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.
Best Answer
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.