The value of `int_(-pi//2)^(pi//2)(x^(3) + x cos x + tan^(5)x + 1)dx` is

Options

A. 0
B. 2
C. `pi`
D. 1

Answer

To solve the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx \), we will analyze the integrand to determine whether it consists of even or odd functions.

### Step-by-Step Solution:

1. **Identify the Function**:
The integrand is \( f(x) = x^3 + x \cos x + \tan^5 x + 1 \).

2. **Check Each Term for Evenness or Oddness**:
- **Term 1: \( x^3 \)**
\( f(-x) = (-x)^3 = -x^3 \)
This is an odd function.

- **Term 2: \( x \cos x \)**
\( f(-x) = (-x) \cos(-x) = -x \cos x \)
This is also an odd function.

- **Term 3: \( \tan^5 x \)**
\( f(-x) = \tan^5(-x) = (-\tan x)^5 = -\tan^5 x \)
This is again an odd function.

- **Term 4: \( 1 \)**
\( f(-x) = 1 \)
This is an even function.

3. **Combine the Results**:
The overall function can be expressed as:
\[ f(-x) = -x^3 - x \cos x - \tan^5 x + 1 \]
Thus, we can separate the odd and even parts:
\[ f(x) = (x^3 + x \cos x + \tan^5 x) + 1 \]
The first three terms are odd, and the last term is even.

4. **Evaluate the Integral**:
The integral of an odd function over a symmetric interval about zero is zero. Therefore:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^3 + x \cos x + \tan^5 x) \, dx = 0 \]
The only term left is the integral of the constant \( 1 \):
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = 1 \cdot \left( \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \right) = \pi \]

5. **Final Result**:
Therefore, the value of the integral is:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx = \pi \]