The value of `int_(-2pi)^(5pi) cot^(-1)(tan x) dx` is equal to

Options

A. `((7pi^(2))/(2))`
B. `-(7pi^(2))`
C. `(3pi)/(2)`
D. `pi^(2)`

Answer

To solve the integral \( I = \int_{-2\pi}^{5\pi} \cot^{-1}(\tan x) \, dx \), we can use the properties of the cotangent and tangent functions.

### Step 1: Simplify the Integral
We know that:
\[ \cot^{-1}(\tan x) = \frac{\pi}{2} - x \quad \text{for } x \in \left(0, \frac{\pi}{2}\right) \]
This property can be extended to the entire range of \( x \) by considering the periodic nature of the tangent function.

### Step 2: Rewrite the Integral
Using the identity, we can rewrite the integral as:
\[ I = \int_{-2\pi}^{5\pi} \left( \frac{\pi}{2} - x \right) \, dx \]

### Step 3: Split the Integral
We can split the integral into two parts:
\[ I = \int_{-2\pi}^{5\pi} \frac{\pi}{2} \, dx - \int_{-2\pi}^{5\pi} x \, dx \]

### Step 4: Calculate Each Integral
1. **Calculate the first integral:**
\[ \int_{-2\pi}^{5\pi} \frac{\pi}{2} \, dx = \frac{\pi}{2} \cdot (5\pi - (-2\pi)) = \frac{\pi}{2} \cdot (5\pi + 2\pi) = \frac{\pi}{2} \cdot 7\pi = \frac{7\pi^2}{2} \]

2. **Calculate the second integral:**
\[ \int_{-2\pi}^{5\pi} x \, dx = \left[ \frac{x^2}{2} \right]_{-2\pi}^{5\pi} = \frac{(5\pi)^2}{2} - \frac{(-2\pi)^2}{2} = \frac{25\pi^2}{2} - \frac{4\pi^2}{2} = \frac{25\pi^2 - 4\pi^2}{2} = \frac{21\pi^2}{2} \]

### Step 5: Combine the Results
Now, substituting back into our expression for \( I \):
\[ I = \frac{7\pi^2}{2} - \frac{21\pi^2}{2} = \frac{7\pi^2 - 21\pi^2}{2} = \frac{-14\pi^2}{2} = -7\pi^2 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{-7\pi^2} \]