The value of `int_(-pi)^(pi) sin^(3) x cos^(2)x dx` is

Explanation

To solve the integral \( \int_{-\pi}^{\pi} \sin^3 x \cos^2 x \, dx \), we will first determine whether the integrand is an odd or even function.

### Step 1: Identify the function
Let \( f(x) = \sin^3 x \cos^2 x \).

### Step 2: Check if the function is odd or even
To determine if \( f(x) \) is odd or even, we will compute \( f(-x) \):

\[ f(-x) = \sin^3(-x) \cos^2(-x) \]

Using the properties of sine and cosine, we know:
- \( \sin(-x) = -\sin(x) \)
- \( \cos(-x) = \cos(x) \)

Thus, we have:

\[ f(-x) = (-\sin x)^3 \cdot (\cos x)^2 = -\sin^3 x \cdot \cos^2 x = -f(x) \]

Since \( f(-x) = -f(x) \), this shows that \( f(x) \) is an odd function.

### Step 3: Apply the property of definite integrals
We can use the property of definite integrals that states:

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

In our case, since \( f(x) \) is odd and we are integrating from \(-\pi\) to \(\pi\):

\[ \int_{-\pi}^{\pi} \sin^3 x \cos^2 x \, dx = 0 \]

### Final Answer
Thus, the value of the integral is:

\[ \int_{-\pi}^{\pi} \sin^3 x \cos^2 x \, dx = 0 \]

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