this answer
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 $$ \int \dfrac{dx}{\sqrt{2-x^2}} = \theta+C = \sin^{-1}\left(\dfrac{x}{\sqrt{2}}\right) + C. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.