The value `int_(0)^(pi//2) sin^(2)xdx` will be `:`

Options

A. 1
B. 0
C. `(pi)/(4)`
D. `(pi)/(2)`

Explanation

To solve the integral \( \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx \), we can follow these steps:

### Step 1: Use the identity for \( \sin^2 x \)
We can use the trigonometric identity:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
This allows us to rewrite the integral:
\[ \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx \]

### Step 2: Separate the integral
Now, we can separate the integral into two parts:
\[ \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \, dx - \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx \]

### Step 3: Evaluate the first integral
The first integral \( \int_{0}^{\frac{\pi}{2}} 1 \, dx \) is straightforward:
\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \]

### Step 4: Evaluate the second integral
Now, we evaluate the second integral \( \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx \):
\[ \int \cos(2x) \, dx = \frac{\sin(2x)}{2} \]
Thus,
\[ \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx = \left[ \frac{\sin(2x)}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{\sin(\pi)}{2} - \frac{\sin(0)}{2} = \frac{0}{2} - \frac{0}{2} = 0 \]

### Step 5: Combine the results
Now we can combine the results:
\[ \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{1}{2} \left( \frac{\pi}{2} \right) - \frac{1}{2} (0) = \frac{\pi}{4} \]

### Final Answer
Thus, the value of the integral \( \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx \) is:
\[ \frac{\pi}{4} \]