The value of `int_(-pi//2)^(199pi//2) sqrt((1+cos 2x))dx` is

Options

A. `50 sqrt2`
B. `100 sqrt2`
C. `150 sqrt2`
D. `200 sqrt2`

Answer

To solve the integral \( I = \int_{-\frac{\pi}{2}}^{\frac{199\pi}{2}} \sqrt{1 + \cos 2x} \, dx \), we will follow these steps:

### Step 1: Simplify the integrand
We know that:
\[ \cos 2x = 2\cos^2 x - 1 \]
Thus:
\[ 1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x \]
So, we can rewrite the integrand:
\[ \sqrt{1 + \cos 2x} = \sqrt{2\cos^2 x} = \sqrt{2} |\cos x| \]

### Step 2: Break the integral into manageable parts
The integral can be split into two parts:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{2} |\cos x| \, dx + \int_{\frac{\pi}{2}}^{\frac{199\pi}{2}} \sqrt{2} |\cos x| \, dx \]

### Step 3: Evaluate the first integral
For the first integral:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{2} |\cos x| \, dx \]
Since \( \cos x \) is non-negative in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), we have:
\[ |\cos x| = \cos x \]
Thus:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{2} \cos x \, dx = \sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx \]
The integral of \( \cos x \) is:
\[ \sin x \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) = 1 - (-1) = 2 \]
So, the first integral evaluates to:
\[ \sqrt{2} \cdot 2 = 2\sqrt{2} \]

### Step 4: Evaluate the second integral
Next, we consider the second integral:
\[ \int_{\frac{\pi}{2}}^{\frac{199\pi}{2}} \sqrt{2} |\cos x| \, dx \]
The function \( \cos x \) is periodic with a period of \( 2\pi \). The integral can be simplified using the periodicity:
\[ \int_{\frac{\pi}{2}}^{\frac{199\pi}{2}} |\cos x| \, dx = \sum_{k=0}^{99} \int_{\frac{\pi}{2} + 2k\pi}^{\frac{3\pi}{2} + 2k\pi} |\cos x| \, dx \]
Each integral from \( \frac{\pi}{2} + 2k\pi \) to \( \frac{3\pi}{2} + 2k\pi \) evaluates to:
\[ \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} |\cos x| \, dx = 2 \]
Thus, for 100 periods:
\[ \int_{\frac{\pi}{2}}^{\frac{199\pi}{2}} |\cos x| \, dx = 100 \cdot 2 = 200 \]
So:
\[ \int_{\frac{\pi}{2}}^{\frac{199\pi}{2}} \sqrt{2} |\cos x| \, dx = 200\sqrt{2} \]

### Step 5: Combine the results
Now we can combine both parts:
\[ I = 2\sqrt{2} + 200\sqrt{2} = 202\sqrt{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{202\sqrt{2}} \]