Answer
$$\dfrac{\sin^2 (4)}{2}$$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$\int^2_0 \int^{4-x^2}_0 \int^x_0 \dfrac{\sin 2z}{4-z} \space dy \space dz \space dx =\int^2_0 \int^{4-x^2}_0 \dfrac{x }{4z} (\sin 2z) \space dz \space dx \\=\int^4_0 \int^{\sqrt{4-z}}_0 (\dfrac{sin 2z}{4-z}) \space dx \space dz \\=\int^4_0 (\dfrac{ \sin (2z) }{4-z}) \times \dfrac{1}{2} (4-z) \space dz \\=[-\dfrac{1}{4} \cos2z]^4_0 \\=\dfrac{\sin^2 (4)}{2}$$