Evaluate the following integrals:
`int frac{sinx+cosx}{sqrt(1+sin2x)}dx`

Answer

To evaluate the integral

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

we will simplify the integrand first.

### Step 1: Simplify the denominator
We know that

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

Thus, we can rewrite the expression inside the square root:

\[ 1 + \sin 2x = 1 + 2 \sin x \cos x. \]

Using the Pythagorean identity, we can express \(1\) as \(\sin^2 x + \cos^2 x\):

\[ 1 + \sin 2x = \sin^2 x + \cos^2 x + 2 \sin x \cos x. \]

This can be factored as:

\[ 1 + \sin 2x = (\sin x + \cos x)^2. \]

### Step 2: Substitute back into the integral
Now substituting this back into the integral, we have:

\[ I = \int \frac{\sin x + \cos x}{\sqrt{(\sin x + \cos x)^2}} \, dx. \]

Since \(\sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|\), we will assume \(\sin x + \cos x\) is positive in the interval we are considering (or we can handle the absolute value later). Thus, we can simplify:

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

### Step 3: Integrate
Now, the integral simplifies to:

\[ I = \int 1 \, dx = x + C, \]

where \(C\) is the constant of integration.

### Final Answer
Thus, the final result of the integral is:

\[ I = x + C. \]

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