To solve the integral
\[
I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin x + \cos x} \, dx,
\]
we can use a symmetry property of definite integrals.
### Step 1: Use the property of definite integrals
We know that:
\[
\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx.
\]
In our case, we can set \( a = \frac{\pi}{2} \). Therefore, we can write:
\[
I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3(\frac{\pi}{2} - x)}{\sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x)} \, dx.
\]
### Step 2: Simplify the expression
Using the identities \( \sin(\frac{\pi}{2} - x) = \cos x \) and \( \cos(\frac{\pi}{2} - x) = \sin x \), we can rewrite the integral as:
\[
I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x}{\cos x + \sin x} \, dx.
\]
### Step 3: Combine the two integrals
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x}{\sin x + \cos x} \, dx \) (Equation 1)
2. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^3 x}{\sin x + \cos x} \, dx \) (Equation 2)
Adding these two equations gives:
\[
2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x + \cos^3 x}{\sin x + \cos x} \, dx.
\]
### Step 4: Use the identity for sum of cubes
Using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), we can write:
\[
\sin^3 x + \cos^3 x = (\sin x + \cos x)(\sin^2 x - \sin x \cos x + \cos^2 x).
\]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[
\sin^3 x + \cos^3 x = (\sin x + \cos x)(1 - \sin x \cos x).
\]
### Step 5: Substitute back into the integral
Substituting this back into our equation for \( 2I \):
\[
2I = \int_{0}^{\frac{\pi}{2}} \frac{(\sin x + \cos x)(1 - \sin x \cos x)}{\sin x + \cos x} \, dx.
\]
The \( \sin x + \cos x \) terms cancel:
\[
2I = \int_{0}^{\frac{\pi}{2}} (1 - \sin x \cos x) \, dx.
\]
### Step 6: Evaluate the integral
Now we can split the integral:
\[
2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx - \int_{0}^{\frac{\pi}{2}} \sin x \cos x \, dx.
\]
The first integral evaluates to:
\[
\int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}.
\]
The second integral can be evaluated using the identity \( \sin x \cos x = \frac{1}{2} \sin(2x) \):
\[
\int_{0}^{\frac{\pi}{2}} \sin x \cos x \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx = \frac{1}{2} \left[-\frac{1}{2} \cos(2x)\right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \left[-\frac{1}{2} (0 - 1)\right] = \frac{1}{4}.
\]
### Step 7: Combine results
Putting it all together:
\[
2I = \frac{\pi}{2} - \frac{1}{4}.
\]
Thus,
\[
2I = \frac{\pi}{2} - \frac{1}{4} = \frac{2\pi - 1}{4}.
\]
Dividing both sides by 2 gives:
\[
I = \frac{2\pi - 1}{8}.
\]
### Final Answer
The value of the integral is
\[
\boxed{\frac{2\pi - 1}{8}}.
\]
