`int_(0)^(32pi//3) sqrt(1+cos 2x) dx`

Options

A. `20 sqrt2- sqrt((3)/(2))`
B. `22 sqrt2 + sqrt((3)/(2))`
C. `22 sqrt2- sqrt((3)/(2))`
D. None

Answer

To solve the integral \( I = \int_{0}^{\frac{32\pi}{3}} \sqrt{1 + \cos 2x} \, dx \), we can follow these steps:

### Step 1: Simplify the Integrand
We start by using the identity for cosine:
\[ \cos 2x = 2\cos^2 x - 1 \]
Thus, we can rewrite the integrand:
\[ \sqrt{1 + \cos 2x} = \sqrt{1 + (2\cos^2 x - 1)} = \sqrt{2\cos^2 x} = \sqrt{2} |\cos x| \]

### Step 2: Determine the Period of the Function
The function \( |\cos x| \) has a period of \( \pi \). Therefore, we can break the integral from \( 0 \) to \( \frac{32\pi}{3} \) into segments of length \( \pi \):
\[ I = \int_{0}^{\frac{32\pi}{3}} \sqrt{2} |\cos x| \, dx = \sqrt{2} \left( \int_{0}^{\pi} |\cos x| \, dx \right) \times n \]
where \( n \) is the number of complete periods of \( \pi \) in \( \frac{32\pi}{3} \).

### Step 3: Calculate the Number of Periods
The length of the interval \( \frac{32\pi}{3} \) can be divided by \( \pi \):
\[ n = \frac{\frac{32\pi}{3}}{\pi} = \frac{32}{3} \approx 10.67 \]
This means there are 10 complete periods of \( \pi \) and a remainder.

### Step 4: Calculate the Integral Over One Period
Next, we compute:
\[ \int_{0}^{\pi} |\cos x| \, dx \]
Since \( \cos x \) is non-negative in \( [0, \frac{\pi}{2}] \) and non-positive in \( [\frac{\pi}{2}, \pi] \), we can split this integral:
\[ \int_{0}^{\pi} |\cos x| \, dx = \int_{0}^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx \]
Calculating these integrals:
\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{0}^{\frac{\pi}{2}} = 1 - 0 = 1 \]
\[ \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx = -[\sin x]_{\frac{\pi}{2}}^{\pi} = -(-1 - 0) = 1 \]
Thus,
\[ \int_{0}^{\pi} |\cos x| \, dx = 1 + 1 = 2 \]

### Step 5: Calculate the Total Integral
Now we can substitute back into our expression for \( I \):
\[ I = \sqrt{2} \cdot 10 \cdot 2 + \sqrt{2} \cdot \int_{0}^{\frac{2\pi}{3}} |\cos x| \, dx \]
The remaining integral \( \int_{0}^{\frac{2\pi}{3}} |\cos x| \, dx \) can be computed as follows:
\[ \int_{0}^{\frac{2\pi}{3}} |\cos x| \, dx = \int_{0}^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\frac{2\pi}{3}} -\cos x \, dx \]
Calculating:
\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx = 1 \]
For \( \int_{\frac{\pi}{2}}^{\frac{2\pi}{3}} -\cos x \, dx \):
\[ -\left[\sin x\right]_{\frac{\pi}{2}}^{\frac{2\pi}{3}} = -\left(\sin\frac{2\pi}{3} - \sin\frac{\pi}{2}\right) = -\left(\frac{\sqrt{3}}{2} - 1\right) = 1 - \frac{\sqrt{3}}{2} \]
Thus,
\[ \int_{0}^{\frac{2\pi}{3}} |\cos x| \, dx = 1 + \left(1 - \frac{\sqrt{3}}{2}\right) = 2 - \frac{\sqrt{3}}{2} \]

### Final Calculation
Putting everything together:
\[ I = \sqrt{2} \cdot 20 + \sqrt{2} \left(2 - \frac{\sqrt{3}}{2}\right) = 20\sqrt{2} + \sqrt{2} \cdot 2 - \frac{\sqrt{2} \cdot \sqrt{3}}{2} \]
\[ = 22\sqrt{2} - \frac{\sqrt{6}}{2} \]

### Conclusion
Thus, the final answer is:
\[ I = 22\sqrt{2} - \frac{\sqrt{6}}{2} \]