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