Evaluate the integral $\int_{0}^{9} \int_{0}^{3} \int_{4 y}^{12} \frac{5 \cos \left(x^{2}\right)}{6 \sqrt{z}} d x d y d z$ by changing the order of integration in an appropriate way. t: 0 \[ \int_{0}^{9} \int_{0}^{3} \int_{4 y}^{12} \frac{5 \cos \left(x^{2}\right)}{6 \sqrt{z}} d x d y d z= \]

Step 1 ：The given triple integral is \(\int_{0}^{9} \int_{0}^{3} \int_{4 y}^{12} \frac{5 \cos \left(x^{2}\right)}{6 \sqrt{z}} d x d y d z\).

Step 2 ：The limits of integration for x are functions of y, and the integrand is a function of x and z.

Step 3 ：To change the order of integration, we need to consider the region of integration in the xyz-space. The region is defined by the inequalities 0 ≤ y ≤ 3, 0 ≤ z ≤ 9, and 4y ≤ x ≤ 12.

Step 4 ：We can change the order of integration to dy dz dx by considering the projection of the region onto the yz-plane. The new limits of integration for y will be from 0 to x/4, for z from 0 to 9, and for x from 0 to 12.

Step 5 ：The integrand is \(\frac{5\cos(x^{2})}{6\sqrt{z}}\).

Step 6 ：The integral of the integrand is \(\frac{5\sin(144)}{8}\).

Step 7 ：Final Answer: The value of the integral is \(\boxed{\frac{5\sin(144)}{8}}\).