`int_(0)^(pi)(x sin x)/(1-sin x)dx=`

Options

A. `pi.log4`
B. `pi log 2`
C. 2 log 2
D. None of these

Answer

To solve the integral \( I = \int_{0}^{\pi} \frac{x \sin x}{1 - \sin x} \, dx \), we will use the property of definite integrals.

### Step 1: Apply the property of definite integrals
Using the property:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \]
we can rewrite our integral:
\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 - \sin(\pi - x)} \, dx \]
Since \( \sin(\pi - x) = \sin x \), we can simplify this to:
\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 - \sin x} \, dx \]

### Step 2: Combine the two integrals
Now we have two expressions for \( I \):
\[ I = \int_{0}^{\pi} \frac{x \sin x}{1 - \sin x} \, dx \]
\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 - \sin x} \, dx \]
Adding these two equations gives:
\[ 2I = \int_{0}^{\pi} \frac{(\pi \sin x)}{1 - \sin x} \, dx \]
Thus,
\[ I = \frac{1}{2} \int_{0}^{\pi} \frac{\pi \sin x}{1 - \sin x} \, dx \]

### Step 3: Simplify the integral
Now we need to evaluate the integral:
\[ \int_{0}^{\pi} \frac{\sin x}{1 - \sin x} \, dx \]
To do this, we will rationalize the integrand by multiplying the numerator and denominator by \( 1 + \sin x \):
\[ \int_{0}^{\pi} \frac{\sin x (1 + \sin x)}{(1 - \sin x)(1 + \sin x)} \, dx = \int_{0}^{\pi} \frac{\sin x (1 + \sin x)}{\cos^2 x} \, dx \]

### Step 4: Separate the integral
This can be separated into two integrals:
\[ \int_{0}^{\pi} \frac{\sin x}{\cos^2 x} \, dx + \int_{0}^{\pi} \frac{\sin^2 x}{\cos^2 x} \, dx \]
The first integral can be rewritten using the substitution \( t = \cos x \), where \( dt = -\sin x \, dx \):
\[ \int_{0}^{\pi} \frac{\sin x}{\cos^2 x} \, dx = -\int_{1}^{-1} \frac{1}{t^2} \, dt = \left[ \frac{1}{t} \right]_{1}^{-1} = 1 - (-1) = 2 \]

### Step 5: Evaluate the second integral
The second integral is:
\[ \int_{0}^{\pi} \tan^2 x \, dx \]
This integral can be evaluated as:
\[ \int_{0}^{\pi} \tan^2 x \, dx = \int_{0}^{\pi} \sec^2 x \, dx - \int_{0}^{\pi} 1 \, dx = [\tan x]_{0}^{\pi} - [x]_{0}^{\pi} = 0 - \pi = -\pi \]

### Step 6: Combine results
Now we can combine the results:
\[ \int_{0}^{\pi} \frac{\sin x}{1 - \sin x} \, dx = 2 - \pi \]
Thus,
\[ I = \frac{1}{2} \cdot \pi (2 - \pi) = \pi - \frac{\pi^2}{2} \]

### Final Answer
Therefore, the value of the integral is:
\[ I = \pi - \frac{\pi^2}{2} \]