$\int_0^4\frac{\log x}{\sqrt{4x-x^2}} dx=0$ I am having trouble proving that it is equal to zero analytically. I have tried plotting and know that for $0<x<1$ the integrand is negative and positive otherwise. I have tried substitution $u\to \sqrt{x}$ but I cannot proceed further. 
 A: $$4x-x^2=2^2-(x-2)^2$$
Let $x-2=2\cos2t$
$$\int_0^4\dfrac{\ln(x)}{\sqrt{4x-x^2}}dx=-4\int_{\pi/2}^0\ln(2+2\cos2t)dt$$
$$=2\ln2\int_0^{\pi/2}dt+4\int_0^{\pi/2}\ln(\cos t)dt$$
See Evaluate $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$
A: Let us consider the integral $$I\left(s\right)=\int_{0}^{4}\frac{x^{s}}{\sqrt{4x-x^{2}}}dx,\, s>-1/2.
 $$ We note thet $$ \int_{0}^{4}\frac{x^{s}}{\sqrt{4x-x^{2}}}dx\overset{x=4y}{=}4^{s+1}\int_{0}^{1}\frac{y^{s}}{\sqrt{16y-16y^{2}}}dy
 $$ $$=4^{s}\int_{0}^{1}y^{s-1/2}\left(1-y\right)^{-1/2}dy=4^{s}\frac{\Gamma\left(s+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(s+1\right)}
 $$ so $$\frac{d}{ds}I\left(s\right)_{s=0}=\int_{0}^{4}\frac{\log\left(x\right)}{\sqrt{4x-x^{2}}}dx=\frac{d}{ds}\left(4^{s}\frac{\Gamma\left(s+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(s+1\right)}\right)_{s=0}
 $$ and note that $$\begin{align}
\frac{d}{ds}\left(4^{s}\frac{\Gamma\left(s+\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(s+1\right)}\right)_{s=0}= & \left(\frac{4^{s}\Gamma\left(\frac{1}{2}\right)\Gamma\left(s+\frac{1}{2}\right)\left(\psi\left(s+\frac{1}{2}\right)-\psi\left(s+1\right)\log\left(4\right)\right)}{\Gamma\left(s+1\right)}\right)_{s=0} \\ = &0.
\end{align}
$$
A: It is possible to show that 
$$
\int_{0}^{4}\frac{\ln x}{\sqrt{4x-x^2}}\,dx=2\int_{0}^{\pi/2}\ln(2\sin u)\,du
$$
Indeed, 
$$
\int_{0}^{4}\frac{\ln x}{\sqrt{4x-x^2}}\,dx=
\int_{0}^{4}\frac{\ln x}{2\sqrt{1-(\frac{x-2}{2})^2}}\,dx
\overset{t=\frac{x-2}{2}}{=}\int_{-1}^{1}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt\\
=\int_{-1}^{0}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt+\int_{0}^{1}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt\\
\overset{u=-t}{=}\int_{1}^{0}\frac{\ln(2-2u)}{\sqrt{1-u^2}}\cdot-du+\int_{0}^{1}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt\\
\int_{0}^{1}\frac{\ln(2-2t)}{\sqrt{1-t^2}}\,dt +\int_{0}^{1}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt\\
$$
Now,
$$
\int_{0}^{1}\frac{\ln(2t+2)}{\sqrt{1-t^2}}\,dt\overset{t=\cos u}{=}
\int_{\pi/2}^{0}\frac{\ln(2\cos u+2)}{\sin u}\cdot-\sin u du
=\int_{0}^{\pi/2}\ln(2\cos u+2)\,du\\
=\int_{0}^{\pi/2}\ln(4\cos^2\frac{u}{2})\,du
=2\int_{0}^{\pi/2}\ln(2\cos\frac{u}{2})\,du
$$
similarly
$$
\int_{0}^{1}\frac{\ln(2-2t)}{\sqrt{1-t^2}}\,dt
=2\int_{0}^{\pi/2}\ln(2\sin\frac{u}{2})\,du
$$
so
$$
\int_{0}^{4}\frac{\ln x}{\sqrt{4x-x^2}}\,dx
=2\int_{0}^{\pi/2}\ln(2\sin u)\,du
$$
