How to calculate $\int_{-a}^{a} \sqrt{a^2-x^2}\ln(\sqrt{a^2-x^2})\mathrm{dx}$ Well,this is a homework problem.
I need to calculate the differential entropy of random variable 
$X\sim f(x)=\sqrt{a^2-x^2},\quad -a<x<a$ and $0$ otherwise. Just how to calculate
$$
\int_{-a}^a \sqrt{a^2-x^2}\ln(\sqrt{a^2-x^2})\,\mathrm{d}x
$$
I can get the result with Mathematica,but failed to calculate it by hand.Please give me some idea.

[Some ideas]
You can rewrite it as follows:
$$\int_{-a}^a \sqrt{a^2-x^2} f(x) dx$$
where $f(x)$ is the logarithm.
Note that the integral, sans $f(x)$, is simply a semicircle of radius $a$. In other words, we can write,
$$\int_{-a}^a \int_0^{\sqrt{a^2-x^2}} f(x) dy dx=\int_{-a}^a \int_0^{\sqrt{a^2-x^2}} \ln{\sqrt{a^2-x^2}} dy dx$$
Edit: Found a mistake. Thinking it through. :-)
