The value of `int_(-pi)^(pi)(1-x^(2)) sin x cos^(2) x" dx"`, is

Options

A. 0
B. `pi-(pi^(3))/(3)`
C. `2pi-pi^(3)`
D. `(pi)/(2)-2pi^(3)`

Answer

To solve the integral \( I = \int_{-\pi}^{\pi} (1 - x^2) \sin x \cos^2 x \, dx \), we will first check whether the integrand is an even or odd function.

### Step 1: Determine the nature of the function

The integrand is given by:
\[ f(x) = (1 - x^2) \sin x \cos^2 x \]

To check if \( f(x) \) is even or odd, we compute \( f(-x) \):
\[ f(-x) = (1 - (-x)^2) \sin(-x) \cos^2(-x) \]
Using the properties of sine and cosine:
\[ \sin(-x) = -\sin x \quad \text{and} \quad \cos(-x) = \cos x \]
Thus,
\[ f(-x) = (1 - x^2)(-\sin x) \cos^2 x = -(1 - x^2) \sin x \cos^2 x = -f(x) \]

Since \( f(-x) = -f(x) \), the function \( f(x) \) is odd.

### Step 2: Evaluate the integral

Since \( f(x) \) is an odd function, we can use the property of definite integrals:
\[ \int_{-a}^{a} f(x) \, dx = 0 \quad \text{for odd functions} \]
Thus,
\[ I = \int_{-\pi}^{\pi} (1 - x^2) \sin x \cos^2 x \, dx = 0 \]

### Final Answer:
The value of the integral is:
\[ \boxed{0} \]