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

Find the integral:

$$\int \frac{x}{\sqrt{4-x^2}} dx = \int \frac{x}{\sqrt{2^2-x^2}} dx$$

using $$\int \frac1{\sqrt{a^2-x^2}} dx = \arcsin(x/a) + C$$

I get $\displaystyle \frac{x^2}{2} \arcsin \left(\frac{x}{2} \right) + C$.

I'm not sure if the $\dfrac{x^2}{2}$ is right. Any suggestions and help would be great.

share|cite|improve this question
It looks like you antidifferentiated $x$ and $1\over\sqrt{4-x^2}$ separately and then multiplied. You can't do this... – David Mitra May 17 '12 at 23:30
In other words, it looks like you are using the "rule", $\int(f(x)/g(x))\,dx=(\int f(x)\,dx)(\int(1/g(x))\,dx)$. It should not be hard to find a simple example to convince yourself that this is not, in general, true. – Gerry Myerson May 17 '12 at 23:33
They don't mean the same thing when they are separated? – dave5678 May 17 '12 at 23:35
TRY AN EXAMPLE! Is it true that $\int(x/x)=(\int x)(\int(1/x))$? – Gerry Myerson May 18 '12 at 2:18
up vote 4 down vote accepted

Hint: Let $u=4-x^2$. The derivative of $u$ is sitting on top. Sort of.

Note that you can always use differentiation to check whether an indefinite integral has been calculated correctly.

share|cite|improve this answer
If I substitute that, I would get $-1/2$ $\int$$1/\sqrt[]{u}$ $du$ = $-\sqrt[]{4-x^2}$ + $C$ – dave5678 May 17 '12 at 23:47
@dave5678: Yes, you can check easily by differentiating that you are right. – André Nicolas May 18 '12 at 0:19

Another way: $$\int\frac{x}{\sqrt{4-x^2}}dx=-\frac{1}{2}\int\frac{d(4-x^2)}{(4-x^2)^{1/2}}dx=-\frac{1}{2}\frac{\sqrt{4-x^2}}{1/2}+C=-\sqrt{4-x^2}+C$$

share|cite|improve this answer
The OP said he wanted to use the integral of $\frac{1}{\sqrt{a^2-x^2}}$ (but didn't say why) – Stefan Smith May 18 '12 at 23:57

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.