` int _(0)^(pi//2) x sin^(2) x cos^(2) x dx ` is equal to

Options

A. `(pi^(2))/(32)`
B. `(pi^(2))/(16)`
C. `(pi)/(32)`
D. None of these

Explanation

To solve the integral \( I = \int_0^{\frac{\pi}{2}} x \sin^2 x \cos^2 x \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand using the identity for sine and cosine:
\[ \sin^2 x \cos^2 x = \left(\frac{1}{4} \sin(2x)\right)^2 = \frac{1}{4} \sin^2(2x) \]
Thus, we can rewrite the integral as:
\[ I = \int_0^{\frac{\pi}{2}} x \cdot \frac{1}{4} \sin^2(2x) \, dx = \frac{1}{4} \int_0^{\frac{\pi}{2}} x \sin^2(2x) \, dx \]

### Step 2: Use integration by parts
Let \( J = \int_0^{\frac{\pi}{2}} x \sin^2(2x) \, dx \). We will use integration by parts, where we let:
- \( u = x \) ⇒ \( du = dx \)
- \( dv = \sin^2(2x) \, dx \)

To find \( v \), we need to integrate \( \sin^2(2x) \):
\[ \sin^2(2x) = \frac{1 - \cos(4x)}{2} \]
Thus,
\[ v = \int \sin^2(2x) \, dx = \int \frac{1 - \cos(4x)}{2} \, dx = \frac{x}{2} - \frac{\sin(4x)}{8} \]

### Step 3: Apply integration by parts formula
Now we apply the integration by parts formula:
\[ J = uv \bigg|_0^{\frac{\pi}{2}} - \int v \, du \]
Calculating \( uv \bigg|_0^{\frac{\pi}{2}} \):
\[ uv \bigg|_0^{\frac{\pi}{2}} = \left(\frac{\pi}{2} \cdot \left(\frac{\frac{\pi}{2}}{2} - \frac{\sin(4 \cdot \frac{\pi}{2})}{8}\right)\right) - \left(0 \cdot \left(\frac{0}{2} - \frac{\sin(0)}{8}\right)\right) \]
Since \( \sin(2\pi) = 0 \):
\[ = \frac{\pi}{2} \cdot \left(\frac{\pi}{4} - 0\right) = \frac{\pi^2}{8} \]

### Step 4: Calculate the remaining integral
Now we need to calculate:
\[ \int_0^{\frac{\pi}{2}} \left(\frac{x}{2} - \frac{\sin(4x)}{8}\right) \, dx \]
This gives us:
\[ \int_0^{\frac{\pi}{2}} \frac{x}{2} \, dx - \int_0^{\frac{\pi}{2}} \frac{\sin(4x)}{8} \, dx \]
Calculating the first integral:
\[ \int_0^{\frac{\pi}{2}} \frac{x}{2} \, dx = \frac{1}{2} \cdot \left[\frac{x^2}{2}\right]_0^{\frac{\pi}{2}} = \frac{1}{2} \cdot \frac{\left(\frac{\pi}{2}\right)^2}{2} = \frac{\pi^2}{8} \]
Calculating the second integral:
\[ \int_0^{\frac{\pi}{2}} \frac{\sin(4x)}{8} \, dx = \frac{1}{8} \cdot \left[-\frac{\cos(4x)}{4}\right]_0^{\frac{\pi}{2}} = \frac{1}{8} \cdot \left[-\frac{\cos(2\pi)}{4} + \frac{\cos(0)}{4}\right] = \frac{1}{8} \cdot \left[-\frac{1}{4} + \frac{1}{4}\right] = 0 \]

### Step 5: Combine results
Thus, we have:
\[ J = \frac{\pi^2}{8} - 0 = \frac{\pi^2}{8} \]
Finally, substituting back into our expression for \( I \):
\[ I = \frac{1}{4} J = \frac{1}{4} \cdot \frac{\pi^2}{8} = \frac{\pi^2}{32} \]

### Final Result
Thus, the value of the integral is:
\[ \boxed{\frac{\pi^2}{32}} \]