`int_(0)^(pi//4) sin ^(2) x dx`

Answer

To solve the integral \( \int_{0}^{\frac{\pi}{4}} \sin^2 x \, dx \), we can use a trigonometric identity to simplify the expression. Here’s a step-by-step solution:

### Step 1: Use the trigonometric identity
We know that:
\[ \cos 2x = 1 - 2\sin^2 x \]
From this, we can express \(\sin^2 x\) in terms of \(\cos 2x\):
\[ \sin^2 x = \frac{1 - \cos 2x}{2} \]

### Step 2: Substitute into the integral
Substituting \(\sin^2 x\) into the integral gives:
\[ \int_{0}^{\frac{\pi}{4}} \sin^2 x \, dx = \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos 2x}{2} \, dx \]

### Step 3: Factor out the constant
We can factor out the \(\frac{1}{2}\):
\[ = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} (1 - \cos 2x) \, dx \]

### Step 4: Separate the integral
Now we can separate the integral into two parts:
\[ = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{4}} 1 \, dx - \int_{0}^{\frac{\pi}{4}} \cos 2x \, dx \right) \]

### Step 5: Evaluate the first integral
The first integral is straightforward:
\[ \int_{0}^{\frac{\pi}{4}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} - 0 = \frac{\pi}{4} \]

### Step 6: Evaluate the second integral
For the second integral, we have:
\[ \int \cos 2x \, dx = \frac{1}{2} \sin 2x \]
Thus,
\[ \int_{0}^{\frac{\pi}{4}} \cos 2x \, dx = \left[ \frac{1}{2} \sin 2x \right]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \left( \sin \frac{\pi}{2} - \sin 0 \right) = \frac{1}{2} (1 - 0) = \frac{1}{2} \]

### Step 7: Combine the results
Now substituting back into our expression:
\[ = \frac{1}{2} \left( \frac{\pi}{4} - \frac{1}{2} \right) \]
\[ = \frac{1}{2} \cdot \frac{\pi}{4} - \frac{1}{2} \cdot \frac{1}{2} = \frac{\pi}{8} - \frac{1}{4} \]

### Final Answer
Thus, the final result is:
\[ \int_{0}^{\frac{\pi}{4}} \sin^2 x \, dx = \frac{\pi}{8} - \frac{1}{4} \]