Evaluate `:-`
`int_(0)^(pi//2)(sinx+cosx)dx`

Explanation

To evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx, \]

we can break it down into two separate integrals:

\[ I = \int_{0}^{\frac{\pi}{2}} \sin x \, dx + \int_{0}^{\frac{\pi}{2}} \cos x \, dx. \]

### Step 1: Evaluate the integral of \(\sin x\)

The integral of \(\sin x\) is:

\[ \int \sin x \, dx = -\cos x. \]

Now we need to evaluate this from \(0\) to \(\frac{\pi}{2}\):

\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = \left[-\cos x\right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)). \]

### Step 2: Substitute the limits

Now we substitute the limits:

\[ -\cos\left(\frac{\pi}{2}\right) + \cos(0) = -0 + 1 = 1. \]

### Step 3: Evaluate the integral of \(\cos x\)

The integral of \(\cos x\) is:

\[ \int \cos x \, dx = \sin x. \]

Now we evaluate this from \(0\) to \(\frac{\pi}{2}\):

\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx = \left[\sin x\right]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0). \]

### Step 4: Substitute the limits

Now we substitute the limits:

\[ \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1. \]

### Step 5: Combine the results

Now we can combine the results from both integrals:

\[ I = \int_{0}^{\frac{\pi}{2}} \sin x \, dx + \int_{0}^{\frac{\pi}{2}} \cos x \, dx = 1 + 1 = 2. \]

Thus, the value of the integral is

\[ \boxed{2}. \]