面白い積分の問題

$$\begin{align} &\int_{0}^{4}\frac{\ln{x}}{\sqrt{4x-x^2}}dx\\[5pt] =&\int_{0}^{4}\frac{\ln{x}}{\sqrt{4-(x-2)^2}}dx\\[5pt] =&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\ln(2\sin\theta+2)}{\sqrt{4-4\sin^2{\theta}}}\cdot2\cos\theta\,d\theta\qquad x-2\mapsto 2\sin\theta\\[5pt] =&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\ln(2\sin\theta+2)}{2\cos\theta}\cdot2\cos\theta\,d\theta\qquad\because\;\sqrt{1-\sin^2\theta}=\cos\theta\quad\left(-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\right)\\[5pt] =&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ln(2\sin\theta+2)\,d\theta\\[5pt] =&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ln{2}\,d\theta+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ln(1+\sin\theta)\,d\theta\\[5pt] =&\pi\ln{2}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ln(1+\sin\theta)\,d\theta\\[5pt] =&\pi\ln2+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ln(1-\sin\theta)\,d\theta\qquad\theta\mapsto-\theta\quad\text{King Property}\\[5pt] =&\frac{1}{2}\left(2\pi\ln2+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ln(1+\sin\theta)+\ln(1-\sin\theta)\,d\theta\right)\qquad\because\;上の二つの式を足して半分にする。\\[5pt] =&\pi\ln2+\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ln(1-\sin^2\theta)\,d\theta\\[5pt] =&\pi\ln2+2\int_{0}^{\frac{\pi}{2}}\ln(\cos\theta)\,d\theta\qquad\because対数の性質と偶関数の性質 \\[5pt] =&\pi\ln2+\int_{0}^{\frac{\pi}{2}}\ln(\cos\theta)\,d\theta+\int_{0}^{\frac{\pi}{2}}\ln(\cos\theta)\,d\theta\\[5pt] =&\pi\ln2+\int_{0}^{\frac{\pi}{2}}\ln(\sin\theta)\,d\theta+\int_{0}^{\frac{\pi}{2}}\ln(\cos\theta)\,d\theta\qquad \theta\mapsto\frac{\pi}{2}-\theta\quad\text{King Property}\\[5pt] =&\pi\ln2+\int_{0}^{\frac{\pi}{2}}\ln(\sin\theta\cos\theta)\,d\theta\qquad\because対数の性質\\[5pt] =&\pi\ln2+\int_{0}^{\frac{\pi}{2}}\ln(\frac{\sin2\theta}{2})\,d\theta\qquad\because半角の公式\\[5pt] =&\pi\ln2+\int_{0}^{\frac{\pi}{2}}\ln(\sin2\theta)\,d\theta-\int_{0}^{\frac{\pi}{2}}\ln2\,d\theta\qquad\because対数の性質\\[5pt] =&\frac{\pi}{2}\ln2+\frac{1}{2}\int_{0}^{\pi}\ln(\sin{t})\,dt\qquad 2\theta\mapsto t\\[5pt] =&\frac{\pi}{2}\ln2+\frac{1}{2}\left(\int_{0}^{\frac{\pi}{2}}\ln(\sin{t})\,dt+\int_{\frac{\pi}{2}}^{\pi}\ln(\sin{t})\,dt\right)\\[5pt] =&\frac{\pi}{2}\ln2+\frac{1}{2}\left(\int_{0}^{\frac{\pi}{2}}\ln(\sin{t})\,dt+\int_{\frac{\pi}{2}}^{0}\ln(\sin(\pi-t))\,(-dt)\right)\qquad t\mapsto \pi-t\\[5pt] =&\frac{\pi}{2}\ln2+\int_{0}^{\frac{\pi}{2}}\ln(\sin{t})\,dt \end{align}$$
同型出現したので13行目の式と比較して、
$$ \int_{0}^{\frac{\pi}{2}}\ln(\cos{\theta})\,d\theta=\int_{0}^{\frac{\pi}{2}}\ln(\sin\theta)\,d\theta=-\frac{\pi}{2}\ln2 $$
なので、求めたい積分は、
$$ \int_{0}^{4}\frac{\ln{x}}{\sqrt{4x-x^2}}dx=0 $$
と計算できました。