`int_0^(pi//4)log(1+tanx)dx` is equal to

Options

A. `pi/8log_e2`
B. `pi/4log_2e`
C. `pi/4log_e2`
D. `pi/8log_e(1/2)`

Answer

To solve the integral \( I = \int_0^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \), we can use a property of definite integrals. Here’s a step-by-step solution:

### Step 1: Define the Integral
Let
\[ I = \int_0^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \]

### Step 2: Use the Property of Logarithms
Using the property of logarithms, we can express the integral as:
\[ I = \int_0^{\frac{\pi}{4}} \log(1 + \tan x) \, dx = \int_0^{\frac{\pi}{4}} \log\left(1 + \tan\left(\frac{\pi}{4} - x\right)\right) \, dx \]
Since \( \tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x} \), we can rewrite the integral.

### Step 3: Simplify the Integral
Thus, we have:
\[ I = \int_0^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan x}{1 + \tan x}\right) \, dx \]
This simplifies to:
\[ I = \int_0^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan x}\right) \, dx \]

### Step 4: Split the Integral
Now we can split the logarithm:
\[ I = \int_0^{\frac{\pi}{4}} \log 2 \, dx - \int_0^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \]
This gives us:
\[ I = \frac{\pi}{4} \log 2 - I \]

### Step 5: Solve for \( I \)
Now, we can solve for \( I \):
\[ 2I = \frac{\pi}{4} \log 2 \]
Thus,
\[ I = \frac{\pi}{8} \log 2 \]

### Final Answer
Therefore, the value of the integral is:
\[ \int_0^{\frac{\pi}{4}} \log(1 + \tan x) \, dx = \frac{\pi}{8} \log 2 \]