`int_(0)^(pi)(xdx)/(1+sinx)` is equal to

Options

A. `-pi`
B. `(pi)/(2)`
C. `pi`
D. none of these

Explanation

To solve the integral \( I = \int_{0}^{\pi} \frac{x \, dx}{1 + \sin x} \), we can use a property of definite integrals. This property states that:

\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]

In our case, \( a = 0 \) and \( b = \pi \). Therefore, we can rewrite the integral as follows:

1. **Step 1: Apply the property of definite integrals.**
\[ I = \int_{0}^{\pi} \frac{x \, dx}{1 + \sin x} = \int_{0}^{\pi} \frac{\pi - x \, dx}{1 + \sin(\pi - x)} \]

Since \( \sin(\pi - x) = \sin x \), we have:
\[ I = \int_{0}^{\pi} \frac{\pi - x \, dx}{1 + \sin x} \]

2. **Step 2: Rewrite the integral.**
We can express this as:
\[ I = \int_{0}^{\pi} \frac{\pi \, dx - x \, dx}{1 + \sin x} = \int_{0}^{\pi} \frac{\pi \, dx}{1 + \sin x} - \int_{0}^{\pi} \frac{x \, dx}{1 + \sin x} \]

Let’s denote the second integral as \( I \) again:
\[ I = \int_{0}^{\pi} \frac{\pi \, dx}{1 + \sin x} - I \]

3. **Step 3: Solve for \( I \).**
Adding \( I \) to both sides gives:
\[ 2I = \int_{0}^{\pi} \frac{\pi \, dx}{1 + \sin x} \]
Therefore,
\[ I = \frac{1}{2} \int_{0}^{\pi} \frac{\pi \, dx}{1 + \sin x} \]

4. **Step 4: Evaluate the integral \( \int_{0}^{\pi} \frac{dx}{1 + \sin x} \).**
To evaluate this integral, we can use the substitution:
\[ \int_{0}^{\pi} \frac{dx}{1 + \sin x} = \int_{0}^{\pi} \frac{1 - \sin x}{\cos^2 x} \, dx \]
This can be simplified using trigonometric identities.

5. **Step 5: Final evaluation.**
After evaluating, we find:
\[ \int_{0}^{\pi} \frac{dx}{1 + \sin x} = \pi \]
Thus,
\[ I = \frac{1}{2} \cdot \pi = \frac{\pi}{2} \]

6. **Final answer:**
Therefore, the value of the integral \( I \) is:
\[ \boxed{\frac{\pi^2}{4}} \]