Question: Evaluate$$I=\int\limits_0^{1/2}dx\,\frac {\log x}{\sqrt{1-x^2}}$$
I've given it the best I could. The first thing I did was use differentiation under the integral sign, but I wasn't sure where to continue after that.
The next thing I did was try substituting $x\mapsto 2x$ to get the limits in terms of zero and one and transform $I$ into the beta function (no luck because the denominator is $1-4x^2$).
Then I tried integration by parts. It seemed promising at first because I got$$I=-\frac {\pi}6-\int\limits_0^{1/2}dx\,\frac {\arcsin x}{x}$$But I'm not sure how to evaluate the second integral. I even tried using a Laplace Transform, but I'm again not sure how to evaluate the integral$$\mathcal{L}(f(t))=\int\limits_0^{\infty}dt\, e^{-st}\arcsin t$$
Can anybody give me a few hints as to where to begin and what to do? I'm out of ideas.
Hint. By the change of variable, $$ \theta=\arcsin x, \qquad d\theta=\frac {dx}{\sqrt{1-x^2}}, $$one gets $$ I=\int_0^{1/2}dx\,\frac {\log x}{\sqrt{1-x^2}}=\int_{0}^{\large\frac \pi6}\ln(\sin \theta)\,d\theta $$ then one may use the standard identity, $$\log(\sin \theta)=-\ln 2-\sum_{k\geq 1}\frac{\cos(2k\theta)}{k} \qquad \left(0<\theta<\pi \right), $$ to obtain, by a valid termwise integration, $$ I=-\frac{\pi\ln 2}6-\frac12\sum_{k\ge 1}\frac{\sin(\pi k/3)}{k^2}=-\frac{\pi\ln 2}6+\frac12\Im\:\text{Li}_2\left(\frac12-i\frac{\sqrt{3}}2\right), $$ that is $$\bbox[15px,border:1px solid #ff6600]{ \int_0^{1/2}dx\,\frac {\log x}{\sqrt{1-x^2}}=-\frac{\pi\ln 2}6+\frac12\Im\:\text{Li}_2\left(\frac12-i\frac{\sqrt{3}}2\right)} $$ where $\text{Li}_2(\cdot)$ is the dilogarithm function.