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To evaluate the integral \(\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx\), let's proceed with the following steps.

First, simplify the expression inside the integral. We know that:

\[
\sin 2x = 2 \sin x \cos x
\]

So, the denominator becomes:

\[
\sqrt{1 + \sin 2x} = \sqrt{1 + 2 \sin x \cos x}
\]

Next, rewrite the integral:

\[
I = \int \frac{\sin x + \cos x}{\sqrt{1 + 2 \sin x \cos x}} \, dx
\]

Let \(u = x + \frac{\pi}{4}\), then:

\[
\sin u = \sin \left(x + \frac{\pi}{4}\right) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \frac{\sin x + \cos x}{\sqrt{2}}
\]

Thus:

\[
\sin x + \cos x = \sqrt{2} \sin u
\]

Also, observe that:

\[
1 + 2 \sin x \cos x = 1 + \sin 2x = 1 + \sin(2(x + \frac{\pi}{4}) - \frac{\pi}{2}) = 1 + \cos(2u)
\]

We know:

\[
1 + \cos(2u) = 2 \cos^2(u)
\]

So, the integral becomes:

\[
I = \int \frac{\sqrt{2} \sin u}{\sqrt{2 \cos^2 u}} \, dx = \int \frac{\sqrt{2} \sin u}{\sqrt{2} |\cos u|} \, du
\]

This simplifies to:

\[
I = \int \frac{\sin u}{|\cos u|} \, du
\]

However, since \(|\cos u| = \cos u\) for \(u\) in the range of integration (assuming \(\cos u\) is non-negative), we get:

\[
I = \int \tan u \, du
\]

The integral of \(\tan u\) is:

\[
\int \tan u \, du = -\ln|\cos u| + C
\]

Substituting back \(u = x + \frac{\pi}{4}\), we obtain:

\[
I = -\ln|\cos(x + \frac{\pi}{4})| + C
\]

So, the evaluated integral is:

\[
\boxed{-\ln|\cos(x + \frac{\pi}{4})| + C}
\]

Would you like any additional details or have any questions?

Here are some related questions:
1. How would you evaluate \(\int \frac{\cos x - \sin x}{\sqrt{1 + \sin 2x}} \, dx\)?
2. What substitution would you use to solve \(\int \frac{\sin x + \cos x}{1 + \sin 2x} \, dx\)?
3. Can you solve \(\int \frac{\sin x}{\sqrt{1 + \sin 2x}} \, dx\) using a different method?
4. How does the integral \(\int \frac{1}{\sqrt{1 + \sin 2x}} \, dx\) compare?
5. What would be the result of integrating \(\int \frac{\sin x + \cos x}{\sqrt{1 + 2 \cos x \sin x}} \, dx\)?
6. How would you verify the result of \(\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx\) using numerical methods?
7. What are the steps to simplify the integrand before solving \(\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} \, dx\)?
8. Can you find the integral \(\int \frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} \, dx\) using a similar approach?

**Tip**: Always check for trigonometric identities that can simplify the integrand before proceeding with more complicated substitutions.

Learn how to evaluate the integral of (sin x + cos x) / sqrt(1 + sin 2x) step-by-step. This problem involves using trigonometric identities and substitution methods in calculus. Suitable for advanced college-level mathematics students.

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