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.