Evaluate `int_(0)^(pi//4) log (1+tan theta)d theta`

Answer

To evaluate the integral \( I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan \theta) \, d\theta \), we will use a symmetry property of definite integrals.

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

### Step 2: Use the property of integrals
Using the property of integrals, we can express \( I \) as:
\[ I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan\left(\frac{\pi}{4} - \theta\right)) \, d\theta \]

### Step 3: Simplify \( \tan\left(\frac{\pi}{4} - \theta\right) \)
Using the tangent subtraction formula:
\[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta} \]
Thus,
\[ 1 + \tan\left(\frac{\pi}{4} - \theta\right) = 1 + \frac{1 - \tan \theta}{1 + \tan \theta} = \frac{(1 + \tan \theta) + (1 - \tan \theta)}{1 + \tan \theta} = \frac{2}{1 + \tan \theta} \]

### Step 4: Substitute back into the integral
Now substituting this back into the integral:
\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan \theta}\right) \, d\theta \]

### Step 5: Break down the logarithm
Using the property of logarithms:
\[ \log\left(\frac{2}{1 + \tan \theta}\right) = \log(2) - \log(1 + \tan \theta) \]
Thus, we can write:
\[ I = \int_{0}^{\frac{\pi}{4}} \log(2) \, d\theta - \int_{0}^{\frac{\pi}{4}} \log(1 + \tan \theta) \, d\theta \]
This simplifies to:
\[ I = \log(2) \cdot \frac{\pi}{4} - I \]

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

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