Using integration, find the area enclosed by the curve `y = cos, y = sin x` and x - axis in the interval `[0, pi//2]`.

Answer

To find the area enclosed by the curves \( y = \cos x \), \( y = \sin x \), and the x-axis in the interval \([0, \frac{\pi}{2}]\), we will follow these steps:

### Step 1: Find the Points of Intersection
We start by finding the points where the curves intersect. We set \( \cos x = \sin x \).

\[ \cos x = \sin x \implies \tan x = 1 \implies x = \frac{\pi}{4} \]

Thus, the curves intersect at \( x = \frac{\pi}{4} \).

### Step 2: Determine the Area Segments
We will divide the area into two parts:
- Area \( A_1 \) under the curve \( y = \sin x \) from \( x = 0 \) to \( x = \frac{\pi}{4} \).
- Area \( A_2 \) under the curve \( y = \cos x \) from \( x = \frac{\pi}{4} \) to \( x = \frac{\pi}{2} \).

### Step 3: Calculate Area \( A_1 \)
The area \( A_1 \) can be calculated using the integral:

\[ A_1 = \int_0^{\frac{\pi}{4}} \sin x \, dx \]

Calculating the integral:

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

Now, we evaluate the definite integral:

\[ A_1 = \left[-\cos x\right]_0^{\frac{\pi}{4}} = -\cos\left(\frac{\pi}{4}\right) - (-\cos(0)) = -\frac{1}{\sqrt{2}} + 1 = 1 - \frac{1}{\sqrt{2}} \]

### Step 4: Calculate Area \( A_2 \)
The area \( A_2 \) can be calculated using the integral:

\[ A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \, dx \]

Calculating the integral:

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

Now, we evaluate the definite integral:

\[ A_2 = \left[\sin x\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{4}\right) = 1 - \frac{1}{\sqrt{2}} \]

### Step 5: Total Area
The total area \( A \) is the sum of \( A_1 \) and \( A_2 \):

\[ A = A_1 + A_2 = \left(1 - \frac{1}{\sqrt{2}}\right) + \left(1 - \frac{1}{\sqrt{2}}\right) = 2 - 2\cdot\frac{1}{\sqrt{2}} = 2 - \sqrt{2} \]

### Final Answer
Thus, the area enclosed by the curves \( y = \cos x \), \( y = \sin x \), and the x-axis in the interval \([0, \frac{\pi}{2}]\) is:

\[ \text{Area} = 2 - \sqrt{2} \text{ square units.} \]