A triple integral has to be calculated step-by-step - one variable at a time:
#int_(z=0)^oo int_(y=0)^(pi/2) int_(x=0)^pi f(x,y,z) dx dy dz equiv int_(z=0)^oo {int_(y=0)^(pi/2) (int_(x=0)^pi f(x,y,z) dx )dy} dz#
We carry out the #x# integral first (here we treat #y# and #z# as constants, at least for the time being)
#int_{x=0}^pi sin(x+y)ye^{-2z}dx = ye^{-2z}int_{x=0}^pi sin(x+y)dx#
#qquad = -ye^{-2z}cos(x+y)|_{x=0}^pi =- ye^{-2z}(cos(pi+y)-cos y)#
#qquad = 2ye^{-2z}cos y#
Next, we carry out the #y# integral
#int_{y=0}^{pi/2}(2ye^{-2z}cos y)dy = 2e^{-2z}int_{y=0}^{pi/2}ycos ydy#
We evaluate this integral by parts
# 2 e^{-2z} (y siny |_{y=0}^{pi/2} - int_{y=0}^{pi/2} sin y dy) #
# qquad = 2e^{-2z} (pi/2+cosy| _{y=0}^{pi/2}) = 2e^{-2z} (pi/2-1)#
Finally, we carry out the final, #z# integral:
# int_{z=0}^oo 2(pi/2-1)e^{-2z}dz = (pi-2) int_{z=0}^oo e^{-2z}dz#
#qquad = -(pi-2)(-1/2e^{-2z})_0^oo=(pi-2)/2#