This integral is little tricky to handle directly. But with some manual help (double angle formula + Feynman integration technique), Mathematica can carry out the computation symbolically.
Numeric check
In[]:= NIntegrate[Sin[x] Cos[y] Cos[z]/(1 + Cos[x] Cos[y] Cos[z]), {x, 0, Pi/2}, {y, 0,Pi/2}, {z, 0, Pi/2}]
Out = 0.78665
Integrate about x and let g(z) = Log[1+Cos[y] Cos[z]]
In[]:= Integrate[Sin[x] Cos[y] Cos[z]/(1+Cos[x] Cos[y] Cos[z]),{x,0,Pi/2},Assumptions->0<y<Pi/2&&0<z<Pi/2]
Out= Log[1+Cos[y] Cos[z]]
Use Feynman's differentiation to take derivative of g[z] about z and then integrate about y. Call it g'[z]:
In[]:= D[Log[1 + Cos[y] Cos[z]], z]
Out= -((Cos[y] Sin[z])/(1 + Cos[y] Cos[z]))
In[]:= Integrate[%,{y,0,\[Pi]/2},Assumptions->0<z<\[Pi]/2]
Out= -(1/2) (\[Pi]+4 ArcTan[Cot[z]-Csc[z]] Csc[z]) Tan[z]
Then use integral by parts to bring it back:
$\int g(z) dz= z \cdot g - \int z \cdot g' $, you can check that the first term is 0 at
$z=0$ and
$z = \pi /2$:
In[]:= -Integrate[%*z,{z,0,Pi/2}]
Out= -1/2 (Pi^2 Log[2]-7 Zeta[3])
The numeric value is
In[]:= N[%]
Out= 0.78665
I think the place where Wolfram can improve is to automatically further simplify this part: ArcTan[Cot[z]-Csc[z]]. If we replace z with z=2u, we have:
In[]:= FullSimplify[ArcTan[Cot[2u]-Csc[2u]],Assumptions->0<u<Pi/4]
Out= -u
This step is correct because
$cos(2 \cdot u) = 1 - 2 \cdot sin(u)^2 $ and
$ sin (2 \cdot u) = 2 \cdot sin (u) \cdot cos (u) $. Then the expression inside ArcTan is simply -Tan[u]. Recall ArcTan is an odd function so this compound part with ArcTan[-Tan[u]] reduces to -u = -z/2.
