The value of `int_(0)^((pi)/(3))log(1+sqrt3 tanx)dx` is equal to

Options

A. `pilog2`
B. `(pi)/(2)log2`
C. `(pi)/(3)log2`
D. `(pi)/(4)log2`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{3}} \log(1 + \sqrt{3} \tan x) \, dx \), we can use a property of definite integrals.

### Step 1: Use the property of definite integrals
We can use the property:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
In our case, \( a = 0 \) and \( b = \frac{\pi}{3} \). Thus, we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{3}} \log(1 + \sqrt{3} \tan\left(\frac{\pi}{3} - x\right)) \, dx \]

### Step 2: Simplify \( \tan\left(\frac{\pi}{3} - x\right) \)
Using the tangent subtraction formula:
\[ \tan\left(\frac{\pi}{3} - x\right) = \frac{\tan\frac{\pi}{3} - \tan x}{1 + \tan\frac{\pi}{3} \tan x} = \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} \]
Now substitute this back into the integral:
\[ I = \int_{0}^{\frac{\pi}{3}} \log\left(1 + \sqrt{3} \cdot \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x}\right) \, dx \]

### Step 3: Simplify the logarithm
Now simplify the argument of the logarithm:
\[ 1 + \sqrt{3} \cdot \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} = \frac{(1 + \sqrt{3} \tan x) + \sqrt{3}(\sqrt{3} - \tan x)}{1 + \sqrt{3} \tan x} = \frac{1 + 3}{1 + \sqrt{3} \tan x} = \frac{4}{1 + \sqrt{3} \tan x} \]
Thus, we have:
\[ I = \int_{0}^{\frac{\pi}{3}} \log\left(\frac{4}{1 + \sqrt{3} \tan x}\right) \, dx \]

### Step 4: Split the logarithm
Using the property of logarithms:
\[ \log\left(\frac{a}{b}\right) = \log a - \log b \]
We can split the integral:
\[ I = \int_{0}^{\frac{\pi}{3}} \log(4) \, dx - \int_{0}^{\frac{\pi}{3}} \log(1 + \sqrt{3} \tan x) \, dx \]
The first integral evaluates to:
\[ \int_{0}^{\frac{\pi}{3}} \log(4) \, dx = \log(4) \cdot \frac{\pi}{3} \]
Thus:
\[ I = \frac{\pi}{3} \log(4) - I \]

### Step 5: Solve for \( I \)
Combining the terms gives:
\[ 2I = \frac{\pi}{3} \log(4) \]
Thus:
\[ I = \frac{\pi}{6} \log(4) \]

### Step 6: Simplify \( \log(4) \)
Since \( \log(4) = 2 \log(2) \), we have:
\[ I = \frac{\pi}{6} \cdot 2 \log(2) = \frac{\pi}{3} \log(2) \]

### Final Result
The value of the integral is:
\[ \boxed{\frac{\pi}{3} \log(2)} \]