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.

This integral can be performed via differentiation under the integral sign. First note that for $|x|\leq1$ we have $\ln \sqrt{1-x^2} = \frac12\ln (1-x^2)$. Moreover, simple application of the chain rule yields
$$ \frac{d}{d\alpha} (1-x^2)^\alpha = (1-x^2)^\alpha \ln(1-x^2) .$$
The remaining integral is a special case of the beta function with $x=1/2$ and $y=\alpha+1$. Thus, we have
$$\int_0^1\!dx\,(1-x^2)^\alpha = 
\frac12\int_0^1\!dy\,y^{1/2} (1-y)^\alpha= \frac{\sqrt{\pi} \Gamma(1+\alpha)}{2 \Gamma(\frac{3}{2} + \alpha)}.$$
The original integral, we obtain by taking the derivative with respect to $\alpha$ and afterwards setting $\alpha=1/2$;
$$
\begin{align}\int_{-a}^a\!dx\,\sqrt{a^2-x^2} \ln\sqrt{a^2-x^2}
&=a^{2} \int_{0}^1\!dx\,\sqrt{1-x^2} [\ln a^2 + \ln(1-x^2)]\\
&= \frac{a^2 \pi \log a}{2}+ a^{2} \frac{d}{d\alpha} \int_0^1\!dx\,(1-x^2)^\alpha \Big|_{\alpha=1/2}\\
&= \frac{a^2 \pi \log a}{2}+ a^{2} \frac{d}{d\alpha} \frac{\sqrt{\pi} \Gamma(1+\alpha)}{2 \Gamma(\frac{3}{2} + \alpha)}  \Big|_{\alpha=1/2} \\
&= \frac{a^2 \pi \log a}{2}+ a^{2}  \frac{\Gamma(1+\alpha)}{\Gamma(3/2+\alpha)} [\psi(1+\alpha)-\psi(\tfrac{3}{2}+\alpha)]\Big|_{\alpha=1/2} \\
&=\frac{a^2 \pi \log a}{2}+  \frac{a^{2}\pi}4(1-2 \ln 2). 
\end{align}$$
Where we used the special values of the $\Gamma$ and the $\psi = (\log \Gamma)'$ at integer and half-integer values.
