`int_(0)^(10pi) sqrt(1- sin^(2) x ) dx` is equal to

Options

A. `0`
B. `10`
C. `10 pi`
D. `20`

Answer

To solve the integral \( \int_{0}^{10\pi} \sqrt{1 - \sin^2 x} \, dx \), we can follow these steps:

### Step 1: Simplify the integrand
We start with the expression inside the integral:
\[ \sqrt{1 - \sin^2 x} \]
Using the Pythagorean identity, we know that:
\[ 1 - \sin^2 x = \cos^2 x \]
Thus, we can rewrite the integrand:
\[ \sqrt{1 - \sin^2 x} = \sqrt{\cos^2 x} = |\cos x| \]

### Step 2: Determine the intervals where \(\cos x\) is positive or negative
The function \(\cos x\) oscillates between positive and negative values. We need to determine where \(\cos x\) is positive and where it is negative over the interval from \(0\) to \(10\pi\).

- \(\cos x\) is positive in the intervals:
- \( [0, \frac{\pi}{2}] \)
- \( [\frac{3\pi}{2}, 2\pi] \)
- \( [2\pi, \frac{5\pi}{2}] \)
- \( [\frac{7\pi}{2}, 4\pi] \)
- \( [4\pi, \frac{9\pi}{2}] \)
- \( [\frac{11\pi}{2}, 5\pi] \)
- \( [5\pi, \frac{13\pi}{2}] \)
- \( [\frac{15\pi}{2}, 6\pi] \)
- \( [6\pi, \frac{17\pi}{2}] \)
- \( [\frac{19\pi}{2}, 10\pi] \)

- \(\cos x\) is negative in the intervals:
- \( [\frac{\pi}{2}, \frac{3\pi}{2}] \)
- \( [2\pi, \frac{5\pi}{2}] \)
- \( [\frac{5\pi}{2}, \frac{7\pi}{2}] \)
- \( [4\pi, \frac{9\pi}{2}] \)
- \( [\frac{9\pi}{2}, \frac{11\pi}{2}] \)
- \( [5\pi, \frac{13\pi}{2}] \)
- \( [\frac{13\pi}{2}, \frac{15\pi}{2}] \)
- \( [6\pi, \frac{17\pi}{2}] \)
- \( [\frac{17\pi}{2}, 8\pi] \)
- \( [8\pi, 10\pi] \)

### Step 3: Break the integral into segments
We can break the integral into segments where \(\cos x\) is positive and negative. However, since we are interested in the absolute value, we can consider:
\[ \int_{0}^{10\pi} |\cos x| \, dx \]

### Step 4: Calculate the integral over one period
The function \(\cos x\) has a period of \(2\pi\). Therefore, we can calculate the integral over one period and then multiply by the number of periods in \(10\pi\).

Calculating the integral over one period:
\[ \int_{0}^{2\pi} |\cos x| \, dx = \int_{0}^{\pi} \cos x \, dx + \int_{\pi}^{2\pi} -\cos x \, dx \]

Calculating each part:
1. \( \int_{0}^{\pi} \cos x \, dx = [\sin x]_{0}^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0 \)
2. \( \int_{\pi}^{2\pi} -\cos x \, dx = -[\sin x]_{\pi}^{2\pi} = -(\sin(2\pi) - \sin(\pi)) = -0 + 0 = 0 \)

Thus, the integral over one period is:
\[ \int_{0}^{2\pi} |\cos x| \, dx = 0 + 0 = 0 \]

### Step 5: Multiply by the number of periods
Since \(10\pi\) contains \(5\) full periods of \(2\pi\):
\[ \int_{0}^{10\pi} |\cos x| \, dx = 5 \cdot 0 = 0 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{10\pi} \sqrt{1 - \sin^2 x} \, dx = 0 \]