Evaluate: ∫₀⁽ π/4⁾ log(1 + tan x) dx?

To evaluate the integral ∫₀^π/4 log(1 + tan x) dx, we can use a clever substitution and properties of definite integrals.

Step 1: Substitution

Let’s use the substitution u = π/4 - x. Then, the differential changes as follows:

• When x = 0, u = π/4.
• When x = π/4, u = 0.
• dx = -du.

Step 2: Transform the Integral

Now, we can rewrite the integral:

∫₀^π/4 log(1 + tan x) dx = ∫_π/4⁰ log(1 + tan(π/4 - u)) (-du).

This simplifies to:

∫₀^π/4 log(1 + tan(π/4 - u)) du.

Step 3: Simplifying tan(π/4 - u)

Using the identity for tangent, we have:

tan(π/4 - u) = (1 - tan u) / (1 + tan u).

Thus, we can express the integral as:

∫₀^π/4 log(1 + (1 - tan u)/(1 + tan u)) du.

Step 4: Combining Integrals

Now, we can combine the two integrals:

Let I = ∫₀^π/4 log(1 + tan x) dx.

Then, we have:

I = ∫₀^π/4 log(1 + tan(π/4 - x)) dx.

Adding these two expressions gives:

2I = ∫₀^π/4 (log(1 + tan x) + log(1 + tan(π/4 - x))) dx.

Step 5: Final Evaluation

Using the property of logarithms, we can combine the logs:

log(1 + tan x) + log(1 + tan(π/4 - x)) = log((1 + tan x)(1 + tan(π/4 - x))).

Evaluating this product leads to a constant value, which can be integrated over the interval. After some calculations, we find:

I = (π/4) log(2).

Result

The final answer is:

∫₀^π/4 log(1 + tan x) dx = (π/4) log(2).