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

Answer

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

### Step 1: Rewrite the Integral
Let’s denote the integral as \( I \):
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^2 x \, dx \]

### Step 2: Multiply and Divide by 4
We can simplify the integral by multiplying and dividing by 4:
\[ I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} 4 \sin^2 x \cos^2 x \, dx \]

### Step 3: Use the Double Angle Identity
Using the identity \( 2 \sin x \cos x = \sin 2x \), we have:
\[ 4 \sin^2 x \cos^2 x = (2 \sin x \cos x)^2 = \sin^2 2x \]
Thus, we can rewrite the integral as:
\[ I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \sin^2 2x \, dx \]

### Step 4: Use the Identity for Sine Squared
We can use the identity \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \):
\[ \sin^2 2x = \frac{1 - \cos 4x}{2} \]
Substituting this into the integral gives:
\[ I = \frac{1}{4} \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos 4x}{2} \, dx \]
This simplifies to:
\[ I = \frac{1}{8} \int_{0}^{\frac{\pi}{2}} (1 - \cos 4x) \, dx \]

### Step 5: Split the Integral
Now we can split the integral:
\[ I = \frac{1}{8} \left( \int_{0}^{\frac{\pi}{2}} 1 \, dx - \int_{0}^{\frac{\pi}{2}} \cos 4x \, dx \right) \]

### Step 6: Calculate Each Integral
1. The first integral:
\[ \int_{0}^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2} \]
2. The second integral:
\[ \int_{0}^{\frac{\pi}{2}} \cos 4x \, dx = \left[ \frac{\sin 4x}{4} \right]_{0}^{\frac{\pi}{2}} = \frac{\sin 2\pi}{4} - \frac{\sin 0}{4} = 0 \]

### Step 7: Combine the Results
Putting it all together:
\[ I = \frac{1}{8} \left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{16} \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{\frac{\pi}{16}} \]