Consider `I = int_(0)^(pi) (xdx)/(1+sinx)`
What is `int_(0)^(pi)((pi-x)dx)/(1+sinx)` equal to ?

Options

A. `pi`
B. `pi//2`
C. `0`
D. `2pi`

Explanation

To solve the integral \( I = \int_{0}^{\pi} \frac{x \, dx}{1 + \sin x} \) and find the value of \( \int_{0}^{\pi} \frac{\pi - x \, dx}{1 + \sin x} \), we can follow these steps:

### Step 1: Set up the integral
Let:
\[ I = \int_{0}^{\pi} \frac{x \, dx}{1 + \sin x} \]

### Step 2: Transform the second integral
We need to evaluate:
\[ J = \int_{0}^{\pi} \frac{\pi - x \, dx}{1 + \sin x} \]

### Step 3: Use the property of definite integrals
Using the property of definite integrals, we can express \( J \) as:
\[ J = \int_{0}^{\pi} \frac{\pi - x}{1 + \sin x} \, dx = \int_{0}^{\pi} \frac{\pi}{1 + \sin x} \, dx - \int_{0}^{\pi} \frac{x}{1 + \sin x} \, dx \]
Thus, we can write:
\[ J = \int_{0}^{\pi} \frac{\pi}{1 + \sin x} \, dx - I \]

### Step 4: Evaluate \( \int_{0}^{\pi} \frac{\pi}{1 + \sin x} \, dx \)
To evaluate \( \int_{0}^{\pi} \frac{\pi}{1 + \sin x} \, dx \), we can factor out \( \pi \):
\[ \int_{0}^{\pi} \frac{\pi}{1 + \sin x} \, dx = \pi \int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx \]

### Step 5: Use a substitution for \( \int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx \)
Using the substitution \( u = \tan\left(\frac{x}{2}\right) \), we can transform the integral:
\[ \sin x = \frac{2u}{1 + u^2}, \quad dx = \frac{2 \, du}{1 + u^2} \]
The limits change from \( 0 \) to \( \infty \) as \( x \) goes from \( 0 \) to \( \pi \).

Thus:
\[ \int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx = \int_{0}^{\infty} \frac{2}{(1 + u^2)(1 + \frac{2u}{1 + u^2})} \, du \]

### Step 6: Simplify and evaluate the integral
After simplification, we find:
\[ \int_{0}^{\pi} \frac{1}{1 + \sin x} \, dx = \frac{\pi}{2} \]
Thus:
\[ \int_{0}^{\pi} \frac{\pi}{1 + \sin x} \, dx = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2} \]

### Step 7: Substitute back to find \( J \)
Now substituting back into our expression for \( J \):
\[ J = \frac{\pi^2}{2} - I \]

### Step 8: Relate \( J \) to \( I \)
Since we have \( I + J = \frac{\pi^2}{2} \):
\[ I + \left( \frac{\pi^2}{2} - I \right) = \frac{\pi^2}{2} \]
This implies:
\[ 2I = \frac{\pi^2}{2} \implies I = \frac{\pi^2}{4} \]

### Step 9: Find \( J \)
Now substituting back to find \( J \):
\[ J = \frac{\pi^2}{2} - \frac{\pi^2}{4} = \frac{\pi^2}{4} \]

### Final Answer
Thus, we conclude that:
\[ \int_{0}^{\pi} \frac{\pi - x \, dx}{1 + \sin x} = \frac{\pi^2}{4} \]