If `A = int_(0)^((pi)/(2))(sin^(3)x)/(1+cos^(2)s)dx and B=int_(0)^((pi)/(2))(cos^(2)x)/(1+sin^(2)x)dx`, then `(2A)/(B)` is equal to

Options

A. `-2A`
B. 4
C. 2
D. `-4`

Answer

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 \( \sin x (1 - \cos^2 x) \):
\[ A = \int_{0}^{\frac{\pi}{2}} \frac{\sin x (1 - \cos^2 x)}{1 + \cos^2 x} \, dx \]
This can be split into two integrals:
\[ A = \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1 + \cos^2 x} \, dx - \int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos^2 x}{1 + \cos^2 x} \, dx \]

### Step 2: Substitute \( \cos x = t \)

Let \( \cos x = t \), then \( -\sin x \, dx = dt \). The limits change as follows:
- When \( x = 0 \), \( t = 1 \)
- When \( x = \frac{\pi}{2} \), \( t = 0 \)

Thus, we have:
\[ A = \int_{1}^{0} \frac{-1 + t^2}{1 + t^2} \, dt = \int_{0}^{1} \frac{1 - t^2}{1 + t^2} \, dt \]

### Step 3: Evaluate \( B \)

Given:
\[ B = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + \sin^2 x} \, dx \]

Using the same substitution \( \sin x = u \) (thus \( \cos x \, dx = du \)):
- When \( x = 0 \), \( u = 0 \)
- When \( x = \frac{\pi}{2} \), \( u = 1 \)

Thus, we have:
\[ B = \int_{0}^{1} \frac{1 - u^2}{1 + u^2} \, du \]

### Step 4: Compare \( A \) and \( B \)

Notice that both integrals \( A \) and \( B \) are equal:
\[ A = B = \int_{0}^{1} \frac{1 - t^2}{1 + t^2} \, dt \]

### Step 5: Calculate \( \frac{2A}{B} \)

Since \( A = B \):
\[ \frac{2A}{B} = \frac{2B}{B} = 2 \]

### Final Answer:
\[ \frac{2A}{B} = 2 \]