The integral `int_(0)^(pi)sqrt(1+4"sin"^(2)(x)/(2)-4 "sin"(x)/(2))dx`equals ,

Options

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

Explanation

To solve the integral

\[ I = \int_{0}^{\pi} \sqrt{1 + 4 \sin^2\left(\frac{x}{2}\right) - 4 \sin\left(\frac{x}{2}\right)} \, dx, \]

we start by simplifying the expression under the square root.

### Step 1: Simplify the expression under the square root

We can rewrite the expression \(1 + 4 \sin^2\left(\frac{x}{2}\right) - 4 \sin\left(\frac{x}{2}\right)\) as follows:

\[ 1 + 4 \sin^2\left(\frac{x}{2}\right) - 4 \sin\left(\frac{x}{2}\right) = 1 - 4 \sin\left(\frac{x}{2}\right) + 4 \sin^2\left(\frac{x}{2}\right). \]

This can be factored as:

\[ 1 - 2 \sin\left(\frac{x}{2}\right)^2 = (1 - 2 \sin\left(\frac{x}{2}\right))^2. \]

Thus, we have:

\[ \sqrt{1 + 4 \sin^2\left(\frac{x}{2}\right) - 4 \sin\left(\frac{x}{2}\right)} = |1 - 2 \sin\left(\frac{x}{2}\right)|. \]

### Step 2: Determine the intervals for the absolute value

Next, we need to find where \(1 - 2 \sin\left(\frac{x}{2}\right)\) is positive or negative.

- \(1 - 2 \sin\left(\frac{x}{2}\right) \geq 0\) implies \(\sin\left(\frac{x}{2}\right) \leq \frac{1}{2}\).
- This occurs when \(\frac{x}{2} \leq \frac{\pi}{6}\) or \(x \leq \frac{\pi}{3}\).

Thus, we can break the integral into two parts:

\[ I = \int_{0}^{\frac{\pi}{3}} (1 - 2 \sin\left(\frac{x}{2}\right)) \, dx + \int_{\frac{\pi}{3}}^{\pi} (2 \sin\left(\frac{x}{2}\right) - 1) \, dx. \]

### Step 3: Evaluate the first integral

Now we evaluate the first integral:

\[ \int_{0}^{\frac{\pi}{3}} (1 - 2 \sin\left(\frac{x}{2}\right)) \, dx. \]

This can be split into two parts:

\[ \int_{0}^{\frac{\pi}{3}} 1 \, dx - 2 \int_{0}^{\frac{\pi}{3}} \sin\left(\frac{x}{2}\right) \, dx. \]

Calculating the first part:

\[ \int_{0}^{\frac{\pi}{3}} 1 \, dx = \left[x\right]_{0}^{\frac{\pi}{3}} = \frac{\pi}{3}. \]

For the second part, we use the substitution \(u = \frac{x}{2}\), \(dx = 2 du\):

\[ \int_{0}^{\frac{\pi}{3}} \sin\left(\frac{x}{2}\right) \, dx = 2 \int_{0}^{\frac{\pi}{6}} \sin(u) \, du = 2 \left[-\cos(u)\right]_{0}^{\frac{\pi}{6}} = 2 \left[-\cos\left(\frac{\pi}{6}\right) + 1\right] = 2 \left[-\frac{\sqrt{3}}{2} + 1\right] = 2 \left(1 - \frac{\sqrt{3}}{2}\right) = 2 - \sqrt{3}. \]

Thus, the first integral evaluates to:

\[ \frac{\pi}{3} - 2(2 - \sqrt{3}) = \frac{\pi}{3} - 4 + 2\sqrt{3}. \]

### Step 4: Evaluate the second integral

Now we evaluate the second integral:

\[ \int_{\frac{\pi}{3}}^{\pi} (2 \sin\left(\frac{x}{2}\right) - 1) \, dx. \]

This can also be split:

\[ 2 \int_{\frac{\pi}{3}}^{\pi} \sin\left(\frac{x}{2}\right) \, dx - \int_{\frac{\pi}{3}}^{\pi} 1 \, dx. \]

Calculating the second part:

\[ \int_{\frac{\pi}{3}}^{\pi} 1 \, dx = \left[x\right]_{\frac{\pi}{3}}^{\pi} = \pi - \frac{\pi}{3} = \frac{2\pi}{3}. \]

For the first part, using the same substitution \(u = \frac{x}{2}\):

\[ 2 \int_{\frac{\pi}{3}}^{\pi} \sin\left(\frac{x}{2}\right) \, dx = 4 \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \sin(u) \, du = 4 \left[-\cos(u)\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}} = 4 \left[-0 + \frac{\sqrt{3}}{2}\right] = 2\sqrt{3}. \]

Thus, the second integral evaluates to:

\[ 2\sqrt{3} - \frac{2\pi}{3}. \]

### Step 5: Combine results

Now we combine the results of both integrals:

\[ I = \left(\frac{\pi}{3} - 4 + 2\sqrt{3}\right) + \left(2\sqrt{3} - \frac{2\pi}{3}\right). \]

Combining like terms gives:

\[ I = \left(\frac{\pi}{3} - \frac{2\pi}{3}\right) + (2\sqrt{3} + 2\sqrt{3}) - 4 = -\frac{\pi}{3} + 4\sqrt{3} - 4. \]

### Final Result

Thus, the value of the integral is:

\[ I = 4\sqrt{3} - \frac{\pi}{3} - 4. \]