Evaluate `int _(0)^(pi//2) sqrt(1+ cos x dx)`

Answer

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

### Step 1: Rewrite the integrand
We know that \( 1 + \cos x \) can be expressed using the double angle formula:
\[ 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \]
Thus, we can rewrite the integral:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{1 + \cos x} \, dx = \int_{0}^{\frac{\pi}{2}} \sqrt{2 \cos^2\left(\frac{x}{2}\right)} \, dx \]

### Step 2: Simplify the square root
The square root can be simplified:
\[ \sqrt{2 \cos^2\left(\frac{x}{2}\right)} = \sqrt{2} \cdot \cos\left(\frac{x}{2}\right) \]
So the integral becomes:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{2} \cdot \cos\left(\frac{x}{2}\right) \, dx \]

### Step 3: Factor out the constant
We can factor out \( \sqrt{2} \) from the integral:
\[ \sqrt{2} \int_{0}^{\frac{\pi}{2}} \cos\left(\frac{x}{2}\right) \, dx \]

### Step 4: Change of variables
Let \( u = \frac{x}{2} \), then \( dx = 2 \, du \). The limits change as follows:
- When \( x = 0 \), \( u = 0 \)
- When \( x = \frac{\pi}{2} \), \( u = \frac{\pi}{4} \)

Thus, we can rewrite the integral:
\[ \sqrt{2} \int_{0}^{\frac{\pi}{4}} \cos(u) \cdot 2 \, du = 2\sqrt{2} \int_{0}^{\frac{\pi}{4}} \cos(u) \, du \]

### Step 5: Evaluate the integral
The integral of \( \cos(u) \) is \( \sin(u) \):
\[ 2\sqrt{2} \left[ \sin(u) \right]_{0}^{\frac{\pi}{4}} = 2\sqrt{2} \left( \sin\left(\frac{\pi}{4}\right) - \sin(0) \right) \]
Since \( \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \) and \( \sin(0) = 0 \):
\[ 2\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 2 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \sqrt{1 + \cos x} \, dx = 2 \]