Answer
$\dfrac{-1}{3}$
Work Step by Step
$ \int_{0}^{\pi/2} \int_{0}^{y} \int_{0}^{x} \cos(x+y+z) dz dx dy= \int_{0}^{\pi/2} \int_{0}^{y} [\sin(x+y+z)]_{0}^{x} dx dy$
or, $=\int_{0}^{\pi/2} (-1/2) \cos 3y+\cos 2y+(1/2) \cos y-\cos y dy$
or, $=\int_{0}^{\pi/2} (-1/2) \cos 3y+\cos 2y-(1/2) \cos y dy$
or, $= [(-1/6) \sin 3y-(1/2) \sin (2y) -(1/2) \sin y]_0^{\dfrac{\pi}{2}}$
or, $= [(-1/6) \sin (\dfrac{3\pi}{2})+(1/2) \sin (\pi) -(1/2) \sin (\dfrac{\pi}{2})$
or, $=\dfrac{-1}{3}$
