3
$\begingroup$

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.

  • 1
    TRY AN EXAMPLE! Is it true that $\int(x/x)=(\int x)(\int(1/x))$?2012-05-18

2 Answers 2

4

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.

  • 0
    @dave5678: Yes, you can check easily by differentiating that you are right.2012-05-18
1

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

  • 0
    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