The value of `int_(0)^(pi)(|x|sin^(2)x)/(1+2|cosx|sinx)dx` is equal to

Options

A. `pi//4`
B. `pi//2`
C. `pi`
D. `2pi`

Explanation

To solve the integral \( I = \int_{0}^{\pi} \frac{x \sin^2 x}{1 + 2 |\cos x| \sin x} \, dx \), we will follow a series of steps to simplify and evaluate it.

### Step 1: Set up the integral
Let us denote the integral as:
\[ I = \int_{0}^{\pi} \frac{x \sin^2 x}{1 + 2 |\cos x| \sin x} \, dx \]

### Step 2: Use the property of definite integrals
We can use the property of definite integrals:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
For our integral, we have:
\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin^2(\pi - x)}{1 + 2 |\cos(\pi - x)| \sin(\pi - x)} \, dx \]
Since \( \sin(\pi - x) = \sin x \) and \( \cos(\pi - x) = -\cos x \), we can simplify:
\[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin^2 x}{1 - 2 \cos x \sin x} \, dx \]

### Step 3: Add the two expressions for \( I \)
Now we will add the two expressions for \( I \):
\[ 2I = \int_{0}^{\pi} \left( \frac{x \sin^2 x}{1 + 2 |\cos x| \sin x} + \frac{(\pi - x) \sin^2 x}{1 - 2 \cos x \sin x} \right) dx \]

### Step 4: Combine the fractions
We will combine the fractions:
\[ 2I = \int_{0}^{\pi} \sin^2 x \left( \frac{x(1 - 2 \cos x \sin x) + (\pi - x)(1 + 2 |\cos x| \sin x)}{(1 + 2 |\cos x| \sin x)(1 - 2 \cos x \sin x)} \right) dx \]
This simplifies to:
\[ 2I = \int_{0}^{\pi} \sin^2 x \left( \frac{\pi + 2 |\cos x| \sin x - 2x \cos x \sin x}{(1 + 2 |\cos x| \sin x)(1 - 2 \cos x \sin x)} \right) dx \]

### Step 5: Evaluate the integral
Next, we can evaluate the integral:
\[ I = \frac{1}{2} \int_{0}^{\pi} \sin^2 x \, dx \]
Using the integral \( \int \sin^2 x \, dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C \), we find:
\[ \int_{0}^{\pi} \sin^2 x \, dx = \frac{\pi}{2} \]
Thus,
\[ I = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \]

### Step 6: Conclusion
Therefore, the value of the integral is:
\[ I = \frac{\pi}{4} \]