Evaluate the definite integral.
\[
\int_{0}^{\pi} \sin (5 x) \sin (9 x) d x
\]

Step 1 ：Given the definite integral \(\int_{0}^{\pi} \sin (5 x) \sin (9 x) d x\)

Step 2 ：We can use the product-to-sum identities, which state that \(\sin(a) \sin(b) = \frac{1}{2} [\cos(a - b) - \cos(a + b)]\)

Step 3 ：Rewrite the integral as \(\frac{1}{2} \int_{0}^{\pi} [\cos((5 - 9)x) - \cos((5 + 9)x)] dx\)

Step 4 ：This simplifies to \(\frac{1}{2} \int_{0}^{\pi} [\cos(-4x) - \cos(14x)] dx\)

Step 5 ：The integral of cosine is sine, and the limits of integration are 0 and pi. So, we can calculate the integral directly

Step 6 ：The integral evaluates to zero. This is because the integral of the cosine function over a complete period is zero, and both cosine functions in the integrand have periods that are integer multiples of pi. Therefore, the integral over the interval [0, pi] is zero for both terms

Step 7 ：Final Answer: The definite integral is \(\boxed{0}\)