IF m and n are integers then what is the value of `int_0^pi sin mx sin nx dx` if `m ne n`

Options

A. 0
B. `1/(m+n)`
C. `1/(m-n)`
D. mn

Answer

To solve the integral \( I = \int_0^{\pi} \sin(mx) \sin(nx) \, dx \) where \( m \neq n \) and \( m, n \) are integers, we can use a trigonometric identity to simplify the expression.

### Step-by-Step Solution:

1. **Use the Product-to-Sum Identity**:
We can use the identity for the product of sines:
\[ 2 \sin A \sin B = \cos(A - B) - \cos(A + B) \]
In our case, let \( A = mx \) and \( B = nx \). Therefore,
\[ 2 \sin(mx) \sin(nx) = \cos((m-n)x) - \cos((m+n)x) \]

2. **Rewrite the Integral**:
Now, we can express the integral \( I \) as:
\[ I = \frac{1}{2} \int_0^{\pi} \left( \cos((m-n)x) - \cos((m+n)x) \right) \, dx \]

3. **Split the Integral**:
This can be split into two separate integrals:
\[ I = \frac{1}{2} \left( \int_0^{\pi} \cos((m-n)x) \, dx - \int_0^{\pi} \cos((m+n)x) \, dx \right) \]

4. **Evaluate Each Integral**:
The integral of \( \cos(kx) \) over \( [0, \pi] \) is given by:
\[ \int_0^{\pi} \cos(kx) \, dx = \left[ \frac{\sin(kx)}{k} \right]_0^{\pi} = \frac{\sin(k\pi)}{k} - \frac{\sin(0)}{k} = 0 \]
This holds true for any integer \( k \neq 0 \).

5. **Apply the Result**:
Since \( m \neq n \), both \( m-n \) and \( m+n \) are non-zero integers. Therefore:
\[ \int_0^{\pi} \cos((m-n)x) \, dx = 0 \quad \text{and} \quad \int_0^{\pi} \cos((m+n)x) \, dx = 0 \]

6. **Combine the Results**:
Thus, we have:
\[ I = \frac{1}{2} (0 - 0) = 0 \]

### Final Answer:
The value of the integral \( \int_0^{\pi} \sin(mx) \sin(nx) \, dx \) when \( m \neq n \) is:
\[ \boxed{0} \]