If `I=int_(0)^(pi//4) log(1+tan x)dx`, then I=

Options

A. `(pi)/(8)log_(e )2`
B. `(pi)/(4)log_(e )2`
C. `-(pi)/(8)log_(e )2`
D. `-(pi)/(4) log_(e ) 2`

Explanation

To solve the integral \( I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \), we can use the property of definite integrals. Here are the steps to find the value of \( I \):

### Step 1: Use the property of definite integrals
We know that for any function \( f(x) \), the following property holds:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, let \( a = \frac{\pi}{4} \). Thus, we can write:
\[ I = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan x) \, dx = \int_{0}^{\frac{\pi}{4}} \log(1 + \tan\left(\frac{\pi}{4} - x\right)) \, dx \]

### Step 2: 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 rewrite \( I \) as:
\[ I = \int_{0}^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan x}{1 + \tan x}\right) \, dx \]

### Step 3: Simplify the logarithm
Now simplify the expression inside 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 4: 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:
\[ I = \frac{\pi}{4} \log(2) - I \]

### Step 5: Solve for \( I \)
Now, add \( I \) to both sides:
\[ 2I = \frac{\pi}{4} \log(2) \]
Thus, we find:
\[ I = \frac{\pi}{8} \log(2) \]

### Final Answer:
\[ I = \frac{\pi}{8} \log(2) \]