To solve the problem, we need to evaluate the integrals \( A \) and \( B \) and then find the value of \( \frac{2A}{B} \).
### Step 1: Evaluate \( A \)
Given:
\[
A = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3 x}{1 + \cos^2 x} \, dx
\]
We can rewrite \( \sin^3 x \) as \( \sin x \cdot \sin^2 x \). Using the identity \( \sin^2 x = 1 - \cos^2 x \), we have:
\[
\sin^3 x = \sin x (1 - \cos^2 x)
\]
Thus, we can express \( A \) as:
\[
A = \int_{0}^{\frac{\pi}{2}} \frac{\sin x (1 - \cos^2 x)}{1 + \cos^2 x} \, dx
\]
### Step 2: Change of Variable for \( A \)
Let \( \cos x = t \). Then, \( -\sin x \, dx = dt \) or \( \sin x \, dx = -dt \). The limits change as follows:
- When \( x = 0 \), \( t = 1 \)
- When \( x = \frac{\pi}{2} \), \( t = 0 \)
Thus, we can rewrite \( A \):
\[
A = \int_{1}^{0} \frac{(1 - t^2)}{1 + t^2} (-dt) = \int_{0}^{1} \frac{1 - t^2}{1 + t^2} \, dt
\]
### Step 3: Evaluate \( B \)
Now, we evaluate \( B \):
\[
B = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + \sin^2 x} \, dx
\]
We can rewrite \( \cos^2 x \) as \( \cos x \cdot \cos x \):
\[
B = \int_{0}^{\frac{\pi}{2}} \frac{\cos x (1 - \sin^2 x)}{1 + \sin^2 x} \, dx
\]
### Step 4: Change of Variable for \( B \)
Let \( \sin x = t \). Then, \( \cos x \, dx = dt \). The limits change as follows:
- When \( x = 0 \), \( t = 0 \)
- When \( x = \frac{\pi}{2} \), \( t = 1 \)
Thus, we can rewrite \( B \):
\[
B = \int_{0}^{1} \frac{(1 - t^2)}{1 + t^2} \, dt
\]
### Step 5: Compare \( A \) and \( B \)
Now we observe that both integrals \( A \) and \( B \) are equal:
\[
A = \int_{0}^{1} \frac{1 - t^2}{1 + t^2} \, dt
\]
\[
B = \int_{0}^{1} \frac{1 - t^2}{1 + t^2} \, dt
\]
Thus, we have:
\[
A = B
\]
### Step 6: Calculate \( \frac{2A}{B} \)
Now we can calculate:
\[
\frac{2A}{B} = \frac{2A}{A} = 2
\]
### Final Answer
Thus, the value of \( \frac{2A}{B} \) is:
\[
\boxed{2}
\]
