Value of \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\displaystyle\log(\tan x)\,dx\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0
Let I be the integral. Here \(a + b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}\). By the property \(\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx\):
\( \displaystyle I = \int_{\pi/6}^{\pi/3}\log\left(\tan\left(\frac{\pi}{2} - x\right)\right)dx = \int_{\pi/6}^{\pi/3}\log(\cot x)\,dx \)
Since \(\log(\cot x) = \log\left(\frac{1}{\tan x}\right) = -\log(\tan x)\):
\(I = -I\)
\(2I = 0,\quad I = 0\)