If `int_(0)^(100pi) sqrt(1-cos 2x)d x=200k`, then k is equal to

Options

A. `2 sqrt2`
B. `pi`
C. `sqrt3`
D. `sqrt2`

Answer

To solve the integral \( I = \int_{0}^{100\pi} \sqrt{1 - \cos(2x)} \, dx \) and find the value of \( k \) such that \( I = 200k \), we will follow these steps:

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

### Step 2: Rewrite the integral
Now we can rewrite the integral:
\[ I = \int_{0}^{100\pi} \sqrt{2} |\sin(x)| \, dx \]

### Step 3: Determine the period of the integrand
The function \( |\sin(x)| \) has a period of \( \pi \). Therefore, we can use the property of definite integrals:
\[ \int_{0}^{a} f(x) \, dx = n \int_{0}^{T} f(x) \, dx \quad \text{if } a = nT \]
where \( T \) is the period of \( f(x) \). Here, \( a = 100\pi \) and \( T = \pi \), so \( n = 100 \).

### Step 4: Calculate the integral over one period
Now we calculate:
\[ \int_{0}^{\pi} |\sin(x)| \, dx \]
Since \( |\sin(x)| = \sin(x) \) in the interval \( [0, \pi] \), we have:
\[ \int_{0}^{\pi} \sin(x) \, dx = [-\cos(x)]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 1 + 1 = 2 \]

### Step 5: Use the periodicity to find \( I \)
Now we can substitute this back into our expression for \( I \):
\[ I = \sqrt{2} \int_{0}^{100\pi} |\sin(x)| \, dx = \sqrt{2} \cdot 100 \int_{0}^{\pi} |\sin(x)| \, dx = \sqrt{2} \cdot 100 \cdot 2 = 200\sqrt{2} \]

### Step 6: Relate \( I \) to \( 200k \)
According to the problem, we have:
\[ I = 200k \]
Substituting the value of \( I \):
\[ 200\sqrt{2} = 200k \]

### Step 7: Solve for \( k \)
Dividing both sides by 200:
\[ k = \sqrt{2} \]

Thus, the value of \( k \) is:
\[ \boxed{\sqrt{2}} \]