`int sqrt(1+sin 2x)dx=`

Options

A. `sinx+cosx+c AA x in R`
B. `sinx-cos x+cAA x in R`
C. `sin x-cosx+c, x in [(-pi)/(4),(3pi)/(4)]`
D. `cosx-sinx+c, x in [(3pi)/(4),(7pi)/(4)]`

Answer

To solve the integral \( \int \sqrt{1 + \sin 2x} \, dx \), we will follow these steps:

### Step 1: Simplify the expression inside the square root
We know that \( \sin 2x = 2 \sin x \cos x \). Thus, we can rewrite the integral as:
\[ \int \sqrt{1 + 2 \sin x \cos x} \, dx \]

### Step 2: Use the identity for sine
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can express \( 1 \) as \( \cos^2 x + \sin^2 x \). Therefore, we have:
\[ 1 + 2 \sin x \cos x = \cos^2 x + \sin^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2 \]
So, we can rewrite the integral as:
\[ \int \sqrt{(\sin x + \cos x)^2} \, dx \]

### Step 3: Remove the square root
Since the square root of a square gives us the absolute value, we have:
\[ \int |\sin x + \cos x| \, dx \]

### Step 4: Determine the intervals for the absolute value
To evaluate the integral, we need to determine where \( \sin x + \cos x \) is positive or negative. We can find the zeros of \( \sin x + \cos x = 0 \):
\[ \tan x = -1 \implies x = \frac{3\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]
This means \( \sin x + \cos x \) is positive in the intervals:
- \( (-\frac{\pi}{4}, \frac{3\pi}{4}) \)
- \( (\frac{7\pi}{4}, \frac{11\pi}{4}) \)

And negative in:
- \( (\frac{3\pi}{4}, \frac{7\pi}{4}) \)

### Step 5: Split the integral based on intervals
Now we can split the integral based on these intervals. For example, if we consider the interval from \( 0 \) to \( 2\pi \):
\[ \int_0^{\frac{3\pi}{4}} (\sin x + \cos x) \, dx + \int_{\frac{3\pi}{4}}^{\frac{7\pi}{4}} -(\sin x + \cos x) \, dx + \int_{\frac{7\pi}{4}}^{2\pi} (\sin x + \cos x) \, dx \]

### Step 6: Evaluate each integral
1. **First Integral:**
\[ \int_0^{\frac{3\pi}{4}} (\sin x + \cos x) \, dx = \left[-\cos x + \sin x\right]_0^{\frac{3\pi}{4}} = \left(-\cos \frac{3\pi}{4} + \sin \frac{3\pi}{4}\right) - \left(-\cos 0 + \sin 0\right) \]
\[ = \left(-(-\frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}}\right) - (1 + 0) = 2\cdot\frac{1}{\sqrt{2}} - 1 = \sqrt{2} - 1 \]

2. **Second Integral:**
\[ \int_{\frac{3\pi}{4}}^{\frac{7\pi}{4}} -(\sin x + \cos x) \, dx = -\left[-\cos x + \sin x\right]_{\frac{3\pi}{4}}^{\frac{7\pi}{4}} = \left(-\cos \frac{7\pi}{4} + \sin \frac{7\pi}{4}\right) - \left(-\cos \frac{3\pi}{4} + \sin \frac{3\pi}{4}\right) \]
\[ = \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) - \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = -2\cdot\frac{1}{\sqrt{2}} = -\sqrt{2} \]

3. **Third Integral:**
\[ \int_{\frac{7\pi}{4}}^{2\pi} (\sin x + \cos x) \, dx = \left[-\cos x + \sin x\right]_{\frac{7\pi}{4}}^{2\pi} = \left(-\cos 2\pi + \sin 2\pi\right) - \left(-\cos \frac{7\pi}{4} + \sin \frac{7\pi}{4}\right) \]
\[ = (1 - 0) - \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = 1 \]

### Step 7: Combine the results
Now we can combine the results of the three integrals:
\[ \text{Total Integral} = (\sqrt{2} - 1) + (-\sqrt{2}) + 1 = 0 \]

### Final Result
Thus, the value of the integral \( \int \sqrt{1 + \sin 2x} \, dx \) over the interval \( [0, 2\pi] \) is:
\[ \int \sqrt{1 + \sin 2x} \, dx = 0 + C \]