`int_(-pi)^(pi) (x^(3) + x cos x+ tan^(5) x +2)`=

Options

A. `4pi`
B. `2pi`
C. `pi`
D. none

Answer

To solve the integral

\[ I = \int_{-\pi}^{\pi} \left( x^3 + x \cos x + \tan^5 x + 2 \right) \, dx, \]

we can break it down into two parts:

\[ I = \int_{-\pi}^{\pi} \left( x^3 + x \cos x + \tan^5 x \right) \, dx + \int_{-\pi}^{\pi} 2 \, dx. \]

### Step 1: Evaluate the integral of the constant term

The integral of the constant term \(2\) is straightforward:

\[ \int_{-\pi}^{\pi} 2 \, dx = 2 \cdot \left[ x \right]_{-\pi}^{\pi} = 2 \cdot (\pi - (-\pi)) = 2 \cdot 2\pi = 4\pi. \]

### Step 2: Evaluate the first integral

Now we need to evaluate

\[ \int_{-\pi}^{\pi} \left( x^3 + x \cos x + \tan^5 x \right) \, dx. \]

We will check if the integrand is an odd function. A function \(f(x)\) is odd if \(f(-x) = -f(x)\).

1. **For \(x^3\)**:
\[ f(-x) = (-x)^3 = -x^3 \quad \text{(odd)} \]

2. **For \(x \cos x\)**:
\[ f(-x) = -x \cos(-x) = -x \cos x \quad \text{(odd)} \]

3. **For \(\tan^5 x\)**:
\[ f(-x) = \tan^5(-x) = (-\tan x)^5 = -\tan^5 x \quad \text{(odd)} \]

Since all three terms \(x^3\), \(x \cos x\), and \(\tan^5 x\) are odd functions, their sum is also an odd function:

\[ f(x) = x^3 + x \cos x + \tan^5 x \quad \text{is odd.} \]

### Step 3: Use the property of definite integrals

By the property of definite integrals, the integral of an odd function over a symmetric interval around zero is zero:

\[ \int_{-\pi}^{\pi} (x^3 + x \cos x + \tan^5 x) \, dx = 0. \]

### Step 4: Combine results

Now we can combine the results:

\[ I = 0 + 4\pi = 4\pi. \]

### Final Answer

Thus, the value of the integral is

\[ \boxed{4\pi}. \]