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 will use a property of definite integrals that allows us to transform the integral.

### Step 1: Use the property of definite integrals
We know that:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, we can set \( a = \frac{\pi}{3} \). Thus, we can write:
\[ I = \int_{0}^{\frac{\pi}{3}} \log(1 + \sqrt{3} \tan(\frac{\pi}{3} - x)) \, dx \]

### Step 2: Simplify \( \tan(\frac{\pi}{3} - x) \)
Using the tangent subtraction formula:
\[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \]
where \( a = \frac{\pi}{3} \) and \( b = x \), we have:
\[ \tan\left(\frac{\pi}{3} - x\right) = \frac{\tan\frac{\pi}{3} - \tan x}{1 + \tan\frac{\pi}{3} \tan x} = \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} \]

### Step 3: Substitute back into the integral
Now substituting this back into our 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 4: Simplify the logarithm
We simplify the expression inside the logarithm:
\[ 1 + \sqrt{3} \cdot \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} = \frac{(1 + \sqrt{3} \tan x) + \sqrt{3}(\sqrt{3} - \tan x)}{1 + \sqrt{3} \tan x} = \frac{1 + 3}{1 + \sqrt{3} \tan x} = \frac{4}{1 + \sqrt{3} \tan x} \]

### Step 5: Rewrite the integral
Now we can express \( I \) as:
\[ I = \int_{0}^{\frac{\pi}{3}} \log\left(\frac{4}{1 + \sqrt{3} \tan x}\right) \, dx \]
This can be split using properties of logarithms:
\[ I = \int_{0}^{\frac{\pi}{3}} \log(4) \, dx - \int_{0}^{\frac{\pi}{3}} \log(1 + \sqrt{3} \tan x) \, dx \]

### Step 6: Combine the integrals
Let \( J = \int_{0}^{\frac{\pi}{3}} \log(1 + \sqrt{3} \tan x) \, dx \). Then we have:
\[ I = \frac{\pi}{3} \log(4) - I \]
Thus:
\[ 2I = \frac{\pi}{3} \log(4) \]
\[ I = \frac{\pi}{6} \log(4) \]

### Step 7: Simplify the result
Since \( \log(4) = \log(2^2) = 2 \log(2) \):
\[ I = \frac{\pi}{6} \cdot 2 \log(2) = \frac{\pi}{3} \log(2) \]

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