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 can follow these steps:

### Step 1: Identify the nature of the function
We need to determine whether the function \( f(x) = \sin^3 x \cos^2 x \) is an odd or even function.

### Step 2: Check for odd/even function
To check if \( f(x) \) is odd, we evaluate \( f(-x) \):
\[ f(-x) = \sin^3(-x) \cos^2(-x) \]
Using the properties of sine and cosine:
\[ \sin(-x) = -\sin(x) \quad \text{and} \quad \cos(-x) = \cos(x) \]
Thus,
\[ f(-x) = (-\sin(x))^3 \cdot (\cos(x))^2 = -\sin^3(x) \cos^2(x) = -f(x) \]
Since \( f(-x) = -f(x) \), we conclude that \( f(x) \) is an odd function.

### Step 3: Evaluate the integral
The integral of an odd function over a symmetric interval around zero is zero:
\[ \int_{-a}^{a} f(x) \, dx = 0 \]
In our case, since the interval is 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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