面白い積分の問題

$$\begin{array}{rl}& {\int }_0^4\frac{\ln x}{\sqrt{4x-x^2}}dx\\ =& {\int }_0^4\frac{\ln x}{\sqrt{4-(x-2{)}^2}}dx\\ =& {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\ln (2\sin \theta +2)}{\sqrt{4-4{\sin }^2\theta }}\cdot 2\cos \theta d\theta x-2\mapsto 2\sin \theta \\ =& {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\ln (2\sin \theta +2)}{2\cos \theta }\cdot 2\cos \theta d\theta \because \sqrt{1-{\sin }^2\theta }=\cos \theta (-\frac{\pi }{2}\le \theta \le \frac{\pi }{2})\\ =& {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln (2\sin \theta +2)d\theta \\ =& {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln 2d\theta +{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln (1+\sin \theta )d\theta \\ =& \pi \ln 2+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln (1+\sin \theta )d\theta \\ =& \pi \ln 2+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln (1-\sin \theta )d\theta \theta \mapsto -\theta \text{King Property}\\ =& \frac{1}{2}(2\pi \ln 2+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln (1+\sin \theta )+\ln (1-\sin \theta )d\theta )\because 上の二つの式を足して半分にする。\\ =& \pi \ln 2+\frac{1}{2}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\ln (1-{\sin }^2\theta )d\theta \\ =& \pi \ln 2+2{\int }_0^{\frac{\pi }{2}}\ln (\cos \theta )d\theta \because 対数の性質と偶関数の性質\\ =& \pi \ln 2+{\int }_0^{\frac{\pi }{2}}\ln (\cos \theta )d\theta +{\int }_0^{\frac{\pi }{2}}\ln (\cos \theta )d\theta \\ =& \pi \ln 2+{\int }_0^{\frac{\pi }{2}}\ln (\sin \theta )d\theta +{\int }_0^{\frac{\pi }{2}}\ln (\cos \theta )d\theta \theta \mapsto \frac{\pi }{2}-\theta \text{King Property}\\ =& \pi \ln 2+{\int }_0^{\frac{\pi }{2}}\ln (\sin \theta \cos \theta )d\theta \because 対数の性質\\ =& \pi \ln 2+{\int }_0^{\frac{\pi }{2}}\ln \left(\frac{\sin 2\theta }{2}\right)d\theta \because 半角の公式\\ =& \pi \ln 2+{\int }_0^{\frac{\pi }{2}}\ln (\sin 2\theta )d\theta -{\int }_0^{\frac{\pi }{2}}\ln 2d\theta \because 対数の性質\\ =& \frac{\pi }{2}\ln 2+\frac{1}{2}{\int }_0^{\pi }\ln (\sin t)dt2\theta \mapsto t\\ =& \frac{\pi }{2}\ln 2+\frac{1}{2}({\int }_0^{\frac{\pi }{2}}\ln (\sin t)dt+{\int }_{\frac{\pi }{2}}^{\pi }\ln (\sin t)dt)\\ =& \frac{\pi }{2}\ln 2+\frac{1}{2}({\int }_0^{\frac{\pi }{2}}\ln (\sin t)dt+{\int }_{\frac{\pi }{2}}^0\ln (\sin (\pi -t))(-dt))t\mapsto \pi -t\\ =& \frac{\pi }{2}\ln 2+{\int }_0^{\frac{\pi }{2}}\ln (\sin t)dt\end{array}$$
同型出現したので13行目の式と比較して、
$${\int }_0^{\frac{\pi }{2}}\ln (\cos \theta )d\theta ={\int }_0^{\frac{\pi }{2}}\ln (\sin \theta )d\theta =-\frac{\pi }{2}\ln 2$$
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。