If m and n are integers , then `int_0^pi sin mx sin nx dx = 0` , if `m != n` and `pi` if `m = n` .

Answer

To solve the integral \( I = \int_0^{\pi} \sin(mx) \sin(nx) \, dx \), we will analyze two cases: when \( m \neq n \) and when \( m = n \).

### Step 1: Use the Product-to-Sum Formula
We can use the product-to-sum identities for sine functions:
\[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
Applying this to our integral, we have:
\[ \sin(mx) \sin(nx) = \frac{1}{2} [\cos((m-n)x) - \cos((m+n)x)] \]
Thus, the integral becomes:
\[ I = \int_0^{\pi} \sin(mx) \sin(nx) \, dx = \frac{1}{2} \int_0^{\pi} [\cos((m-n)x) - \cos((m+n)x)] \, dx \]

### Step 2: Evaluate the Integral
Now, we can evaluate 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: Integrate Each Cosine Term
1. For \( \int_0^{\pi} \cos((m-n)x) \, dx \):
- If \( m \neq n \):
\[ \int_0^{\pi} \cos((m-n)x) \, dx = \left[ \frac{\sin((m-n)x)}{m-n} \right]_0^{\pi} = \frac{\sin((m-n)\pi)}{m-n} - \frac{\sin(0)}{m-n} = 0 \]
- If \( m = n \):
\[ \int_0^{\pi} \cos(0) \, dx = \int_0^{\pi} 1 \, dx = \pi \]

2. For \( \int_0^{\pi} \cos((m+n)x) \, dx \):
- Regardless of \( m \) and \( n \):
\[ \int_0^{\pi} \cos((m+n)x) \, dx = \left[ \frac{\sin((m+n)x)}{m+n} \right]_0^{\pi} = \frac{\sin((m+n)\pi)}{m+n} - \frac{\sin(0)}{m+n} = 0 \]

### Step 4: Combine Results
Combining these results, we find:
- If \( m \neq n \):
\[ I = \frac{1}{2} (0 - 0) = 0 \]
- If \( m = n \):
\[ I = \frac{1}{2} (\pi - 0) = \frac{\pi}{2} \]

### Conclusion
Thus, we conclude:
- \( I = 0 \) if \( m \neq n \)
- \( I = \frac{\pi}{2} \) if \( m = n \)

### Final Result
Therefore, the final result is:
\[ \int_0^{\pi} \sin(mx) \sin(nx) \, dx = \begin{cases} 0 & \text{if } m \neq n \\ \frac{\pi}{2} & \text{if } m = n \end{cases} \]