`int_(-pi//2)^(pi//2) sin^(2)x cos^(2) x(sin xcos 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 use the property of even and odd functions.

### Step 1: Identify the function

Let

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

### Step 2: Check if the function is odd or even

To determine if \( f(x) \) is odd or even, we will compute \( f(-x) \):

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

Using the properties of sine and cosine:

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

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

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

### Step 3: Conclusion about the function

Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 4: Use the property of odd functions in definite integrals

The property of definite integrals states that the integral of an odd function over a symmetric interval about zero is zero:

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

### Step 5: Apply the property to our integral

Thus, we have:

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

### Final Answer

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

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