Help with integrating triple integral xz^3*cos(xyz)

trying to integrate the following function

xz^3*cos(xyz)
bounds are 0<=x<=1
0<=y<=1
0<=z=1

I think it would be easiest to first integrate the following way dydxdz...
for dy use u substition. I got u=xyz ..... du=xz.
so the integral becomes
z^2cos(u) when you integrate you get z^2sin(xyz)dxdz.
then integrate by parts maybe?? with
u=z^2 and dv=sin(xyz)
Could someone please help me. Soooo confused about how to integrate these triple integrals!

∫(z = 0 to 1) ∫(x = 0 to 1) ∫(y = 0 to 1) xz^3 cos(xyz) dy dx dz
= ∫(z = 0 to 1) ∫(x = 0 to 1) xz^3 sin(xyz)/(xz) {for y = 0 to 1} dx dz
= ∫(z = 0 to 1) ∫(x = 0 to 1) z^2 sin(xz) dx dz
= ∫(z = 0 to 1) z^2 * -cos(xz)/z {for x = 0 to 1} dz
= ∫(z = 0 to 1) (z - z cos z) dz
= z^2/2 - (z sin z + cos z) {for z = 0 to 1}, by using integration by parts
= 3/2 - sin 1 - cos 1.

I hope this helps!