`int_(0)^(pi//2) sin^(2) x dx`

Answer

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

### Step 1: Use the half-angle identity
We know that:
\[ \sin^2 x = \frac{1 - \cos(2x)}{2} \]
So we can rewrite the integral as:
\[ \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx \]

### Step 2: Simplify the integral
This can be simplified to:
\[ \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} (1 - \cos(2x)) \, dx \]
This can be separated into two integrals:
\[ = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{2}} 1 \, dx - \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx \right) \]

### Step 3: Evaluate the first integral
The first integral 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
For the second integral, we need to integrate \( \cos(2x) \):
\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) \]
Thus,
\[ \int_{0}^{\frac{\pi}{2}} \cos(2x) \, dx = \left[ \frac{1}{2} \sin(2x) \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} (\sin(\pi) - \sin(0)) = \frac{1}{2} (0 - 0) = 0 \]

### Step 5: Combine the results
Now substituting back into our expression:
\[ \frac{1}{2} \left( \frac{\pi}{2} - 0 \right) = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \]

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

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