`int_(-pi//2)^(pi//2) sin^(2)x cos^(2) x(sin x+cos x)dx=`

Options

A. `2//15`
B. `4//15`
C. `2//15`
D. 0

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx, \]

we will analyze the integrand to determine if it is an odd or even function.

### Step 1: Identify the function
The integrand is

\[ f(x) = \sin^2 x \cos^2 x (\sin x + \cos x). \]

### Step 2: Check for odd/even function
To check if \( f(x) \) is odd, we need to compute \( f(-x) \):

\[ f(-x) = \sin^2(-x) \cos^2(-x) (\sin(-x) + \cos(-x). \]

Using the properties of sine and cosine, we have:

- \( \sin(-x) = -\sin x \)
- \( \cos(-x) = \cos x \)

Substituting these into \( f(-x) \):

\[ f(-x) = \sin^2 x \cos^2 x (-\sin x + \cos x). \]

### Step 3: Simplify \( f(-x) \)
Now, simplify \( f(-x) \):

\[ f(-x) = \sin^2 x \cos^2 x (-\sin x + \cos x) = -\sin^2 x \cos^2 x (\sin x - \cos x). \]

### Step 4: Compare \( f(-x) \) and \( f(x) \)
Now we compare \( f(-x) \) with \( f(x) \):

\[ f(-x) \neq f(x) \quad \text{and} \quad f(-x) = -f(x). \]

This shows that \( f(x) \) is an odd function.

### Step 5: Use the property of integrals
Since \( f(x) \) is odd, we can use the property of definite integrals:

\[ \int_{-a}^{a} f(x) \, dx = 0 \quad \text{for odd functions}. \]

Thus,

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(x) \, dx = 0. \]

### Conclusion
The value of the integral is

\[ \boxed{0}. \]