`int(sinx+cosx)/(sqrt(1+sin2x))dx`

Explanation

To solve the integral

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

we can follow these steps:

### Step 1: Rewrite the denominator
First, we note that \( \sin 2x = 2 \sin x \cos x \). Thus, we can rewrite the expression under the square root:

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

### Step 2: Use the Pythagorean identity
We can use the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) to rewrite 1:

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

### Step 3: Substitute into the integral
Now substituting this back into the integral gives us:

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

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

Thus, we have:

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

### Step 4: Integrate
Now, integrating \( 1 \) with respect to \( x \):

\[ I = x + C, \]

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

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

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

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