Solve the following integration
`int (sin x+cos x)/(sqrt(1+sin 2x))*dx`, here `(sin x + cos x) gt 0`

Answer

To solve the integral

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

we will follow these steps:

### Step 1: Use Trigonometric Identities
We know that:
- \(\sin^2 x + \cos^2 x = 1\)
- \(\sin 2x = 2 \sin x \cos x\)

We can rewrite the integral using these identities:

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

### Step 2: Simplify the Denominator
Using the identity \(\sin 2x = 2 \sin x \cos x\), we can express the denominator:

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

Thus, we can rewrite the integral as:

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

### Step 3: Simplify the Integral
Since \(\sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|\), we can express the integral as:

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

### Step 4: Apply the Given Condition
Given the condition \(\sin x + \cos x > 0\), we can simplify the absolute value:

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

### Step 5: Integrate
The integral of 1 with respect to \(x\) is simply:

\[ I = x + C \]

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

### Final Answer
Thus, the solution to the integral is:

\[ I = x + C \]