For integration of $\log(\tan x)$, we generally consider the definite integral over the interval from 0 to $\pi/2$

To calculate the definite integral of $\log(\tan x)$ from (0) to $\pi/2$, we use a technique involving symmetry and the properties of logarithms. The integral is:

\[

\int_{0}^{\pi/2} \log(\tan x) \, dx

\]

Let

$I=\int_{0}^{\pi/2} \log(\tan x) \, dx$ –(1)

Now we know

$\int_{0}^{a} f(x) dx=\int_{0}^{a} f((a-x)) dx$

Therefore

$I=\int_{0}^{\pi/2} \log(\tan (\pi/2 – x) \, dx=\int_{0}^{\pi/2} \log(\cot x) \, dx $

$I=\int_{0}^{\pi/2} \log \frac {1}{\tan x} \, dx $

$I= – \int_{0}^{\pi/2} \log(\tan x) \, dx$

$I=-I$

or $ I =0$

Therefore

\[

\int_{0}^{\pi/2} \log(\tan x) \, dx = 0

\]

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