If m and n are integers , then what is the value of `int_(0)^(pi)" sin mxsinnx dx "`, if `m ne n` ?

Options

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

Explanation

To solve the integral \( I = \int_{0}^{\pi} \sin(mx) \sin(nx) \, dx \) for integers \( m \) and \( n \) where \( m \neq n \), we can use the product-to-sum identities and properties of definite integrals.

### Step-by-Step Solution:

**Step 1: Use the product-to-sum identity.**

We know that:
\[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
Applying this to our integral:
\[ I = \int_{0}^{\pi} \sin(mx) \sin(nx) \, dx = \frac{1}{2} \int_{0}^{\pi} [\cos((m-n)x) - \cos((m+n)x)] \, dx \]

**Step 2: Split the integral.**

Now we can split the integral:
\[ I = \frac{1}{2} \left( \int_{0}^{\pi} \cos((m-n)x) \, dx - \int_{0}^{\pi} \cos((m+n)x) \, dx \right) \]

**Step 3: Evaluate the first integral.**

For the first integral:
\[ \int_{0}^{\pi} \cos((m-n)x) \, dx \]
If \( m \neq n \), then \( m-n \) is non-zero. The integral of cosine over one full period is zero:
\[ \int_{0}^{\pi} \cos((m-n)x) \, dx = 0 \]

**Step 4: Evaluate the second integral.**

For the second integral:
\[ \int_{0}^{\pi} \cos((m+n)x) \, dx \]
Similarly, since \( m+n \) is also non-zero, we have:
\[ \int_{0}^{\pi} \cos((m+n)x) \, dx = 0 \]

**Step 5: Combine the results.**

Substituting back into our expression for \( I \):
\[ I = \frac{1}{2} (0 - 0) = 0 \]

### Final Result:
Thus, the value of the integral is:
\[ \int_{0}^{\pi} \sin(mx) \sin(nx) \, dx = 0 \quad \text{for } m \neq n \]