`int_(0)^(pi//4) log (1+tan x) dx =?`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \), we can use a property of definite integrals. Let's go through the steps:

### Step 1: Define the integral
Let
\[ I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \]

### Step 2: Use the property of definite integrals
Using the property of definite integrals, we have:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
In our case, \( a = 0 \) and \( b = \frac{\pi}{4} \), so we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan\left(\frac{\pi}{4} - x\right)) \, dx \]

### Step 3: Simplify \( \tan\left(\frac{\pi}{4} - x\right) \)
Using the tangent subtraction formula:
\[ \tan\left(\frac{\pi}{4} - x\right) = \frac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\frac{\pi}{4} \tan x} = \frac{1 - \tan x}{1 + \tan x} \]
Thus, we can substitute this into the integral:
\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan x}{1 + \tan x}\right) \, dx \]

### Step 4: Simplify the logarithmic expression
Now simplify the argument of the logarithm:
\[ 1 + \frac{1 - \tan x}{1 + \tan x} = \frac{(1 + \tan x) + (1 - \tan x)}{1 + \tan x} = \frac{2}{1 + \tan x} \]
So we have:
\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan x}\right) \, dx \]

### Step 5: Split the logarithm
Using the property of logarithms, we can split the integral:
\[ I = \int_{0}^{\frac{\pi}{4}} \log(2) \, dx - \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \]
This gives us:
\[ I = \log(2) \cdot \frac{\pi}{4} - I \]

### Step 6: Solve for \( I \)
Now, we can solve for \( I \):
\[ 2I = \log(2) \cdot \frac{\pi}{4} \]
Thus,
\[ I = \frac{\log(2) \cdot \frac{\pi}{4}}{2} = \frac{\pi}{8} \log(2) \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx = \frac{\pi}{8} \log(2) \]