How to Integrate log1 tan x from 0 to π/4?

To solve the integral ∫₀^π/4 log(tan x) dx, we can use the property of symmetry in trigonometric functions.

First, we notice that:

• tan(π/4 – x) = cot(x)
• log(tan(π/4 – x)) = log(cot(x)) = -log(tan(x))

Now, we can express the integral from 0 to π/4 in terms of our original integral:

I = ∫0π/4 log(tan x) dx

Using the substitution u = π/4 – x, the limits will remain the same:

I = ∫0π/4 log(tan(π/4 - u)) (-du)
  = -∫0π/4 log(cot u) du
  = -∫0π/4 -log(tan u) du
  = ∫0π/4 log(tan u) du = I

This leads us to:

2I = ∫0π/4 log(tan x) dx - ∫0π/4log(tan x) dx = 0

Thus, we find that:

I = - I

This implies:

2I = 0 
  => I = 0

Therefore, the integral of log(tan x) from 0 to π/4 is:

∫0π/4 log(tan x) dx = 0