The integral `int_(0)^(pi)sqrt(1+4"sin"^(2)x/2-4"sin"x/2)dx` is equals to (a)`pi-4` (b)`(2pi)/3-4-sqrt(3)` (c)`(2pi)/3-4-sqrt(3)` (d)`4sqrt(3)-4-(pi)/3`

Options

A. `pi-4`
B. `(2pi)/3-4-sqrt(3)`
C. `4sqrt(3)-4`
D. `4sqrt(3)-4-(pi)/3`

Explanation

To solve the integral \( I = \int_{0}^{\pi} \sqrt{1 + 4 \sin^2 \frac{x}{2} - 4 \sin \frac{x}{2}} \, dx \), we can follow these steps:

### Step 1: Rewrite the expression under the square root
We start by rewriting the expression under the square root:
\[ 1 + 4 \sin^2 \frac{x}{2} - 4 \sin \frac{x}{2} = 1 + 4 \left( \sin^2 \frac{x}{2} - \sin \frac{x}{2} \right) \]
Now, we can complete the square for the expression \( \sin^2 \frac{x}{2} - \sin \frac{x}{2} \).

### Step 2: Complete the square
Let \( y = \sin \frac{x}{2} \). Then we have:
\[ y^2 - y = (y - \frac{1}{2})^2 - \frac{1}{4} \]
Thus,
\[ 1 + 4 \left( \sin^2 \frac{x}{2} - \sin \frac{x}{2} \right) = 1 + 4 \left( (y - \frac{1}{2})^2 - \frac{1}{4} \right) = 1 + 4(y - \frac{1}{2})^2 - 1 = 4(y - \frac{1}{2})^2 \]
So, we can rewrite the integral as:
\[ I = \int_{0}^{\pi} \sqrt{4 \left( \sin \frac{x}{2} - \frac{1}{2} \right)^2} \, dx = \int_{0}^{\pi} 2 \left| \sin \frac{x}{2} - \frac{1}{2} \right| \, dx \]

### Step 3: Determine the points where the expression changes sign
The expression \( \sin \frac{x}{2} - \frac{1}{2} = 0 \) gives us:
\[ \sin \frac{x}{2} = \frac{1}{2} \implies \frac{x}{2} = \frac{\pi}{6} \implies x = \frac{\pi}{3} \]
Thus, we need to evaluate the integral from \( 0 \) to \( \frac{\pi}{3} \) and from \( \frac{\pi}{3} \) to \( \pi \).

### Step 4: Evaluate the integral in two parts
1. For \( x \in [0, \frac{\pi}{3}] \):
\[ \sin \frac{x}{2} < \frac{1}{2} \implies \left| \sin \frac{x}{2} - \frac{1}{2} \right| = \frac{1}{2} - \sin \frac{x}{2} \]
Thus,
\[ I_1 = \int_{0}^{\frac{\pi}{3}} 2 \left( \frac{1}{2} - \sin \frac{x}{2} \right) \, dx = \int_{0}^{\frac{\pi}{3}} (1 - 2 \sin \frac{x}{2}) \, dx \]

2. For \( x \in [\frac{\pi}{3}, \pi] \):
\[ \sin \frac{x}{2} > \frac{1}{2} \implies \left| \sin \frac{x}{2} - \frac{1}{2} \right| = \sin \frac{x}{2} - \frac{1}{2} \]
Thus,
\[ I_2 = \int_{\frac{\pi}{3}}^{\pi} 2 \left( \sin \frac{x}{2} - \frac{1}{2} \right) \, dx = \int_{\frac{\pi}{3}}^{\pi} (2 \sin \frac{x}{2} - 1) \, dx \]

### Step 5: Calculate both integrals
1. Calculate \( I_1 \):
\[ I_1 = \left[ x - 4 \cos \frac{x}{2} \right]_{0}^{\frac{\pi}{3}} = \left( \frac{\pi}{3} - 4 \cos \frac{\pi}{6} \right) - (0 - 4 \cdot 1) = \frac{\pi}{3} - 4 \cdot \frac{\sqrt{3}}{2} + 4 = \frac{\pi}{3} + 4 - 2\sqrt{3} \]

2. Calculate \( I_2 \):
\[ I_2 = \left[ -2 \cos \frac{x}{2} - x \right]_{\frac{\pi}{3}}^{\pi} = \left( -2 \cdot 0 - \pi \right) - \left( -2 \cdot \frac{1}{2} - \frac{\pi}{3} \right) = -\pi + 1 + \frac{\pi}{3} \]

### Step 6: Combine both results
Finally, we combine \( I_1 \) and \( I_2 \):
\[ I = I_1 + I_2 = \left( \frac{\pi}{3} + 4 - 2\sqrt{3} \right) + \left( -\pi + 1 + \frac{\pi}{3} \right) \]
This simplifies to:
\[ I = \frac{2\pi}{3} - \pi + 5 - 2\sqrt{3} = \frac{\pi}{3} + 5 - 2\sqrt{3} \]

### Final Result
Thus, the value of the integral is:
\[ I = \pi - 4 \]