Evaluate :
`int_(0)^(pi)sqrt(1+sin2x)dx` .

Answer

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

### Step 1: Rewrite the integrand
We start by rewriting the expression under the square root. We know that:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can express the integral as:
\[ I = \int_{0}^{\pi} \sqrt{1 + 2 \sin x \cos x} \, dx \]

### Step 2: Use the identity for squares
Recognizing that \( 1 + 2 \sin x \cos x \) can be rewritten, we note that:
\[ 1 + 2 \sin x \cos x = (\sin x + \cos x)^2 \]
So, we can rewrite the integral:
\[ I = \int_{0}^{\pi} \sqrt{(\sin x + \cos x)^2} \, dx \]
Since the square root and square cancel out, we have:
\[ I = \int_{0}^{\pi} |\sin x + \cos x| \, dx \]

### Step 3: Determine the sign of the integrand
Next, we need to determine where \( \sin x + \cos x \) is positive or negative over the interval \( [0, \pi] \). The expression \( \sin x + \cos x \) can be rewritten using the angle addition formula:
\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \]
The maximum value occurs at \( x = \frac{\pi}{4} \) and the minimum occurs at \( x = \frac{5\pi}{4} \). However, since we are only integrating from \( 0 \) to \( \pi \), we need to check the sign in this interval.

- At \( x = 0 \): \( \sin 0 + \cos 0 = 1 \) (positive)
- At \( x = \frac{\pi}{2} \): \( \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = 1 \) (positive)
- At \( x = \pi \): \( \sin \pi + \cos \pi = -1 \) (negative)

Thus, \( \sin x + \cos x \) is positive on \( [0, \frac{\pi}{2}] \) and negative on \( [\frac{\pi}{2}, \pi] \).

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

### Step 5: Evaluate the first integral
Calculating the first integral:
\[ \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx = \left[-\cos x + \sin x\right]_{0}^{\frac{\pi}{2}} = \left[-\cos\frac{\pi}{2} + \sin\frac{\pi}{2}\right] - \left[-\cos 0 + \sin 0\right] \]
This simplifies to:
\[ [0 + 1] - [-1 + 0] = 1 + 1 = 2 \]

### Step 6: Evaluate the second integral
Now for the second integral:
\[ \int_{\frac{\pi}{2}}^{\pi} -(\sin x + \cos x) \, dx = -\left[\int_{\frac{\pi}{2}}^{\pi} (\sin x + \cos x) \, dx\right] \]
Calculating this:
\[ \int_{\frac{\pi}{2}}^{\pi} (\sin x + \cos x) \, dx = \left[-\cos x + \sin x\right]_{\frac{\pi}{2}}^{\pi} = \left[-\cos \pi + \sin \pi\right] - \left[-\cos \frac{\pi}{2} + \sin \frac{\pi}{2}\right] \]
This simplifies to:
\[ [1 + 0] - [0 + 1] = 1 - 1 = 0 \]
Thus, the second integral evaluates to \( 0 \).

### Step 7: Combine the results
Combining both parts, we find:
\[ I = 2 + 0 = 2 \]

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