Evaluate :
`int_(0)^(pi//2) sqrt(1+ sin 2x dx)`

Answer

To evaluate the integral \( I = \int_{0}^{\frac{\pi}{2}} \sqrt{1 + \sin 2x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{1 + \sin 2x} \, dx \]
We know that \( \sin 2x = 2 \sin x \cos x \). Thus, we can rewrite the expression inside the square root:
\[ 1 + \sin 2x = 1 + 2 \sin x \cos x \]

### Step 2: Use the identity \( \sin^2 x + \cos^2 x = 1 \)
We can express \( 1 \) as \( \cos^2 x + \sin^2 x \):
\[ 1 + \sin 2x = \cos^2 x + \sin^2 x + 2 \sin x \cos x = (\sin x + \cos x)^2 \]
So, we have:
\[ \sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2} = \sin x + \cos x \]

### Step 3: Substitute back into the integral
Now we can substitute this back into our integral:
\[ I = \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx \]

### Step 4: Separate the integral
We can separate the integral into two parts:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin x \, dx + \int_{0}^{\frac{\pi}{2}} \cos x \, dx \]

### Step 5: Evaluate the integrals
Now we evaluate each integral separately:
1. For \( \int \sin x \, dx \):
\[ \int \sin x \, dx = -\cos x \quad \text{(evaluated from 0 to } \frac{\pi}{2}\text{)} \]
\[ = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1 \]

2. For \( \int \cos x \, dx \):
\[ \int \cos x \, dx = \sin x \quad \text{(evaluated from 0 to } \frac{\pi}{2}\text{)} \]
\[ = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]

### Step 6: Combine the results
Now we can combine the results of the two integrals:
\[ I = 1 + 1 = 2 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{2} \]