[再寄小读者之数学篇](2015-06-08 一个有意思的定积分计算)

$$\beex \bea \int_0^\frac{\pi}{4}\ln (1+\tan x)\rd x &=\int_0^\frac{\pi}{4} \ln \frac{\cos x+\sin x}{\cos x}\rd x\\ &=\int_0^\frac{\pi}{4} \ln \sez{\sqrt{2}\sin \sex{x+\frac{\pi}{4}}}-\ln \cos x\rd x\\ &=\frac{\pi}{8}\ln 2 +\int_\frac{\pi}{4}^\frac{\pi}{2}\ln \sin \sex{x+\frac{\pi}{4}}\rd x -\int_0^\frac{\pi}{4}\ln \cos x\rd x\\ &=\frac{\pi}{8}\ln 2 +\int_\frac{\pi}{4}^\frac{\pi}{2}\ln \sin t\rd t -\int_{-\frac{\pi}{4}}^0 \ln \cos s\rd s\quad\sex{x+\frac{\pi}{4}=t,\ x=-s}\\ &=\frac{\pi}{8}\ln 2\quad\sex{s+\frac{\pi}{2}=t}. \eea \eeex$$ 今天上课有个学生问的. 当时写出了第一步, 没写下去了...