[ The value of `int_(-pi)^( pi)sum_(r=0)^(999)cos rx(1+sum_(r=1)^(999)sin rx)`dx, is [ (1) 2 pi, (2) 999 pi, (3) 0]]

Options

A. `2pi`
B. `999pi`
C. `0`
D. `pi`

Explanation

To solve the integral

\[ I = \int_{-\pi}^{\pi} \sum_{r=0}^{999} \cos(rx) \left(1 + \sum_{r=1}^{999} \sin(rx)\right) dx, \]

we will break it down step by step.

### Step 1: Rewrite the Integral
We can express the integral as:

\[ I = \int_{-\pi}^{\pi} \sum_{r=0}^{999} \cos(rx) \, dx + \int_{-\pi}^{\pi} \sum_{r=1}^{999} \cos(rx) \sin(rx) \, dx. \]

### Step 2: Evaluate the First Integral
The first integral is:

\[ \int_{-\pi}^{\pi} \sum_{r=0}^{999} \cos(rx) \, dx. \]

Using the property of definite integrals, we know that:

\[ \int_{-\pi}^{\pi} \cos(rx) \, dx = 0 \quad \text{for } r \neq 0, \]
and for \( r = 0 \):

\[ \int_{-\pi}^{\pi} \cos(0 \cdot x) \, dx = \int_{-\pi}^{\pi} 1 \, dx = 2\pi. \]

Thus, the first integral evaluates to:

\[ \int_{-\pi}^{\pi} \sum_{r=0}^{999} \cos(rx) \, dx = 2\pi. \]

### Step 3: Evaluate the Second Integral
Now consider the second integral:

\[ \int_{-\pi}^{\pi} \sum_{r=1}^{999} \cos(rx) \sin(rx) \, dx. \]

Using the identity \( \cos(rx) \sin(rx) = \frac{1}{2} \sin(2rx) \), we can rewrite the integral as:

\[ \int_{-\pi}^{\pi} \sum_{r=1}^{999} \frac{1}{2} \sin(2rx) \, dx = \frac{1}{2} \sum_{r=1}^{999} \int_{-\pi}^{\pi} \sin(2rx) \, dx. \]

Since \( \int_{-\pi}^{\pi} \sin(2rx) \, dx = 0 \) for all \( r \), the second integral evaluates to 0:

\[ \int_{-\pi}^{\pi} \sum_{r=1}^{999} \cos(rx) \sin(rx) \, dx = 0. \]

### Step 4: Combine Results
Now we combine the results from the two integrals:

\[ I = 2\pi + 0 = 2\pi. \]

### Final Answer
Thus, the value of the integral is:

\[ \boxed{2\pi}. \]