# Finding the integral of $x/\sqrt{4-x^2}$

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.

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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

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