If `m ne n , m n in N` then the value of `int_(0)^(2pi) cos mx cos nx dx` is

Options

A. `0`
B. `2pi`
C. `pi`
D. dependent on `m and n`

Answer

To solve the integral \( I = \int_{0}^{2\pi} \cos(mx) \cos(nx) \, dx \) where \( m \neq n \) and \( m, n \in \mathbb{N} \), we will use a trigonometric identity to simplify the expression.

### Step-by-step Solution:

1. **Use the Product-to-Sum Formula**:
We can use the trigonometric identity:
\[ \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) \]
Applying this to our integral, we have:
\[ \cos(mx) \cos(nx) = \frac{1}{2} (\cos((m+n)x) + \cos((m-n)x)) \]
Thus, we can rewrite the integral as:
\[ I = \int_{0}^{2\pi} \cos(mx) \cos(nx) \, dx = \frac{1}{2} \int_{0}^{2\pi} (\cos((m+n)x) + \cos((m-n)x)) \, dx \]

2. **Separate the Integral**:
Now we can separate the integral into two parts:
\[ I = \frac{1}{2} \left( \int_{0}^{2\pi} \cos((m+n)x) \, dx + \int_{0}^{2\pi} \cos((m-n)x) \, dx \right) \]

3. **Evaluate Each Integral**:
The integral of cosine over a complete period (from \(0\) to \(2\pi\)) is zero:
\[ \int_{0}^{2\pi} \cos(kx) \, dx = 0 \quad \text{for any } k \neq 0 \]
Since \( m+n \) and \( m-n \) are both non-zero (because \( m \neq n \)), we have:
\[ \int_{0}^{2\pi} \cos((m+n)x) \, dx = 0 \]
and
\[ \int_{0}^{2\pi} \cos((m-n)x) \, dx = 0 \]

4. **Combine the Results**:
Therefore, we can conclude:
\[ I = \frac{1}{2} (0 + 0) = 0 \]

### Final Answer:
The value of the integral \( \int_{0}^{2\pi} \cos(mx) \cos(nx) \, dx \) is \( \boxed{0} \).