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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...2012-05-17
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    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.2012-05-17
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    They don't mean the same thing when they are separated?2012-05-17
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    TRY AN EXAMPLE! Is it true that $\int(x/x)=(\int x)(\int(1/x))$?2012-05-18

2 Answers 2

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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.

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    If I substitute that, I would get $-1/2$ $\int$$1/\sqrt[]{u}$ $du$ = $-\sqrt[]{4-x^2}$ + $C$2012-05-17
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    @dave5678: Yes, you can check easily by differentiating that you are right.2012-05-18
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Another way: $$\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$$

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    The OP said he wanted to use the integral of $\frac{1}{\sqrt{a^2-x^2}}$ (but didn't say why)2012-05-18