To solve the integral analytically:
$$
\int_0^{\pi/2} \int_{1/(\cos d + \sin d)}^{1} r^3 , dr , dd
$$
we continue from the point where we split the integral into two parts:
$$
\int_0^{\pi/2} \left( \frac{1}{4} - \frac{1}{4(\cos d + \sin d)^4} \right) , dd
$$
Step 1: Evaluate the Simple Integral First, we evaluate the simpler part of the integral:
$$
\frac{1}{4} \int_0^{\pi/2} , dd = \frac{1}{4} \cdot \frac{\pi}{2} = \frac{\pi}{8}
$$
Step 2: Evaluate the Complex Integral Now, we need to evaluate the more complex part:
$$
\frac{1}{4} \int_0^{\pi/2} \frac{1}{(\cos d + \sin d)^4} , dd
$$
To solve this, we use the symmetry and properties of trigonometric functions. Let's use the substitution ( t = \tan(d) ):
For ( d \in [0, \pi/2] ):
When ( d = 0 ), ( t = 0 )
When ( d = \pi/2 ), ( t \to \infty )
The differential ( dd ) in terms of ( t ) is ( dd = \frac{dt}{1 + t^2} ).
The integral becomes:
$$
\int_0^\infty \frac{1}{(\cos d + \sin d)^4} \cdot \frac{dt}{1 + t^2}
$$
Since ( \cos d = \frac{1}{\sqrt{1 + t^2}} ) and ( \sin d = \frac{t}{\sqrt{1 + t^2}} ):
$$
\cos d + \sin d = \frac{1 + t}{\sqrt{1 + t^2}}
$$
Thus, the integral transforms to:
$$
\int_0^\infty \frac{1}{\left( \frac{1 + t}{\sqrt{1 + t^2}} \right)^4} \cdot \frac{dt}{1 + t^2} = \int_0^\infty \frac{(1 + t^2)^2}{(1 + t)^4} \cdot \frac{dt}{1 + t^2} = \int_0^\infty \frac{1}{(1 + t)^4} dt
$$
Now, simplify and integrate:
$$
\int_0^\infty \frac{dt}{(1 + t)^4}
$$
This integral can be solved using the formula for integrals of the form:
$$
\int_0^\infty \frac{dt}{(1 + t)^n} = \frac{1}{n - 1}
$$
for ( n > 1 ). Here, ( n = 4 ):
$$
\int_0^\infty \frac{dt}{(1 + t)^4} = \frac{1}{3}
$$
Thus, the complex part of the integral evaluates to:
$$
\frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}
$$
Combining both parts, we get:
$$
\frac{\pi}{8} - \frac{1}{12}
$$
Simplify:
$$
\frac{\pi}{8} - \frac{1}{12} = \frac{3\pi}{24} - \frac{2}{24} = \frac{3\pi - 2}{24}
$$
So, the final result is:
$$
\int_0^{\pi/2} \int_{1/(\cos d + \sin d)}^{1} r^3 , dr , dd = \frac{3\pi - 2}{24}
$$