To evaluate the integral
\[
I = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{\sqrt{1 + \sin 2x}}{|\cos x|} \, dx,
\]
we can start by simplifying the integrand.
### Step 1: Simplify the integrand
We know that
\[
\sin 2x = 2 \sin x \cos x.
\]
Thus, we can rewrite \(1 + \sin 2x\) as:
\[
1 + \sin 2x = 1 + 2 \sin x \cos x = (\sin x + \cos x)^2.
\]
So, we have:
\[
\sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|.
\]
Now, substituting this back into the integral gives:
\[
I = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{|\sin x + \cos x|}{|\cos x|} \, dx.
\]
### Step 2: Analyze the absolute values
Next, we need to determine where \(\sin x + \cos x\) changes sign within the limits of integration. The expression \(\sin x + \cos x = 0\) can be solved as follows:
\[
\tan x = -1 \implies x = -\frac{\pi}{4} \text{ (within the interval)}.
\]
Thus, we can break the integral into two parts:
1. From \(-\frac{\pi}{3}\) to \(-\frac{\pi}{4}\)
2. From \(-\frac{\pi}{4}\) to \(\frac{\pi}{3}\)
### Step 3: Split the integral
We can express \(I\) as:
\[
I = \int_{-\frac{\pi}{3}}^{-\frac{\pi}{4}} \frac{-(\sin x + \cos x)}{|\cos x|} \, dx + \int_{-\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{(\sin x + \cos x)}{|\cos x|} \, dx.
\]
### Step 4: Evaluate each integral
1. **First Integral**:
\[
I_1 = \int_{-\frac{\pi}{3}}^{-\frac{\pi}{4}} \frac{-(\sin x + \cos x)}{|\cos x|} \, dx = -\int_{-\frac{\pi}{3}}^{-\frac{\pi}{4}} \frac{\sin x + \cos x}{\cos x} \, dx = -\int_{-\frac{\pi}{3}}^{-\frac{\pi}{4}} \left( \tan x + 1 \right) \, dx.
\]
Evaluating this integral gives:
\[
-\left[ -\ln |\cos x| + x \right]_{-\frac{\pi}{3}}^{-\frac{\pi}{4}}.
\]
2. **Second Integral**:
\[
I_2 = \int_{-\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{(\sin x + \cos x)}{|\cos x|} \, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{3}} \left( \tan x + 1 \right) \, dx.
\]
Evaluating this integral gives:
\[
\left[ \ln |\sec x| + x \right]_{-\frac{\pi}{4}}^{\frac{\pi}{3}}.
\]
### Step 5: Combine results
Now, we combine \(I_1\) and \(I_2\) to find the total value of \(I\).
### Final Result
After evaluating the limits and simplifying, we find:
\[
I = \frac{\pi}{2} - \ln(\sqrt{2}).
\]
