The value of `int_(0)^(32pi//3) sqrt(1+cos2x)dx` is

Options

A. `(44+sqrt(3))/(sqrt(2))`
B. `(44-sqrt(3))/(sqrt(2))`
C. `(22-sqrt(3))/(sqrt(2))`
D. `(22+sqrt(3))/(sqrt(2))`

Explanation

To solve the integral \( I = \int_0^{\frac{32\pi}{3}} \sqrt{1 + \cos 2x} \, dx \), we will follow these steps:

### Step 1: Simplify the integrand
Using the trigonometric identity \( 1 + \cos 2x = 2 \cos^2 x \), we can rewrite the integral:
\[ I = \int_0^{\frac{32\pi}{3}} \sqrt{1 + \cos 2x} \, dx = \int_0^{\frac{32\pi}{3}} \sqrt{2 \cos^2 x} \, dx = \int_0^{\frac{32\pi}{3}} \sqrt{2} |\cos x| \, dx \]

### Step 2: Factor out the constant
Since \( \sqrt{2} \) is a constant, we can factor it out of the integral:
\[ I = \sqrt{2} \int_0^{\frac{32\pi}{3}} |\cos x| \, dx \]

### Step 3: Determine the period of \( |\cos x| \)
The function \( |\cos x| \) has a period of \( 2\pi \). We need to find how many full periods fit into \( \frac{32\pi}{3} \):
\[ \frac{32\pi/3}{2\pi} = \frac{32}{6} = \frac{16}{3} \text{ periods} \]
This means there are 5 full periods of \( 2\pi \) and a remaining interval of \( \frac{2\pi}{3} \).

### Step 4: Break the integral into parts
We can break the integral into two parts:
\[ I = \sqrt{2} \left( \int_0^{10\pi} |\cos x| \, dx + \int_{10\pi}^{\frac{32\pi}{3}} |\cos x| \, dx \right) \]

### Step 5: Calculate the integral over one period
The integral of \( |\cos x| \) over one period \( [0, 2\pi] \) is:
\[ \int_0^{2\pi} |\cos x| \, dx = 2 \int_0^{\pi} \cos x \, dx = 2 \left[ \sin x \right]_0^{\pi} = 2(0 - 0) = 2 \]

### Step 6: Calculate the integral over the full periods
Since there are 5 full periods in \( [0, 10\pi] \):
\[ \int_0^{10\pi} |\cos x| \, dx = 5 \cdot 2 = 10 \]

### Step 7: Calculate the integral over the remaining interval
Now, we need to calculate \( \int_{10\pi}^{\frac{32\pi}{3}} |\cos x| \, dx \):
- The interval \( [10\pi, \frac{32\pi}{3}] \) is \( [10\pi, 10\pi + \frac{2\pi}{3}] \).
- In this interval, \( \cos x \) is negative from \( 10\pi \) to \( 11\pi/2 \) and positive from \( 11\pi/2 \) to \( \frac{32\pi}{3} \).

Calculating:
1. From \( 10\pi \) to \( 11\pi/2 \):
\[ \int_{10\pi}^{11\pi/2} -\cos x \, dx = -\left[ \sin x \right]_{10\pi}^{11\pi/2} = -\left( \sin(11\pi/2) - \sin(10\pi) \right) = -\left( 1 - 0 \right) = -1 \]
2. From \( 11\pi/2 \) to \( \frac{32\pi}{3} \):
\[ \int_{11\pi/2}^{\frac{32\pi}{3}} \cos x \, dx = \left[ \sin x \right]_{11\pi/2}^{\frac{32\pi}{3}} = \sin\left(\frac{32\pi}{3}\right) - \sin\left(\frac{11\pi}{2}\right) = \left(-\frac{\sqrt{3}}{2}\right) - 1 = -\frac{\sqrt{3}}{2} - 1 \]

### Step 8: Combine the results
Thus, the total integral becomes:
\[ I = \sqrt{2} \left( 10 + (-1) + \left(-\frac{\sqrt{3}}{2} - 1\right) \right) = \sqrt{2} \left( 8 - \frac{\sqrt{3}}{2} \right) \]

### Final Result
The final value of the integral is:
\[ I = \sqrt{2} \left( 8 - \frac{\sqrt{3}}{2} \right) \]