$\displaystyle\int_0^{\pi/2} \log(\tan x)\, dx = ?$
- $-(\pi/2)\ln 2$
- $0$ (by King's symmetry between $\sin$ and $\cos$)
- $\pi\ln 2$
- $\ln 2$
Show answer and solution
Correct answer: B. $0$ (by King's symmetry between $\sin$ and $\cos$)
$I = \int \log\tan x = \int(\log\sin x - \log\cos x)$.
Both pieces equal by King's (previous problem), so the difference is 0.