Integration of log(tanx) from 0 to π/4

Overview:
We need to find the value of the definite integral ∫(0 to π/4) log(tanx) dx.

Step 1: Simplifying the Integral:
To solve this integral, we can use the property of logarithms. The logarithm of a product is equal to the sum of the logarithms of the individual factors. Therefore, we can rewrite log(tanx) as log(sin x / cos x).

Step 2: Applying the Logarithmic Property:
Using the property mentioned above, we can rewrite the integral as ∫(0 to π/4) [log(sin x) - log(cos x)] dx.

Step 3: Breaking the Integral into Two Parts:
Now, we can split the integral into two separate integrals:
∫(0 to π/4) log(sin x) dx - ∫(0 to π/4) log(cos x) dx.

Step 4: Evaluating the Integrals:
To evaluate the first integral, we can use the substitution u = sin x, which gives us du = cos x dx. The limits of integration also change accordingly: when x = 0, u = sin 0 = 0, and when x = π/4, u = sin(π/4) = 1/√2. Therefore, the first integral becomes:
∫(0 to 1/√2) log(u) du.

Using the integration by parts method, we integrate log(u) and differentiate u:
∫ log(u) du = u log(u) - ∫(1/u) du.

Evaluating the integral, we have:
[1/√2 log(1/√2)] - [0 - ∫(0 to 1/√2) (1/u) du]
= [1/√2 log(1/√2)] - ∫(0 to 1/√2) (1/u) du.

Next, we evaluate the second integral ∫(0 to π/4) log(cos x) dx. Similarly, we use the substitution v = cos x, which gives us dv = -sin x dx. The limits of integration change to v = cos(π/4) = 1/√2 and v = cos 0 = 1. The second integral becomes:
- ∫(1/√2 to 1) log(v) dv.

Using the same integration by parts method, we integrate log(v) and differentiate v:
- ∫ log(v) dv = - v log(v) + ∫(1/v) dv.

Evaluating the integral, we have:
- [1/√2 log(1/√2)] + [1 log(1) - 1/√2 log(1/√2)]
= - [1/√2 log(1/√2)] + [0 - 1/√2 log(1/√2)]
= - [1/√2 log(1/√2)] + 1/√2 log(1/√2).

Step 5: Simplifying the Result