The value of : `int_(pi//6)^(5pi//6)sqrt(4-4sin^(2)t) dt`, is

Options

A. 0
B. 2
C. 1
D. none of these

Explanation

To solve the integral \( I = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \sqrt{4 - 4\sin^2 t} \, dt \), we can follow these steps:

### Step 1: Simplify the integrand
We start by factoring out the constant from the square root:
\[ I = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \sqrt{4(1 - \sin^2 t)} \, dt \]
This simplifies to:
\[ I = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \sqrt{4} \sqrt{1 - \sin^2 t} \, dt = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 2 \sqrt{1 - \sin^2 t} \, dt \]

### Step 2: Use the Pythagorean identity
Using the identity \( 1 - \sin^2 t = \cos^2 t \), we can rewrite the integral:
\[ I = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} 2 \sqrt{\cos^2 t} \, dt \]
Since \( \sqrt{\cos^2 t} = |\cos t| \), we have:
\[ I = 2 \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} |\cos t| \, dt \]

### Step 3: Determine the sign of \( \cos t \)
Next, we need to analyze the behavior of \( \cos t \) in the interval \( \left[\frac{\pi}{6}, \frac{5\pi}{6}\right] \):
- At \( t = \frac{\pi}{6} \), \( \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} > 0 \).
- At \( t = \frac{5\pi}{6} \), \( \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} < 0 \).
- The function \( \cos t \) becomes zero at \( t = \frac{\pi}{2} \).

Thus, we can split the integral at \( t = \frac{\pi}{2} \):
\[ I = 2 \left( \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos t \, dt - \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} \cos t \, dt \right) \]

### Step 4: Calculate the integrals
Now we compute the two integrals:
1. For \( \int \cos t \, dt \):
\[ \int \cos t \, dt = \sin t \]
Therefore:
\[ \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos t \, dt = \sin\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{6}\right) = 1 - \frac{1}{2} = \frac{1}{2} \]

2. For the second integral:
\[ \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} \cos t \, dt = \sin\left(\frac{5\pi}{6}\right) - \sin\left(\frac{\pi}{2}\right) = \frac{1}{2} - 1 = -\frac{1}{2} \]

### Step 5: Combine the results
Now substituting back into the expression for \( I \):
\[ I = 2 \left( \frac{1}{2} - \left(-\frac{1}{2}\right) \right) = 2 \left( \frac{1}{2} + \frac{1}{2} \right) = 2 \cdot 1 = 2 \]

### Final Result
Thus, the value of the integral is:
\[ \boxed{2} \]