Evaluate the innermost integral first, where [low limit, high limit]:
(evaluate high limit value and subtract the lower limit value)
∫[0,sec(θ)]sin(2ϕ)dρ=[0,sec(θ)](ρ)sin(2ϕ)=sec(θ)sin(2ϕ) (sin(2ϕ) is considered as a constant in this integral).
Now the second integral:
∫[0,π/4]sec(θ)sin(2ϕ)dϕ=
[0,π/4](-½cos(2ϕ))sec(θ)=
(-½cos(π/2)-(-½cos(0)))sec(θ)=sec(θ)/2 (sec(θ) is considered as a constant in this integral).
Final integral:
∫[0,π](sec(θ)/2)dθ.
To integrate ½sec(θ), multiply by (sec(θ)+tan(θ))/(sec(θ)+tan(θ)):
½(sec²(θ)+sec(θ)tan(θ))/(tan(θ)+sec(θ)).
Integrate wrt θ and we get ½ln(tan(θ)+sec(θ)).
Apply the limits [0,π]:
θ=π: ½ln(0-1) is undefined; θ=0: ½ln(0+1)=0.
Therefore, the triple integral is undefined.