If `int_(0)^(pi)((x)/(1+sinx))^(2) dx=A,` then the value for `int_(0)^(pi)(2x^(2). cos^(2)x//2)/((1+ sin x^(2)))dx` is equal to

Options

A. `A+2pi-pi^(2)`
B. `A-2pi+pi^(2)`
C. `2pi-A-pi^(2)`
D. None of these

Answer

To solve the given problem step by step, we will evaluate the integral \( I = \int_{0}^{\pi} \frac{2x^2 \cos^2 \frac{x}{2}}{(1 + \sin x)^2} \, dx \) based on the information provided about \( A = \int_{0}^{\pi} \frac{x}{(1 + \sin x)^2} \, dx \).

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int_{0}^{\pi} \frac{2x^2 \cos^2 \frac{x}{2}}{(1 + \sin x)^2} \, dx \]

Using the half-angle identity, we can express \( \cos^2 \frac{x}{2} \) in terms of \( \sin x \):
\[ \cos^2 \frac{x}{2} = \frac{1 + \cos x}{2} \]
Thus,
\[ I = \int_{0}^{\pi} \frac{2x^2 \cdot \frac{1 + \cos x}{2}}{(1 + \sin x)^2} \, dx = \int_{0}^{\pi} \frac{x^2 (1 + \cos x)}{(1 + \sin x)^2} \, dx \]

### Step 2: Split the Integral
We can split the integral into two parts:
\[ I = \int_{0}^{\pi} \frac{x^2}{(1 + \sin x)^2} \, dx + \int_{0}^{\pi} \frac{x^2 \cos x}{(1 + \sin x)^2} \, dx \]
Let \( I_1 = \int_{0}^{\pi} \frac{x^2}{(1 + \sin x)^2} \, dx \) and \( I_2 = \int_{0}^{\pi} \frac{x^2 \cos x}{(1 + \sin x)^2} \, dx \).

### Step 3: Evaluate \( I_1 \)
From the problem statement, we know:
\[ I_1 = A \]

### Step 4: Evaluate \( I_2 \)
To evaluate \( I_2 \), we can use the property of definite integrals:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
For \( I_2 \):
\[ I_2 = \int_{0}^{\pi} \frac{(\pi - x)^2 \cos(\pi - x)}{(1 + \sin(\pi - x))^2} \, dx \]
Using the identities \( \cos(\pi - x) = -\cos x \) and \( \sin(\pi - x) = \sin x \):
\[ I_2 = \int_{0}^{\pi} \frac{(\pi - x)^2 (-\cos x)}{(1 + \sin x)^2} \, dx = -\int_{0}^{\pi} \frac{(\pi - x)^2 \cos x}{(1 + \sin x)^2} \, dx \]

### Step 5: Combine the Integrals
Now we can combine \( I_2 \) with \( I_1 \):
\[ 2I_2 = I_2 + (-I_2) = \int_{0}^{\pi} \frac{x^2 \cos x}{(1 + \sin x)^2} \, dx - \int_{0}^{\pi} \frac{(\pi - x)^2 \cos x}{(1 + \sin x)^2} \, dx \]
This simplifies to:
\[ 2I_2 = -\pi^2 \int_{0}^{\pi} \frac{\cos x}{(1 + \sin x)^2} \, dx \]

### Step 6: Final Calculation
Putting everything together:
\[ I = A + I_2 \]
Substituting the values we have:
\[ I = A + \left(-\frac{\pi^2}{2} + 2\pi\right) \]

### Conclusion
Thus, the final value of the integral is:
\[ I = A + 2\pi - \pi^2 \]