The value of
`I=int_(-2)^(0){x^(3)+3x^(2)+3x+3+(x+1)cos(x+1)cos(x+1)}dx`, is

Options

A. 0
B. 3
C. 4
D. 1

Explanation

To solve the integral

\[ I = \int_{-2}^{0} \left( x^3 + 3x^2 + 3x + 3 + (x+1) \cos(x+1) \cos(x+1) \right) dx, \]

we will proceed step by step.

### Step 1: Rewrite the Integral
First, we can rewrite the integral in a more manageable form:

\[ I = \int_{-2}^{0} \left( x^3 + 3x^2 + 3x + 3 + (x+1) \cos^2(x+1) \right) dx. \]

### Step 2: Change of Variable
Next, we will perform a substitution. Let \( t = x + 1 \). Then, \( x = t - 1 \) and \( dx = dt \). The limits change as follows:
- When \( x = -2 \), \( t = -1 \).
- When \( x = 0 \), \( t = 1 \).

Thus, the integral becomes:

\[ I = \int_{-1}^{1} \left( (t-1)^3 + 3(t-1)^2 + 3(t-1) + 3 + t \cos^2(t) \right) dt. \]

### Step 3: Expand the Polynomial
Now, we will expand the polynomial:

1. \( (t-1)^3 = t^3 - 3t^2 + 3t - 1 \)
2. \( 3(t-1)^2 = 3(t^2 - 2t + 1) = 3t^2 - 6t + 3 \)
3. \( 3(t-1) = 3t - 3 \)

Combining these, we have:

\[ (t-1)^3 + 3(t-1)^2 + 3(t-1) + 3 = (t^3 - 3t^2 + 3t - 1) + (3t^2 - 6t + 3) + (3t - 3) + 3. \]

Simplifying this gives:

\[ t^3 + (0)t^2 + (0)t + 2 = t^3 + 2. \]

### Step 4: Substitute Back into the Integral
Now, substituting back into the integral, we have:

\[ I = \int_{-1}^{1} \left( t^3 + 2 + t \cos^2(t) \right) dt. \]

### Step 5: Evaluate the Integral
We can split the integral into three parts:

\[ I = \int_{-1}^{1} t^3 dt + \int_{-1}^{1} 2 dt + \int_{-1}^{1} t \cos^2(t) dt. \]

1. The first integral \( \int_{-1}^{1} t^3 dt = 0 \) (since \( t^3 \) is an odd function).
2. The second integral \( \int_{-1}^{1} 2 dt = 2 \times (1 - (-1)) = 4 \).
3. The third integral \( \int_{-1}^{1} t \cos^2(t) dt = 0 \) (since \( t \cos^2(t) \) is also an odd function).

### Step 6: Combine Results
Combining these results, we find:

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

### Final Answer
Thus, the value of the integral \( I \) is

\[ \boxed{4}. \]