The value of the definite integral `int_(0)^(pi//3) ln (1+ sqrt3tan x )dx` equals

Options

A. `pi/3 ln 2`
B. `pi/3`
C. `(pi^(2))/(6) ln 2`
D. `pi/2 ln 2`

Answer

To solve the definite integral \( I = \int_{0}^{\frac{\pi}{3}} \ln(1 + \sqrt{3} \tan x) \, dx \), we will use the property of definite integrals and some logarithmic identities. Let's go through the steps in detail.

### Step-by-Step Solution:

1. **Define the Integral**:
\[ I = \int_{0}^{\frac{\pi}{3}} \ln(1 + \sqrt{3} \tan x) \, dx \]

2. **Use the Property of Definite Integrals**:
We can use the property that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
Here, we set \( a = \frac{\pi}{3} \):
\[ I = \int_{0}^{\frac{\pi}{3}} \ln(1 + \sqrt{3} \tan(\frac{\pi}{3} - x)) \, dx \]

3. **Simplify \( \tan(\frac{\pi}{3} - x) \)**:
Using the tangent subtraction formula:
\[ \tan(\frac{\pi}{3} - x) = \frac{\tan(\frac{\pi}{3}) - \tan(x)}{1 + \tan(\frac{\pi}{3}) \tan(x)} = \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} \]

4. **Substitute Back into the Integral**:
Substitute this back into the integral:
\[ I = \int_{0}^{\frac{\pi}{3}} \ln\left(1 + \sqrt{3} \cdot \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x}\right) \, dx \]

5. **Simplify the Logarithm**:
Simplifying the argument of the logarithm:
\[ 1 + \sqrt{3} \cdot \frac{\sqrt{3} - \tan x}{1 + \sqrt{3} \tan x} = \frac{(1 + \sqrt{3} \tan x) + \sqrt{3}(\sqrt{3} - \tan x)}{1 + \sqrt{3} \tan x} = \frac{4}{1 + \sqrt{3} \tan x} \]
Thus:
\[ I = \int_{0}^{\frac{\pi}{3}} \ln\left(\frac{4}{1 + \sqrt{3} \tan x}\right) \, dx \]

6. **Split the Logarithm**:
Using the property of logarithms:
\[ I = \int_{0}^{\frac{\pi}{3}} \ln(4) \, dx - \int_{0}^{\frac{\pi}{3}} \ln(1 + \sqrt{3} \tan x) \, dx \]
The first integral evaluates to:
\[ \ln(4) \cdot \frac{\pi}{3} \]

7. **Combine the Integrals**:
So we have:
\[ I = \frac{\pi}{3} \ln(4) - I \]
Adding \( I \) to both sides gives:
\[ 2I = \frac{\pi}{3} \ln(4) \]
Thus:
\[ I = \frac{\pi}{6} \ln(4) \]

8. **Simplify the Result**:
Since \( \ln(4) = 2 \ln(2) \):
\[ I = \frac{\pi}{6} \cdot 2 \ln(2) = \frac{\pi}{3} \ln(2) \]

### Final Answer:
\[ I = \frac{\pi}{3} \ln(2) \]