`int_(0)^(pi//3) [sqrt3 tan x] dx`, where [.] denotes the greatest integer function is

Options

A. `(pi)/(2)- "tan"^(-2) (2)/(sqrt3)`
B. `(5pi)/(6) - "tan"^(-1) (2)/(sqrt3)`
C. `(5pi)/(6)`
D. none

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{3}} [\sqrt{3} \tan x] \, dx \), where \([.]\) denotes the greatest integer function, we will follow these steps:

### Step 1: Determine the range of \( \tan x \) from \( 0 \) to \( \frac{\pi}{3} \)
The function \( \tan x \) is continuous and increasing in the interval \( \left[0, \frac{\pi}{3}\right] \).
- At \( x = 0 \), \( \tan 0 = 0 \).
- At \( x = \frac{\pi}{3} \), \( \tan \frac{\pi}{3} = \sqrt{3} \).

Thus, \( \tan x \) varies from \( 0 \) to \( \sqrt{3} \) as \( x \) goes from \( 0 \) to \( \frac{\pi}{3} \).

### Step 2: Analyze \( \sqrt{3} \tan x \)
Now, we analyze the expression \( \sqrt{3} \tan x \):
- At \( x = 0 \), \( \sqrt{3} \tan 0 = 0 \).
- At \( x = \frac{\pi}{3} \), \( \sqrt{3} \tan \frac{\pi}{3} = 3 \).

Thus, \( \sqrt{3} \tan x \) varies from \( 0 \) to \( 3 \).

### Step 3: Determine the intervals for the greatest integer function
We need to find the intervals where \( [\sqrt{3} \tan x] \) takes different integer values:
- For \( 0 \leq \sqrt{3} \tan x < 1 \): This occurs when \( 0 \leq \tan x < \frac{1}{\sqrt{3}} \), which corresponds to \( 0 \leq x < \frac{\pi}{6} \).
- For \( 1 \leq \sqrt{3} \tan x < 2 \): This occurs when \( \frac{1}{\sqrt{3}} \leq \tan x < \frac{2}{\sqrt{3}} \), which corresponds to \( \frac{\pi}{6} \leq x < \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \).
- For \( 2 \leq \sqrt{3} \tan x < 3 \): This occurs when \( \frac{2}{\sqrt{3}} \leq \tan x < \sqrt{3} \), which corresponds to \( \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \leq x < \frac{\pi}{3} \).

### Step 4: Set up the integral based on intervals
Now we can set up the integral:
\[ \int_{0}^{\frac{\pi}{3}} [\sqrt{3} \tan x] \, dx = \int_{0}^{\frac{\pi}{6}} 0 \, dx + \int_{\frac{\pi}{6}}^{\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)} 1 \, dx + \int_{\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)}^{\frac{\pi}{3}} 2 \, dx \]

### Step 5: Evaluate the integrals
1. **First Integral**:
\[ \int_{0}^{\frac{\pi}{6}} 0 \, dx = 0 \]

2. **Second Integral**:
\[ \int_{\frac{\pi}{6}}^{\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)} 1 \, dx = \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) - \frac{\pi}{6} \]

3. **Third Integral**:
\[ \int_{\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)}^{\frac{\pi}{3}} 2 \, dx = 2 \left( \frac{\pi}{3} - \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \right) \]

### Step 6: Combine the results
Combining all the results:
\[ \int_{0}^{\frac{\pi}{3}} [\sqrt{3} \tan x] \, dx = 0 + \left( \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) - \frac{\pi}{6} \right) + 2 \left( \frac{\pi}{3} - \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \right) \]
\[ = \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) - \frac{\pi}{6} + \frac{2\pi}{3} - 2\tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \]
\[ = -\tan^{-1}\left(\frac{2}{\sqrt{3}}\right) + \frac{2\pi}{3} - \frac{\pi}{6} \]
\[ = -\tan^{-1}\left(\frac{2}{\sqrt{3}}\right) + \frac{4\pi}{6} - \frac{\pi}{6} \]
\[ = -\tan^{-1}\left(\frac{2}{\sqrt{3}}\right) + \frac{3\pi}{6} \]
\[ = \frac{\pi}{2} - \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{3}} [\sqrt{3} \tan x] \, dx = \frac{\pi}{2} - \tan^{-1}\left(\frac{2}{\sqrt{3}}\right) \]