Value of $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \log(\tan x)dx$ is

Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \log(\tan x)dx$

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Using the King Property of definite integrals:

$\int_a^b f(x)dx = \int_a^b f(a+b-x)dx$

Here, $a = \frac{\pi}{6}$ and $b = \frac{\pi}{3}$, so:

$a + b = \frac{\pi}{6} + \frac{\pi}{3}$

$= \frac{\pi}{6} + \frac{2\pi}{6}$

$= \frac{\pi}{2}$

Therefore:

$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \log(\tan(\frac{\pi}{2} - x))dx$

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Using the complementary angle formula $\tan(\frac{\pi}{2} - x) = \cot(x) = \frac{1}{\tan x}$:

$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \log\left(\frac{1}{\tan x}\right)dx$

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Using the logarithm property $\log\left(\frac{1}{a}\right) = -\log(a)$:

$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} -\log(\tan x)dx$

$I = -\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \log(\tan x)dx$

$I = -I$

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$I = -I$

$I + I = 0$

$2I = 0$

$I = 0$

Therefore, the value of the integral is $0$.