The area bounded by the curve `f(x)=||tanx+cotx|-|tanx-cotx||` between the lines `x=0,x=pi/2` and the X-axis is

Options

A. log 4
B. log`sqrt(2)`
C. `2log2`
D. `sqrt(2)log 2`

Answer

To find the area bounded by the curve \( f(x) = ||\tan x + \cot x| - |\tan x - \cot x|| \) between the lines \( x = 0 \) and \( x = \frac{\pi}{2} \) and the X-axis, we will follow these steps:

### Step 1: Analyze the function

The function \( f(x) = ||\tan x + \cot x| - |\tan x - \cot x|| \) can be simplified based on the intervals of \( x \).

### Step 2: Identify intervals

1. **Interval 1:** \( x \in [0, \frac{\pi}{4}] \)
2. **Interval 2:** \( x \in [\frac{\pi}{4}, \frac{\pi}{2}] \)

### Step 3: Evaluate \( f(x) \) in each interval

#### For \( x \in [0, \frac{\pi}{4}] \):
- Here, \( \tan x \leq \cot x \).
- Thus, \( f(x) = |(\tan x + \cot x)| - |(\tan x - \cot x)| \).
- Since \( \tan x + \cot x \) is positive and \( \tan x - \cot x \) is negative, we have:
\[ f(x) = (\tan x + \cot x) - (-(\tan x - \cot x)) = \tan x + \cot x + \tan x - \cot x = 2\tan x \]

#### For \( x \in [\frac{\pi}{4}, \frac{\pi}{2}] \):
- Here, \( \tan x \geq \cot x \).
- Thus, \( f(x) = |(\tan x + \cot x)| - |(\tan x - \cot x)| \).
- Since both \( \tan x + \cot x \) and \( \tan x - \cot x \) are positive, we have:
\[ f(x) = (\tan x + \cot x) - (\tan x - \cot x) = 2\cot x \]

### Step 4: Set up the area integrals

The area \( A \) is given by the sum of the integrals over the two intervals:
\[ A = \int_0^{\frac{\pi}{4}} 2\tan x \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} 2\cot x \, dx \]

### Step 5: Evaluate the integrals

1. **Integral from \( 0 \) to \( \frac{\pi}{4} \)**:
\[ \int 2\tan x \, dx = -2 \ln |\cos x| + C \]
Evaluating from \( 0 \) to \( \frac{\pi}{4} \):
\[ \left[-2 \ln |\cos x|\right]_0^{\frac{\pi}{4}} = -2 \ln \left(\frac{1}{\sqrt{2}}\right) - (-2 \ln(1)) = -2 \ln \left(\frac{1}{\sqrt{2}}\right) = 2 \ln(\sqrt{2}) = \ln(2) \]

2. **Integral from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \)**:
\[ \int 2\cot x \, dx = 2 \ln |\sin x| + C \]
Evaluating from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \):
\[ \left[2 \ln |\sin x|\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = 2 \ln(1) - 2 \ln\left(\frac{1}{\sqrt{2}}\right) = 0 - (-\ln(2)) = \ln(2) \]

### Step 6: Combine the results

Adding both areas:
\[ A = \ln(2) + \ln(2) = 2\ln(2) \]

### Final Result

Thus, the area bounded by the curve \( f(x) \) between the lines \( x = 0 \) and \( x = \frac{\pi}{2} \) and the X-axis is:
\[ \boxed{2\ln(2)} \]