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.

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