Evaluate the definite integral.
$${\int }_0^{\pi }\sin (5x)\sin (9x)dx$$

Step 1 ：Given the definite integral ${\int }_0^{\pi }\sin (5x)\sin (9x)dx$

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 $\boxed{{\displaystyle 0}}$