Find $\displaystyle \int_0^1\frac{\ln x}{\sqrt{x-x^2}}\,dx.$
Another way? What does (s)he want? Hmm ...
Let $\displaystyle I=\int_0^1\frac{\ln x}{\sqrt{x-x^2}}\mathrm dx$ and let $\displaystyle J=\int_0^1\frac{\ln x}{\sqrt{1-x^2}}\mathrm dx$.
The substitution $\displaystyle x=u^2$ shows that $\displaystyle I=4J$.
The substituion $\displaystyle y=1-x$ shows that $\displaystyle I=\int_0^1\frac{\ln (1-x)}{\sqrt{x-x^2}}\mathrm dx$ and so $\displaystyle 2I=\int_0^1\frac{\ln x+\ln(1-x)}{\sqrt{x-x^2}}\mathrm dx=\int_0^1\frac{\ln(x-x^2)}{\sqrt{x-x^2}}\mathrm dx$.
Thus $\displaystyle I=\frac12\int_0^1\frac{\ln(x-x^2)}{\sqrt{x-x^2}}\mathrm dx
=\int_0^{1/2}\frac{\ln(x-x^2)}{\sqrt{x-x^2}}\mathrm dx$ by symmetry.
Now put $\displaystyle u=2\sqrt{x-x^2}$ so that $\displaystyle 1-u^2=(1-2x)^2$, $\displaystyle 1-2x=\sqrt{1-u^2}$ for $\displaystyle 0\leq x\leq{\textstyle\frac12}$ and $\displaystyle 0\leq u\leq 1$, and $\displaystyle \mathrm dx=\frac{u\,\mathrm du}{2\sqrt{1-u^2}}$.
Thus $\displaystyle I=\int_0^1\frac{\ln (u^2/4)}{u/2}\frac{u\,\mathrm du}{2\sqrt{1-u^2}}=\int_0^1\frac{2\ln u-\ln 4}{\sqrt{1-u^2}}\mathrm du=2J-\ln 4\bigl[\arcsin u\bigr]_0^1={\textstyle\frac12}I-\pi\ln 2$.
Thus $\displaystyle I=-2\pi\ln 2$.
Wait a moment, this is merely a disguised version of the $\displaystyle \int_0^{\pi/2}\ln(\sin x)\,\mathrm dx$ method. It's no better than NonCommAlg's rejected solution.
(Thinks: Perhaps I should have used an infinite series, or some kind of limit, or introduced a parameter, or ... (Mind boggles at this point))
An answer that is not really an answer? This is trickier than it looks.
Hmm ...
What about this ?
$\displaystyle \int_0^1 \frac{\ln(x)}{\sqrt{x-x^2}}~dx $
$\displaystyle = \lim_{a\to0} \frac{1}{a} [ \int_0^1 \frac{x^a-1}{\sqrt{x-x^2}}~dx ] $
$\displaystyle = \lim_{a\to0} \frac{1}{a} \sqrt{\pi} [ \frac{ \Gamma(a + \frac{1}{2})}{\Gamma(a + 1) } - \sqrt{\pi} ]$
From the Lengre's Identity
$\displaystyle = \lim_{a\to0} \frac{1}{a} \sqrt{\pi} [ \frac{\sqrt{\pi} \Gamma(2a) 2^{-2a+1}}{ \Gamma(a) \Gamma(a + 1) } - \sqrt{\pi} ] $
$\displaystyle = \pi \lim_{a\to0} \frac{1}{a}[ 2^{-2a+1} / 2 - 1] $
$\displaystyle = \pi \lim_{a\to0} \frac{1}{a}[ 2^{-2a} - 1]$
$\displaystyle = \pi \ln(2) (-2) = - 2 \pi \ln(2)$
I think it is the best method by substituting $\displaystyle x = \sin^2{\theta} $ ..